I need to find the sum of all integer bases b>9 for which the polynomial x2+2x+6 has a friend in the base b. But the problem is cut off. Let me read it again.
"Find the sum of all integer bases b>9 for which"
It seems incomplete. Probably, it's implying that for some condition, but it's not specified. Perhaps it's a common type of problem where we need to find bases where a certain number is represented in a particular way. Maybe it's about the polynomial having roots in that base or something.
Perhaps "has a friend" means that there is some integer or something related. I recall that in some contexts, a "friend" might refer to a number that is friendly or something, but that doesn't make sense.
Another thought: in some problems, a number is said to have a "friend" if it can be written as a sum of consecutive integers or something, but here it's a polynomial.
Perhaps the polynomial is given, and we need to find bases where it has integer roots or something. But the polynomial is monic, so if it has integer roots, they would be integers regardless of base, but the base affects how numbers are represented.
I think I need to interpret the problem. Let me assume that "has a friend" means that there is an integer n such that when we evaluate the polynomial in base b, it gives a certain value, but that seems vague.
Perhaps it's about the constant term or coefficients being valid in that base. But the coefficients are 1,2,6, which are less than b since b>9, so that should be fine.
Another idea: perhaps "friend" refers to the fact that the number represented by the polynomial is a perfect square or something. But again, the polynomial is in x, not a number.
I recall that in some problems, we have a number like 121 in base b, and we find when it's a perfect square, but here it's a polynomial.
Let's look at the polynomial: x^2 + 2x + 6. If we evaluate it at some x, it's a number, but x is a variable.
Perhaps we need to find bases b such that when we write the number 121 in base b, it equals 1b^2 + 2b + 1, and set it equal to something, but here the polynomial is different.
The polynomial is x^2 + 2x + 6, so perhaps we need to find when this polynomial has integer roots for some b, but the roots depend on b? No, the polynomial is given with coefficients, so its roots are fixed: let's calculate the roots.
Discriminant d = 4 - 24 = -20, so roots are (-2 ± sqrt(-20))/2 = -1 ± i√5, so complex, not real. So that can't be it.
Perhaps in base b, the representation of the polynomial or something. I think I need to recall what "has a friend" means in this context.
Another thought: in some contexts, a number is said to be "friendly" if it is a sum of consecutive integers, but here it's a polynomial.
Perhaps the problem is: for which bases b>9, the number 121 in base b is a perfect square. But 121_b = b^2 + 2b + 1 = (b+1)^2, which is always a perfect square for any b, so that would be all b, but the sum would be infinite, which is not possible.
But the polynomial is x^2 + 2x + 6, not x^2 + 2x + 1.
So perhaps it's for x^2 + 2x + 6 in base b, but what does that mean?
Perhaps we need to find when the number represented by "126" in base b is a perfect square or something. "126"_b = 1b^2 + 2b + 6, and we need this to be a perfect square for some b>9.
That makes sense. Let me check the problem: "the polynomial x^2 + 2x + 6", and "has a friend", but perhaps "friend" means that it is a perfect square or something similar.
In some languages, "friend" might be a mistranslation, but let's assume that's the case, as it's a common type of problem.
So, let's set f(b) = b^2 + 2b + 6, and we need this to be a perfect square for integer b>9.
So, b^2 + 2b + 6 = k^2 for some integer k.
Then, k^2 - b^2 = 2b + 6
(k-b)(k+b) = 2b + 6
Set d = k - b, then k + b = (2b + 6)/d
Since k + b and k - b are both integers, and their product is 2b+6, and they have the same parity, but since b is integer, 2b+6 is even, so both must be even.
Set m = k - b, n = k + b, then m n = 2b + 6, and n - m = 2b
From n - m = 2b, and m n = 2b + 6.
But 2b = n - m, so substitute:
m n = (n - m) + 6
m n - n + m = 6
n(m - 1) + m = 6
Better: m n - n + m = 6
n(m - 1) + m = 6
Add and subtract 1: n(m-1) + m = 6
Set t = m-1, then n t + (t+1) = 6 ⇒ n t + t + 1 = 6 ⇒ t(n + 1) = 5
Since t and n+1 are integers whose product is 5.
So possible pairs (t, n+1): (1,5), (5,1), (-1,-5), (-5,-1)
Since b > 9, and k > b probably, but let's see.
Case 1: t=1, n+1=5 ⇒ n=4
Then t = m - 1 = 1 ⇒ m=2
Then n - m = 2b ⇒ 4 - 2 = 2b ⇒ 2 = 2b ⇒ b=1
But b>9, not satisfied.
Case 2: t=5, n+1=1 ⇒ n=0
Then m-1=5 ⇒ m=6
n - m = 2b ⇒ 0 - 6 = 2b ⇒ -6 = 2b ⇒ b=-3, not >9.
Case 3: t=-1, n+1=-5 ⇒ n=-6
m-1=-1 ⇒ m=0
n - m = 2b ⇒ -6 - 0 = 2b ⇒ b=-3, again not.
Case 4: t=-5, n+1=-1 ⇒ n=-2
m-1=-5 ⇒ m=-4
n - m = 2b ⇒ -2 - (-4) = 2b ⇒ 2 = 2b ⇒ b=1, not >9.
So no solutions? But that can't be, probably I made a mistake.
t(n+1)=5, and I listed all integer pairs, but 5 can also be (-1,-5), etc., I think I covered: (1,5),(5,1),(-1,-5),(-5,-1). That's all since 5 is prime.
But no solutions for b>9. But perhaps k could be less than b, but in that case k^2 < b^2, while b^2 +2b+6 > b^2, so k^2 > b^2, so k > b, since b>0. So no solutions.
But that can't be right. Perhaps "has a friend" means something else.
Another interpretation: perhaps "friend" refers to a number that is amicable or something, but that's usually for pairs.
Perhaps the polynomial has an integer friend, meaning that there is an integer x such that the value is integer, but that's always true.
Or perhaps in base b, the number 126_b is equal to some other number.
Another idea: perhaps "has a friend" means that it can be written as a sum of two squares or something, but that seems unlikely.
Let's look back at the problem: "the polynomial x^2 + 2x + 6 has a friend in the base b".
Perhaps it means that there is a value x such that when evaluated in base b, it gives a number that is a friend, but what is a friend?
I recall that in some problems, a number is a "friend" if it is divisible by the sum of its digits or something, but here it's a polynomial.
Perhaps for the polynomial itself, when written in base b, the digits are friendly, but that doesn't make sense.
Another thought: perhaps "friend" is a typo, and it's "friends", meaning that the digits are consecutive or something, but the coefficients are 1,2,6, which are not consecutive.
Perhaps it's about the number 126 in base b being a perfect square, but as we saw, no solution for b>9.
But let's double-check the calculation. b^2 +2b+6 = k^2.
Then k^2 - b^2 = 2b+6
( k-b)(k+b) = 2b+6
Let d = k-b, then k+b = (2b+6)/d
Then from d = k-b, and (2b+6)/d = k+b
Add them: d + (2b+6)/d = 2k
Subtract: (2b+6)/d - d = 2b
So (2b+6)/d - d = 2b
Multiply both sides by d: 2b+6 - d^2 = 2b d
So -d^2 + 2b + 6 = 2b d
Bring all to one side: -d^2 + 2b + 6 - 2b d = 0
2b - 2b d + 6 - d^2 = 0
2b(1 - d) + 6 - d^2 = 0
2b(1 - d) = d^2 - 6
So b = (d^2 - 6)/(2(1 - d)) for d ≠ 1
Simplify: b = (d^2 - 6)/(2-2d) = (d^2 - 6)/[2(1-d)] = - (d^2 - 6)/(2(d-1))
Since d-1 = -(1-d), so b = - (d^2 - 6)/(2(d-1)) = (6 - d^2)/(2(d-1))
Now d = k-b, integer, and since k > b, d > 0.
b > 9, so (6 - d^2)/(2(d-1)) > 9
But 6 - d^2 < 0 for d≥3, since d=2: 6-4=2>0, d=3:6-9=-3<0, etc.
So for d=2: b = (6 - 4)/(2(2-1)) = 2/(2*1) = 2/2 = 1
d=3: b = (6-9)/(2(3-1)) = (-3)/(2*2) = -3/4, not integer.
d=4: b = (6-16)/(2(4-1)) = (-10)/(6) = -5/3, not integer.
d=0: but d=k-b, if d=0, k=b, then from earlier, 0 = 2b+6, b=-3, not >9.
d=1: if d=1, from earlier, when we had division by zero, so check separately.
If d=1, k-b=1, then from (k-b)(k+b) =2b+6, so 1*(k+b) =2b+6, but k=b+1, so b+1 + b =2b+6 ⇒ 2b+1=2b+6 ⇒1=6, contradiction.
d=5: b= (6-25)/(2(5-1)) = (-19)/(8) = -19/8, not integer.
d=6: b= (6-36)/(2(6-1)) = (-30)/10 = -3, not >9.
d=0 already did.
d negative? If d<0, then k<b, but as we saw, b^2 +2b+6 > b^2, so k^2 > b^2, so k>b, so d>0. So no negative d.
So no solutions? But that can't be. Perhaps "has a friend" means that the number is a perfect cube or something else.
Perhaps it's that the polynomial has an integer root in base b, but the roots are complex, so not integer.
Another idea: perhaps "friend" refers to the fact that the number 126_b is a triangular number or something.
Let's calculate b^2 +2b+6 for b>9. b=10, 100+20+6=126, is 126 a perfect square? 11^2=121, 12^2=144, no. b=11, 121+22+6=149, not square. b=12, 144+24+6=174, not. b=13, 169+26+6=201, not. b=14, 196+28+6=230, not. b=15, 225+30+6=261, not. b=16, 256+32+6=294, not. b=17, 289+34+6=329, not. b=18, 324+36+6=366, not. b=19, 361+38+6=405, not square (20^2=400, 21^2=441). b=20, 400+40+6=446, not. b=21, 441+42+6=489, 22^2=484, 23^2=529, 489-484=5, not square. b=22, 484+44+6=534, not. b=23, 529+46+6=581, 24^2=576, 581-576=5, not square. b=24, 576+48+6=630, not. b=25, 625+50+6=681, 26^2=676, 681-676=5, not. I see a pattern; it's often 5 more than a square. b^2 +2b+6 = (b+1)^2 +5, since (b+1)^2 = b^2+2b+1, so b^2+2b+6 = (b+1)^2 +5. So it's always 5 more than a perfect square, so it can never be a perfect square for b>0, since it's strictly between (b+1)^2 and (b+2)^2.
For b>0, (b+1)^2 < b^2+2b+6 < (b+2)^2, because (b+2)^2 = b^2+4b+4, and b^2+2b+6 - (b^2+4b+4) = -2b+2, which for b>1 is negative, so yes, b^2+2b+6 < (b+2)^2 for b>1, and > (b+1)^2, so indeed never a perfect square for b>1.
So that can't be it.
Perhaps "has a friend" means that it is a prime number or something, but that would be many, and sum might be large.
Another common type is when the number is a palindrome or something, but 126 is not palindrome in general.
Perhaps in base b, the representation of the number 6 in base b is something, but that doesn't make sense.
Let's think differently. Perhaps "the polynomial has a friend" means that there is another polynomial that is friends with it, but that's vague.
Perhaps it's about the digits in the base. For example, in base b, the number x^2 +2x+6 might have digits that sum to something, but again, when evaluated.
I recall that in some problems, we have a number like 121 in base b, and we find when it's a square, but here it's 126.
Another idea: perhaps "friend" means that the number is divisible by b or something, but that would be trivial.
Let's search for similar problems. Perhaps "has a friend" is a mistranslation, and it's "is a perfect square" but we saw it's not.
Perhaps it's for the constant term or something.
Let's read the problem again: "Find the sum of all integer bases b>9 for which x^2 +2x+6 has a friend in the base b".
Perhaps "has a friend" means that there is an integer x such that the value is equal to b or something, but that would be x^2+2x+6 = b, which is a quadratic, but for each b, there might be x, but not necessarily integer.
For example, for b>9, x^2+2x+6 = b, then x = [-2 ± sqrt(4 - 4(6-b))]/2 = [-2 ± sqrt(4b - 20)]/2 = -1 ± sqrt(b-5). So for x to be integer, b-5 must be perfect square, say m^2, then b = m^2 +5, and x = -1 ± m. Since b>9, m^2 >4, so m≥3. Then b = m^2 +5 for m=3,4,5,... but then the sum would be infinite, which is not possible for a sum of bases.
So probably not.
Perhaps x is in base b, but x is a number, so it's complicated.
Another thought: perhaps "the polynomial" refers to the representation of a number, but in base b, the number is written with those coefficients, but for different bases, the number is different, but the coefficients are given, so for each b, the number is b^2 +2b+6, as before.
Perhaps "has a friend" means that this number is a perfect power or something.
Perhaps it is a triangular number. A triangular number is of the form n(n+1)/2.
So set b^2 +2b+6 = n(n+1)/2 for some integer n.
Then 2(b^2 +2b+6) = n^2 + n
2b^2 +4b+12 = n^2 + n
n^2 + n - 2b^2 -4b -12 = 0
Treat as quadratic in n: n = [-1 ± sqrt(1 + 8b^2 +32b +48)]/2 = [-1 ± sqrt(8b^2 +32b +49)]/2
Discriminant d = 8b^2 +32b +49. Need this to be perfect square.
d = 8b^2 +32b +49. For b>9, d is always odd, since 8b^2+32b even, +49 odd.
Let d = k^2, so k^2 = 8b^2 +32b +49
k^2 - 8b^2 -32b = 49
Complete the square for b.
8b^2 +32b = 8(b^2 +4b) = 8((b+2)^2 -4) = 8(b+2)^2 -32
So k^2 - 8(b+2)^2 = 49 -32 = 17
So k^2 - 8m^2 = 17, where m = b+2.
So we need to solve k^2 - 8m^2 = 17 in integers.
This is a Pell-like equation.
We can find all solutions to k^2 - 8m^2 = 17.
First, find fundamental solutions.
For m=0, k^2=17, not square.
m=1, k^2=8+17=25, k=5 or -5.
m=2, k^2=32+17=49, k=7 or -7.
m=3, k^2=72+17=89, not square.
m=4, k^2=128+17=145, not square (12^2=144, 13^2=169).
m=5, k^2=200+17=217, not square (14^2=196, 15^2=225).
m=6, k^2=288+17=305, not square (17^2=289, 18^2=324).
m=7, k^2=392+17=409, not square (20^2=400, 21^2=441).
m=8, k^2=512+17=529, k=23 or -23.
m=9, k^2=648+17=665, not square (25^2=625, 26^2=676).
m=10, k^2=800+17=817, not square (28^2=784, 29^2=841).
So far, solutions for m=1,2,8.
m=1: k=5
m=2: k=7
m=8: k=23
Now, the equation is k^2 - 8m^2 = 17.
The fundamental solution to the Pell equation x^2 - 8y^2 = 1 is x=3, y=1, since 3^2 - 8*1^2 = 9-8=1.
To generate all solutions, if (k,m) is a solution to k^2 - 8m^2 = 17, then so is (k x + 8 m y, k y + m x) for (x,y) solution to x^2 - 8y^2 = 1.
The automorphism is given by multiplying by (3 + sqrt(8))^n.
So for each fundamental solution, we can generate others.
First solution: k=5, m=1
Then next solution: k' = 53 + 811 = 15 + 8 = 23, m' = 51 + 1*3 = 5+3=8
k' = k x + d m y, with d=8, x=3, y=1, so k' = k x + d m y = 53 + 811 = 15+8=23, m' = k y + m x = 51 + 1*3 = 8, which is the (23,8) we have.
Next: k'' = 233 + 881 = 69 + 64 = 133, m'' = 231 + 8*3 = 23+24=47
Then k^2 - 8m^2 = 133^2 - 847^2 = 17689 - 82209 = 17689 - 17672 = 17, yes.
Now, also we had (k,m)=(7,2)
Apply the same: k' = 73 + 821 = 21 + 16 = 37, m' = 71 + 2*3 = 7+6=13
Then k^2 - 8m^2 = 37^2 - 813^2 = 1369 - 8169 = 1369 - 1352 = 17, yes.
Then next: k'' = 373 + 8131 = 111 + 104 = 215, m'' = 371 + 13*3 = 37+39=76
And so on.
Also, we have negative k, but since m=b+2>0, we take k positive.
Now, are there more fundamental solutions? We have m=1,2,8, but m=8 came from m=1, so only two fundamental solutions: (5,1) and (7,2).
m=1 and m=2 are both small.
Now, for each solution, m = b+2, and b>9, so m>11.
From (k,m) = (5,1), b+2=1, b=-1, not >9.
(7,2): b+2=2, b=0, not.
(23,8): b+2=8, b=6, not >9.
(37,13): b+2=13, b=11>9, good.
(133,47): b+2=47, b=45>9.
(215,76): b+76? m=76, b=74>9.
From the other chain: (7,2) we have (37,13), same as above? No, (7,2) gave (37,13), but (5,1) gave (23,8) and (133,47), so (37,13) is from (7,2).
Now next for (5,1) chain: after (133,47), next is k=1333 + 8471 = 399 + 376 = 775, m=1331 + 47*3 = 133+141=274, so b=274-2=272.
For (7,2) chain: after (37,13), next is k=373 + 8131 = 111+104=215, m=371 +13*3=37+39=76, so b=76-2=74.
But earlier from (5,1) we have (133,47), b=45, and (775,274), b=272.
From (7,2) we have (37,13), b=11, and (215,76), b=74.
Now, are there more fundamental solutions? Earlier I checked up to m=10, but m=8 gave k=23, which is already in the chain. Is there a solution with m>10 not in these chains?
For example, m=12, k^2 = 8*144 +17=1152+17=1169, is square? 34^2=1156, 35^2=1225, 1156+13=1169? 1169-1156=13, not square.
m=14, k^2=8*196 +17=1568+17=1585, 39^2=1521, 40^2=1600, 1600-1585=15, not square.
m=15, 8*225+17=1800+17=1817, 42^2=1764, 43^2=1849>1817, 1849-1817=32, not square.
m=16, 8*256+17=2048+17=2065, 45^2=2025, 46^2=2116, 2065-2025=40, not square.
m=17, 8*289+17=2312+17=2329, 48^2=2304, 49^2=2401, 2329-2304=25, so k=48. something, 48.2^2=2323.24, 48.3^2=2332.89, too big, 2329 - 2304 =25, so (48 + a)^2 = 2304 + 96a + a^2 =2329, so 96a + a^2=25, no integer a.
m=18, 8*324+17=2592+17=2609, 51^2=2601, 2609-2601=8, not square.
m=19, 8*361+17=2888+17=2905, 53^2=2809, 54^2=2916>2905, 2916-2905=11, not square.
m=20, 8*400+17=3200+17=3217, 56^2=3136, 57^2=3249>3217, 3249-3217=32, not square.
m=21, 8*441+17=3528+17=3545, 59^2=3481, 60^2=3600, 3545-3481=64, which is 8^2, so k=59 + 8? (59 + a)^2 = 3481 + 118a + a^2 =3545, so 118a + a^2=64. a=0.5, not integer. 59.5^2=3540.25, 3545-3540.25=4.75, not integer.
But earlier at m=13,14,... we have solutions at m=13,47,76, etc. from the chains.
m=13 from (37,13), m=47 from (133,47), m=76 from (215,76).
Now, is there a solution not generated by these? For example, m=3 to 10 we checked, no, m=11: k^2=8*121+17=968+17=985, 31^2=961, 32^2=1024, 985-961=24, not square.
m=12: as above, 1169-1156=13, not.
m=15:1817-1764=53? 42^2=1764, 1817-1764=53, not square.
Earlier I did m=16 to 21, no. m=22:8484+17=3872+17=3889, 62^2=3844, 63^2=3969, 3889-3844=45, not square. m=23:8529+17=4232+17=4249, 65^2=4225, 4249-4225=24, not square. m=24:8576+17=4608+17=4625, 68^2=4624, 4625-4624=1, so k=68. something? 68^2=4624, 4625 - 4624=1, so k^2=4625, k=68. something? 68^2=4624, 69^2=4761, so 4625 - 4624=1, not a square? k^2 = 4625, but 68^2=4624, 68.1^2=4637.61, too big? 68^2=4624, 4625 - 4624=1, so if it were 4625, it would be 68. something, but 68.5^2=4692.25, no: (68 + a)^2 = 4624 + 136a + a^2 =4625, so 136a + a^2=1, a=1/136, not integer. But 4625 is 65^2? 60^2=3600, 65^2=4225, 70^2=4900, 68^2=4624, 69^2=4761, so 4625 is not a perfect square? 68.01^2=4625.7601? No, 68^2=4624, so 68.1^2=68^2 + 268*0.1 + 0.1^2=4624 + 13.6 + 0.01=4637.61, too big. So 4625 is not a perfect square. Indeed, 68^2=4624, 69^2=4761, so no.
But earlier for m=24, d=8576 +49? No, for the discriminant, but in the triangular number, we have d=8b^2 +32b +49, and for b=22, m=b+2=24, d=8(24)^2 +32*24 +49? No, in the equation for triangular, we have k^2 = 8b^2 +32b +49, with b the base.
For b=10, k^2=8100 +320 +49=800+320+49=1169, as before for m=12? b=10, m=b+2=12, k^2=8144 +32*10 +49? I think I confused.
Earlier for the discriminant to be square, it was for n, but in the equation k^2 = 8b^2 +32b +49.
So for b=10, 8100=800, 3210=320, total 800+320+49=1169, and sqrt(1169)≈34.19, 34^2=1156, 35^2=1225, 1169-1156=13, not square.
b=11, 8121=968, 3211=352, 968+352=1320, +49=1369, and 37^2=1369, yes! So for b=11, it works.
Similarly, b=45, 82025=16200, 3245=1440, total 16200+1440=17640, +49=17689, and 133^2=17689? 130^2=16900, 133^2=130^2 + 21303 +3^2=16900+780+9=17689, yes.
b=74, 874^2=85476=43808, 3274=2368, total 43808+2368=46176, +49=46225, and 215^2=46225? 200^2=40000, 15^2=225, 2200*15=6000, so 40000+6000+225=46225, yes.
b=272, 8272^2=873984=591872, 32272=8704, total 591872+8704=600576, +49=600625, and 775^2=600625? 700^2=490000, 75^2=5625, 2700*75=105000, so 490000+105000=595000+5625=600625, yes.
Now, are there more? From the chains, we have b=11,45,272 from one chain, and from the other chain: (7,2) gave (37,13), b=13-2=11, same as above. (37,13) b=11, then next is (215,76), b=76-2=74, then next k=2153 + 8761=645+608=1253, m=2151 +76*3=215+228=443, so b=443-2=441.
Then b=441>9.
Similarly, from the first chain, after (775,274), b=272, next k=7753 + 82741=2325 + 2192=4517, m=7751 + 274*3=775+822=1597, b=1597-2=1595.
And so on.
But we also had a solution from (5,1) chain: (5,1),(23,8),(133,47),(775,274),... b=-1,6,45,272,...
And (7,2),(37,13),(215,76),(1253,443),... b=0,11,74,441,...
Now, is there a solution for negative m? But m=b+2>0 since b>9.
Now, are there other fundamental solutions? For example, when I checked m=0 to 10, I had m=1,2,8. m=8 is already in the first chain. But is there a solution with larger m not in these chains?
For example, suppose there is another solution. The equation k^2 - 8m^2 =17.
The minimal solutions are (k,m) = (5,1), (7,2), and also (-5,1), etc., but we consider positive.
Note that (5,1) and (7,2) are both fundamental in the sense that they are not generated from smaller by the automorphism.
Now, the general solution can be found by multiplying by the fundamental solution of the homogeneous equation.
The solutions are generated by taking each fundamental solution and multiplying by (3+sqrt8)^n.
Since the fundamental solution of x^2-8y^2=1 is (3,1), so for each solution (k,m), the next is (3k + 8m, k + 3m)? Earlier we used (k' = k x + d m y, m' = k y + m x) with x=3,y=1,d=8, so k' = k3 + 8m1, m' = k1 + m*3.
Yes, as we did.
So for (5,1): (53 +811, 51 +1*3) = (15+8,5+3)=(23,8)
Then (233 +881, 231 +8*3)=(69+64,23+24)=(133,47)
Then (1333 +8471, 1331 +47*3)=(399+376,133+141)=(775,274)
Then (7753 +82741, 7751 +274*3)=(2325+2192,775+822)=(4517,1597)
And so on. b=m-2, so b=1-2=-1, 8-2=6, 47-2=45, 274-2=272, 1597-2=1595, etc.
For (7,2): (73 +821, 71 +2*3)=(21+16,7+6)=(37,13)
Then (373 +8131, 371 +13*3)=(111+104,37+39)=(215,76)
Then (2153 +8761, 2151 +76*3)=(645+608,215+228)=(1253,443)
Then (12533 +84431, 12531 +443*3)=(3759+3544,1253+1329)=(7303,2582)
b=13-2=11, 76-2=74, 443-2=441, 2582-2=2580, etc.
Now, are there more fundamental solutions? For example, is (23,8) fundamental? But it is generated from (5,1). Similarly, (37,13) from (7,2).
But earlier when I computed, for m=0 to 20, only these, but perhaps there is a solution with larger m.
Note that since |k| must be at least sqrt(8m^2 +17) > 2sqrt2 m ≈2.828m, and we can check if there are solutions not in these sequences.
The general solution to k^2 - 8m^2 =17 can be found by considering the continued fraction of sqrt(8) or by bounding.
Note that k^2 - 8m^2 =17, so (k - 2√2 m)(k + 2√2 m) =17.
Since 17 is prime, the factors are limited.
k - 2√2 m is small, so it must be that k - 2√2 m = ±1, ±17, ±1/2, etc., but since irrational, we can consider the min and max.
The solutions are given by the convergents or by the fundamental solutions we have.
We can list all solutions by solving for each possible k mod something.
Note that k^2 ≡ 17 mod 8. 17 mod 8=1, and squares mod 8 are 0,1,4, so possible.
k^2 ≡17≡1 mod 8, so k odd.
Now, k^2 - 8m^2=17.
For m≥1, k > 2√2 m ≈ 2.828m.
The difference between consecutive solutions in the chain is large, so probably no other solutions for small m, but for larger m, but since we are generating all solutions systematically, and for b>9, we can list all with m>11.
From the two chains:
First chain: b = m - 2, with m from (5,1): m=1,8,47,274,1597,... so b= -1,6,45,272,1595,...
Second chain: m=2,13,76,443,2582,... b=0,11,74,441,2580,...
Now b>9, so from first chain: b=45,272,1595,...
From second chain: b=11,74,441,2580,...
Now, are there more? For example, is there a solution with m between 13 and 47? We checked m=14 to 46, no solutions, as earlier computation showed for m up to 24, and for larger, but since the next in chain are m=47 and m=76, which are larger, and the difference is large, likely no other solutions. To be sure, the minimal solution not in chains would have larger m.
The fundamental solution of the Pell equation is (3,1), and the solutions to k^2 - 8m^2 =17 can be found by the continued fraction expansion of sqrt(8).
sqrt(8)=2√2≈2.8284, continued fraction: 2 + 1/(1/(2.8284-2)) = 2 + 1/(1/0.8284)≈2+1/1.206≈2+0.8284? Better: 2.8284 - 2 = 0.8284, 1/0.8284≈1.2068, so 1 + 0.2068, 1/0.2068≈4.835, so 4 + 0.835, 1/0.835≈1.197, and so on.
Convergents: first 2/1, then 2 + 1/1 =3/1? Standard way: a0=2, then 1/0.8284≈1.2068, a1=1, then 1/0.2068≈4.835, a2=4, etc.
Convergents: p0=2,q0=1; p1=2*1+1=3? Formula: p0=a0=2, q0=1; p1=a1 a0 +1? Standard: for a0,a1,a2,...
p-1=1,q-1=0; p0=a0,q0=1; then p1=a1 p0 + p-1= a1 a0 +1? Better:
Let c0 = a0 = 2
c1 = a0 + 1/a1 = 2 + 1/1 = 3/1
c2 = a0 + 1/(a1 + 1/a2) = 2 + 1/(1 + 1/4) = 2 + 1/(5/4) = 2 + 4/5 = 14/5? a2=4.
c2 = [a0; a1,a2] = [2;1,4] = 2 + 1/(1 + 1/4) = 2 + 1/(5/4) = 2 + 4/5 = 14/5 = 2.8
c3 = [2;1,4,a3] with a3=1, since 1/0.835≈1.197, a3=1, then c3=2+1/(1+1/(4+1/1))=2+1/(1+1/5)=2+1/(6/5)=2+5/6=17/6≈2.8333
c4: a4= term after, 1/0.197≈5.076, a4=5, c4=2+1/(1+1/(4+1/(1+1/5)))=2+1/(1+1/(4+1/(6/5)))=2+1/(1+1/(4+5/6))=2+1/(1+1/(29/6))=2+1/(1+6/29)=2+1/(35/29)=2+29/35=89/35≈2.82857
c5: a5= term, 1/(0.076)≈13.15, a5=13, c5=2+1/(1+1/(4+1/(1+1/(5+1/13)))) and so on.
Now the convergents are 2/1, 3/1, 14/5, 17/6, 89/35, etc.
Now for each convergent p/q, |p^2 - 8 q^2| should be small, and we can see if it equals 17.
c0=2/1, p^2-8q^2=4-8= -4
c1=3/1, 9-8=1
c2=14/5, 196 - 8*25=196-200= -4
c3=17/6, 289 - 8*36=289-288=1
c4=89/35, 89^2=7921, 835^2=81225=9800, 7921-9800= -1879? No, 89^2=7921? 90^2=8100, 89^2=7921, yes. 8*1225=9800, but 7921 - 9800 = -1879, not small. I think I messed up.
The convergent c4 should be [a0;a1,a2,a3,a4]= [2;1,4,1,5] since a3=1, a4=5.
So p4 = a4 p3 + p2, q4 = a4 q3 + q2
Recurrence: p_n = a_n p_{n-1} + p_{n-2}
q_n = a_n q_{n-1} + q_{n-2}
With p_{-2}=0,p_{-1}=1? Standard:
Set n=0: p0=a0=2, q0=1
n=1: p1=a1 p0 + p_{-1}, usually p0=a0, q0=1; p1=a1 a0 +1?
List:
For n=0: p0 = a0 = 2, q0 = 1
n=1: p1 = a1 p0 + p_{-1}, but p_{-1} not defined. Better: initialize with n=-1 and n=0.
Set p_{-2}=0, p_{-1}=1? Common: for n=0, p0=a0, q0=1
But for n=1, p1 = a1 a0 + 1? Example: a0=2, a1=1, then convergent should be 2 + 1/1 =3, so p1=3, q1=1
Generally, p_n = a_n p_{n-1} + p_{n-2} for n≥0, with p_{-2}=0, p_{-1}=1? Check:
For n=0: p0 = a0 p_{-1} + p_{-2} = a0 *1 + 0? But we need p0=a0, so set p_{-1}=1, p_0 = a0 * p_{-1} + p_{-2} ? Standard is p_n = a_n p_{n-1} + p_{n-2} for n≥1, with p0=a0, p1=a1 a0 +1?
For [a0;a1] = a0 + 1/a1 = (a0 a1 +1)/a1, so p1 = a0 a1 +1, q1=a1
Then [a0;a1,a2] = a0 + 1/(a1 + 1/a2) = a0 + a2/(a1 a2 +1) = (a0 (a1 a2 +1) + a2) / (a1 a2 +1)
But a1 a2 +1 is q2? Generally, p_n = a_n p_{n-1} + p_{n-2}, with p0=a0, p1=a1 a0 +1? For a0=2,a1=1, p0=2, p1=12 +1? 3, yes. Then p2 = a2 p1 + p0 =43 +2=14, q2= a2 q1 + q0=4*1 +1? q1 should be denominator. q0=1, q1=a1=1? For n=0, q0=1, n=1, q1=a1? But for n=1, convergent is (a1 a0 +1)/a1, so q1=a1.
So for a0=2,a1=1, p0=2,q0=1; p1=1*2 +1? p1 = a1 p0 + p_{-1}, but usually defined with p0 and p1.
Set: for n=0: p0 = a0 = 2, q0 = 1
n=1: p1 = a1 p0 + p_{-1}, but p_{-1} is not standard. Better: the first convergent is a0/1, second is (a0 a1 +1)/a1
So p0 = a0, q0 = 1
p1 = a0 a1 + 1, q1 = a1
Then p2 = a2 p1 + p0, q2 = a2 q1 + q0
And so on.
So for a0=2, a1=1, a2=4, a3=1, a4=5, etc.
So p0 = 2, q0 = 1
p1 = a1 p0 + p_{-1}? From above, p1 = a0 a1 + 1 = 2*1 + 1 = 3, q1 = a1 = 1
Then p2 = a2 p1 + p0 = 4*3 + 2 = 12 + 2 = 14? But earlier I said 14/5, inconsistency.
q2 = a2 q1 + q0 = 4*1 + 1 = 5, so 14/5, yes.
p3 = a3 p2 + p1 = 114 + 3 = 17, q3 = a3 q2 + q1 = 15 + 1 = 6, so 17/6
p4 = a4 p3 + p2 = 5*17 + 14 = 85 + 14 = 99? But earlier I said 89/35, mistake.
a4=5, p4=517 + 14 = 85+14=99, q4=56 + 5=30+5=35? 99/35
But earlier I computed 89/35, but 89? No, 99/35.
99/35 = 2.82857, while sqrt8≈2.828427, close.
p^2 - 8q^2 for p4/q4=99/35: 99^2=9801, 835^2=81225=9800, so 9801-9800=1, so |p^2-8q^2|=1.
Similarly, p3/q3=17/6, 17^2=289, 8*36=288, 289-288=1.
p2/q2=14/5, 196 - 8*25=196-200=-4
p1/q1=3/1,9-8=1
p0/q0=2/1,4-8=-4
Now for the equation k^2 - 8m^2 =17, we can compute for each convergent.
But |p^2 - 8q^2| is small, for example p1/q1=3/1, |9-8|=1
p3/q3=17/6, |289-288|=1
p4/q4=99/35, |9801-9800|=1
But we need =17, so not directly.
However, the solutions to k^2 - 8m^2 =17 can be found by considering combinations.
Since the fundamental solution to x^2-8y^2=1 is (3,1), and we have solutions (k,m)=(5,1),(7,2), etc., and all solutions are generated by multiplying by (3+2√2)^n for each fundamental solution.
(3+2√2) is the fundamental unit, since (3+2√2)(3-2√2)=9-8=1.
So for each fundamental solution (k0,m0) to k^2 - 8m^2 =17, the general solution is given by k + m √8 = (k0 + m0 √8) (3 + √8)^n for integer n.
Since √8 = 2√2, but same.
So for (k0,m0)=(5,1), then k + m √8 = (5 + √8) (3 + √8)^n
Similarly for (7,2): (7 + 2√8) (3 + √8)^n
Note that √8 = 2√2, but we can write as is.
For n=0: (5,1)
n=1: (5 + √8)(3 + √8) = 5*3 + 5√8 + 3√8 + √8 * √8 = 15 + 5√8 + 3√8 + 8 = 23 + 8√8, so k=23,m=8
n=2: (5+√8)(3+√8)^2, first (3+√8)^2=9 + 6√8 + 8=17+6√8
Then (5+√8)(17+6√8)=517 + 56√8 + √8 *17 + √8 *6√8=85 + 30√8 + 17√8 + 48=133 + 47√8, so k=133,m=47
Same as before.
For (7,2): n=0: (7,2)
n=1: (7+2√8)(3+√8)=73 +7√8 +2√83 +2√8*√8=21 +7√8 +6√8 +16=37+13√8? 2*8=16, yes k=37,m=13
But earlier we had m=13 for b=11.
n=2: (7+2√8)(17+6√8)=717 +76√8 +2√817 +2√86√8=119 +42√8 +34√8 +96=215+76√8, k=215,m=76, b=74.
Now, are there other fundamental solutions? For example, is (23,8) fundamental? But it is generated from (5,1) with n=1.
Similarly, (37,13) from (7,2) n=1.
But what about negative n? For n=-1, (3+√8)^{-1} = 1/(3+√8) = (3-√8)/(9-8) =3-√8
So for (5,1): (5+√8)(3-√8)=53 -5√8 +3√8 -√8√8=15 -5√8 +3√8 -8=7 -2√8, so k=7,m=-2? But m should be positive, or take absolute value, but k=7, m=2, which is the other fundamental solution.
Similarly, for (7,2): (7+2√8)(3-√8)=73 -7√8 +2√83 -2√8*√8=21 -7√8 +6√8 -16=5 - √8, so k=5,m=1, again the other.
So actually, the two fundamental solutions are associated, and all solutions are generated by taking (k,m) = (5,1) (3+√8)^n or (7,2) (3+√8)^n for n integer, but for n negative, we get the same pairs.
Specifically, for each n, we get a solution, but for n=0: (5,1) and (7,2) are different.
For n=1: from (5,1): (23,8), from (7,2):(37,13)
But (23,8) and (37,13) are distinct.
In terms of b, for each solution, b = m - 2, since m = b+2.
So the solutions are:
From first fundamental: for n≥0: m=1,8,47,274,1597,... b=-1,6,45,272,1595,...
For n<0: but n=-1 gives (7,2), which is the other fundamental.
Similarly, from second fundamental: n≥0: m=2,13,76,443,... b=0,11,74,441,...
n<0: n=-1 gives (5,1), same as first.
So all solutions are covered by the sequences: b = m-2 with m from the orbits.
Specifically, the b values are: for the first sequence: b_n for n=0,1,2,...: b_0 = 1-2=-1, b_1=8-2=6, b_2=47-2=45, b_3=274-2=272, b_4=1597-2=1595, etc.
From second: b_0=2-2=0, b_1=13-2=11, b_2=76-2=74, b_3=443-2=441, b_4=2582-2=2580, etc.
Now b>9, so we take b≥10.
From first sequence: b=45,272,1595,...
From second: b=11,74,441,2580,...
Now, are there duplicates? 11,45,74,272,441,1595,2580,... all distinct.
Now, the problem is to sum all such b>9.
But there are infinitely many, since the sequences grow exponentially, but the sum should be finite? Or perhaps not, but the problem says "sum of all", implying finite, but here infinite.
Perhaps for "has a friend" it's not triangular numbers.
Perhaps only for the smallest or something, but the problem says "all".
Another thought: perhaps "has a friend" means that the number b^2 +2b+6 is a perfect square, but we saw it's not.
Or perhaps it's that the polynomial evaluated at x=1 or something is a perfect square in base b.
For example, when x=0, it's 6, which is always a single digit, so in base b>6, it's represented as '6', which is fine, but not interesting.
x=1: 1+2+6=9, which is 9 in base b, always a single digit for b>9, and 9 is a perfect square, but that's true for all b>9, sum infinite.
x=2: 4+4+6=14, is 14 a perfect square? No.
x=3:9+6+6=21, not square.
x=4:16+8+6=30, not.
x=5:25+10+6=41, not.
x=6:36+12+6=54, not.
x=7:49+14+6=69, not.
x=8:64+16+6=86, not.
x=9:81+18+6=105, not.
x=10:100+20+6=126, not square, as before.
So not.
Perhaps "friend" means that it is equal to b itself or something.
Another idea: perhaps "has a friend" means that there is a digit that is friendly, but vague.
Perhaps in the context, "friend" means that the number is divisible by b-1 or something, like for digit sum divisibility.
For example, a number is divisible by b-1 if the sum of digits is divisible by b-1.
For the number 126 in base b, sum of digits is 1+2+6=9.
So if 9 is divisible by b-1, for some b>9.
But b-1 >8 since b>9, b≥10, b-1≥9, and 9 is divisible by b-1 only if b-1 divides 9, so b-1|9, b-1=9, since b-1≥9, so b-1=9, b=10.
Then for b=10, 126_10 = 126, sum of digits 1+2+6=9, 9 div by 9? b-1=9, yes, 9/9=1, so divisible.
But is that "has a friend"? Perhaps, but only one base, b=10, but b>9, and sum is 10, but there might be more.
b-1 could be 3, but b-1≥9? b>9, so b≥10, b-1≥9, and divisors of 9 are 1,3,9, so only b-1=9, b=10.
But perhaps for other values.
Perhaps it's that the number itself is divisible by b-1, not just the sum.
So set b^2 +2b+6 ≡ 0 mod (b-1)
Since b≡1 mod (b-1), so 1 +2*1 +6 =9 ≡0 mod (b-1), so b-1|9.
Same as above, b-1|9, and b>9, so b-1>8, so b-1=9, b=10.
Sum is 10, but probably not, as it's too small, and the problem likely has more.
Perhaps divisible by b, but then b| b^2 +2b+6, so b|6, so b|6, b>9, no solution.
Or by b+1 or other.
Another common type is when the number is a perfect square, but we saw not.
Perhaps "friend" means that it is a palindrome in that base, but 126 in base b is a three-digit number, and for it to be palindrome, the first and last digit must be the same, but 1 and 6 are different, so unless the number is small, but in general, for b>6, it's three digits, 1 and 6 are different, so never palindrome.
So not.
Perhaps for the polynomial, when written in base b, the coefficients are single digits, which they are for b>6, but that's always true for b>6, and b>9 given, so all b, sum infinite.
So not.
Let's go back to the triangular number idea. Perhaps the problem is to find when it is a triangular number, and perhaps they want only the smallest or something, but the problem says "all", and sum, but it's infinite.
Unless for b>9, and perhaps there is a constraint that n is integer, which it is, but still infinite.
Perhaps "has a friend" means that there is a number that is a friend, like in some contexts friend means coprime or something, but unlikely.
Another idea: perhaps "friend" means that the number is a perfect power, like square, cube, etc.
We saw not square. Cube? b^2 +2b+6 = k^3.
For b>9, b=10,126, not cube (5^3=125, 6^3=216). b=11,121+22+6=149, not. b=12,144+24+6=174, not. b=13,169+26+6=201, not. b=14,196+28+6=230, not. b=15,225+30+6=261, not. b=16,256+32+6=294, not. b=17,289+34+6=329, not. b=18,324+36+6=366, not. b=19,361+38+6=405, 7^3=343, 8^3=512, no. b=20,400+40+6=446, not. b=21,441+42+6=489, 7.8^3=474.552, 7.9^3=493.039, not integer. b=22,484+44+6=534, 8.1^3=531.441, 8.2^3=551.368, 534-531.441=2.559, not. b=23,529+46+6=581, 8.3^3=571.787, 8.4^3=592.704, 581-571.787=9.213, not. b=24,576+48+6=630, 8.5^3=614.125, 8.6^3=636.056, 630-614.125=15.875, not. b=25,625+50+6=681, 8.8^3=681.472, close but 8.8^3=681.472 >681, 8.7^3=658.503, 681-658.503=22.497, not integer. So no small solutions, and likely no solutions, but possible.
But probably not.
Perhaps for x=0 or other values.
Let's try a different approach. Perhaps "has a friend" means that the polynomial has an integer root in the ring of integers of some extension, but that's too advanced.
Perhaps in base b, the number 6 is represented as a single digit, which it is for b>6, but again all.
Another idea: perhaps "friend" refers to the fact that the constant term is friendly with the base, but vague.
Perhaps the polynomial can be factored in some way in base b.
For example, in integers, but it's irreducible over Q.
In base b, the coefficients are fixed, so the polynomial is the same.
Perhaps for some b, it has roots that are integers in base b.
But the roots are -1 ± i√5, so not real.
Perhaps we need to find when the discriminant is a perfect square in base b, but discriminant is -20, and -20 in base b is not likely to be a square.
-20 is negative, not a square.
Perhaps the value at some point.
Let's look for similar problems or think differently.
Another thought: perhaps "has a friend" means that there is a digit that is 1,2, or 6, but that's always true for the coefficients.
Perhaps the number b^2 +2b+6 is equal to (b+1)^2 or something, but (b+1)^2 = b^2+2b+1, so b^2+2b+6 = (b+1)^2 +5, as before.
Perhaps it is equal to (b+2)^2 = b^2+4b+4, so b^2+2b+6 = b^2+4b+4 ⇒ 2b+6=4b+4 ⇒ 2b=2, b=1, not >9.
Or (b+3)^2 = b^2+6b+9, so b^2+2b+6 = b^2+6b+9 ⇒ 2b+6=6b+9 ⇒ 4b= -3, not.
Or (b-1)^2 = b^2-2b+1, set equal: b^2+2b+6 = b^2-2b+1 ⇒ 4b= -5, not.
So not.
Perhaps it is a multiple of b, but then b|6, no.
Let's try b=10, number 126. Is 126 a "friendly" number? In some contexts, friendly numbers are amicable, but 126 is not amicable.
Perhaps it is a Harshad number, divisible by sum of digits, which for b=10, sum digits 1+2+6=9, 126/9=14, yes, so divisible by sum of digits.
For general b, the number is 1b^2 +2b +6, sum of digits is 1+2+6=9, and for it to be divisible by 9, but in base b, the sum of digits is 9, but the number is b^2+2b+6, and for it to be divisible by 9, not by the sum of digits in base 10, but in the base b, the sum of digits is 9, and for the number to be divisible by 9, but in base b, the divisibility rule for 9 is not necessarily the same as in base 10.
In base b, a number is divisible by d if some function, but for d=9, it depends on b.
The number N = b^2 +2b+6, and we want N divisible by 9.
So b^2 +2b+6 ≡ 0 mod 9.
So b^2 +2b+6 ≡ 0 mod 9.
For b>9, b mod 9 can be 0 to 8.
Compute for each residue:
b=0 mod 9: 0+0+6=6 not 0
b=1:1+2+6=9≡0, yes
b=2:4+4+6=14≡5 not 0
b=3:9+6+6=21≡3 not 0
b=4:16+8+6=30≡3 not 0
b=5:25+10+6=41≡5 not 0
b=6:36+12+6=54≡0, yes
b=7:49+14+6=69≡6 not 0? 49 mod 9: 4+9=13,1+3=4, or 49/9=59=45, remainder 4. 14:1+4=5, 6:6, so 4+5+6=15,1+5=7? Better: b=7, b^2=49, 49 div 9: 59=45, rem 4. 2b=14, 19=9, rem 5. 6. So 4+5+6=15, 1+5=6 not 0. Or 4+5=9, +6=15, 15-91=6, not 0.
b=8:64+16+6=86, 64/9:79=63, rem 1. 16/9:19=9, rem 7? 2b=16, rem 7? 16-9*1=7. 6. So 1+7+6=14, 1+4=5 not 0.
b=0:0+0+6=6 not 0.
So only when b≡1 or 6 mod 9.
So for b>9, b=10,10 mod 9=1, so yes. b=11,11-9=2, not. b=12,3, not. b=13,4, not. b=14,5, not. b=15,6, yes. b=16,7, not. b=17,8, not. b=18,0, not. b=19,1, yes. b=20,2, not. b=21,3, not. b=22,4, not. b=23,5, not. b=24,6, yes. b=25,7, not. b=26,8, not. b=27,0, not. b=28,1, yes. and so on.
So infinitely many: b=10,15,19,24,28,33,37,42, etc.
Sum is infinite, so probably not.
But if we want b>9, and perhaps only when it is a "friend" in some other way, or perhaps this is a condition, but not sufficient.
Perhaps "has a friend" means that it is a multiple of 9, but in base b, and for the number itself, but again, many.
Perhaps for the sum of digits to be divisible by b-1, which is the divisibility rule for b-1.
In base b, a number is divisible by b-1 if the sum of its digits is divisible by b-1.
So for the number 126_b, sum of digits is 1+2+6=9, so we need b-1 | 9.
As earlier, b-1 | 9, and b>9, so b-1 >8, so b-1=9, since 9 is the only divisor greater than or equal to 9, so b=10.
Then sum is 10.
But perhaps there are other interpretations.
b-1 could be 18, but 18>8, and 9 not divisible by 18.
Or b-1=3, but b-1=3, b=4<9, not.
So only b=10.
But then the sum is 10, but the problem says "bases", plural, so probably more.
Perhaps for other values of x, but the polynomial is given, so likely for the number it represents.
Another idea: perhaps "the polynomial has a friend" means that there is an integer k such that when evaluated at k, it is equal to b in some base, but that doesn't make sense.
Perhaps in base b, the number b^2 +2b+6 is a perfect square, but we know it's not.
Let's calculate for b=10, 126, not square.
b=11,149, not.
etc.
Perhaps it is a square in a different base, but that doesn't make sense.
Another thought: perhaps "in the base b" modifies "friend", but what is a friend in base b? Vague.
Perhaps "friend" is a specific number, like 0 or 1 or b, but not specified.
Perhaps it means that the number is a friend to the base, like coprime or something.
For example, gcd(N,b) =1, where N= b^2+2b+6.
Then gcd(b, b^2+2b+6) = gcd(b,6), since b^2+2b+6 ≡ 6 mod b.
So gcd(b,6) =1, so b coprime to 6, i.e., not divisible by 2 or 3.
But then for b>9, infinitely many, sum infinite.
So not.
Perhaps gcd(N, b-1) =1 or something.
But again, likely not.
Let's try to search online or think of common problems.
Perhaps "has a friend" means that it is a palindrome, but as said, not for this number.
For the number to be a palindrome in base b.
So 126_b is a three-digit number, so for it to be palindrome, the first digit must equal the last digit, but 1 and 6 are different, so unless the number is not three-digit, but for b>6, it is three-digit, since 1* b^2 > b, so for b>6, it is three-digit number with digits 1,2,6, and 1≠6, so never palindrome.
So not.
Perhaps for other representations, but the polynomial is given as x^2+2x+6, so likely the number is 126_b.
Another idea: perhaps "friend" means that it can be written as a sum of two consecutive integers or something.
For example, N = k(k+1) for some k, but that's triangular, which we did, and we have solutions, but infinite.
And for those b, it is triangular.
But perhaps the problem is to find when it is triangular, and sum all such b>9, but since there are infinitely many, and the sum diverges, probably not.
Unless the problem has a typo, and it's for b<10 or something, but it says b>9.
Perhaps "has a friend" means that it is a perfect square, and we need to find when, but we know never.
But for b=0,1, but b>9.
Perhaps in the context, "base b" includes b=1,2, but b>9.
Another thought: perhaps the polynomial is to be evaluated at a specific x, but not specified.
Perhaps x is the base, but then it's the same.
Let's read the problem again: "Find the sum of all integer bases b>9 for which x^2 +2x+6 has a friend in the base b".
Perhaps "has a friend" means that there is a number that is a friend, and friend is defined as in number theory.
In some contexts, a number is a friend if it is not in some set, but vague.
Perhaps it means that the number is composite or prime, but 126 is composite, but for b=10, but for other b, it may be prime or composite.
For example, b=10, N=126, composite.
b=11,121+22+6=149, is 149 prime? Yes.
b=12,144+24+6=174, composite.
b=13,169+26+6=201, composite.
b=14,196+28+6=230, composite.
b=15,225+30+6=261, composite.
b=16,256+32+6=294, composite.
b=17,289+34+6=329, 7*47, composite.
b=18,324+36+6=366, composite.
b=19,361+38+6=405, composite.
b=20,400+40+6=446, composite.
b=21,441+42+6=489, 3*163, composite.
b=22,484+44+6=534, composite.
b=23,529+46+6=581, 7*83, composite.
b=24,576+48+6=630, composite.
b=25,625+50+6=681, 3*227, composite.
So for b=11, it is prime, others composite, but not clear what "friend" means.
Perhaps when it is prime, but then b=11, and perhaps others, but b=11 is the only one in this list, but for b=26,484+52+6=542, composite, b=27,729+54+6=789, composite, b=28,784+56+6=846, composite, b=29,841+58+6=905, 5181, composite, b=30,900+60+6=966, composite, b=31,961+62+6=1029, 3343, composite, b=32,1024+64+6=1094, composite, b=33,1089+66+6=1161, 3387, composite, b=34,1156+68+6=1230, composite, b=35,1225+70+6=1301, 1301 div by 1301/1301, is it prime? 1301 / 1301, check divisibility: by 2,3,5,7,11,13,17,19,23,29,31: 3141=1271, 1301-1271=30, not divisible. 3735=1295, 1301-1295=6, not. 4131=1271, as above. 4330=1290, 1301-1290=11, not. 4727=1269, 1301-1269=32, not. 5324=1272, 1301-1272=29, not. 5922=1298, 1301-1298=3, not. 6121=1281, 1301-1281=20, not. 6719=1273, 1301-1273=28, not. 7118=1278, 1301-1278=23, not. 7317=1241, 1301-1241=60, not. 7916=1264, 1301-1264=37, not. 8315=1245, 1301-1245=56, not. 8914=1246, 1301-1246=55, not. 9713=1261, 1301-1261=40, not. 10112=1212, 1301-1212=89, not. 10312=1236, 1301-1236=65, not. 10712=1284, 1301-1284=17, not. 10911=1199, 1301-1199=102, not. 11311=1243, 1301-1243=58, not. so perhaps 1301 is prime? 1301 = 1300+1, 36^2=1296, 37^2=1369>1301, so not square, but for prime, 1301 / 1301, and since 36.5^2=1332.25>1301, and we checked up to sqrt(1301)≈36.6, and 37>36.6, and we checked up to 36, but 31,33, etc, but 31<36.6, 36.6, and 31,33,35,37, but 33 and 35 not prime, we checked the primes: 2,3,5,7,11,13,17,19,23,29,31, and next is 37>36.6, so yes, 1301 is prime. So for b=35, N=35^2 +235+6=1225+70+6=1301, prime.
So there are infinitely many b where N is prime, so sum infinite.
So not.
Perhaps when it is a perfect number or something, but 126 is not perfect.
Another idea: perhaps "friend" means that it is a powerful number or something, but unlikely.
Perhaps in the context of the problem, "has a friend" means that it is a quadratic residue or something, but for what modulus.
Let's try to consider that for some b, the number N = b^2 +2b+6 is a perfect square in the integers, but we know it's not for b>1.
Perhaps in base b, it is written as a square, but that doesn't make sense.
Perhaps the polynomial has a root that is an integer in base b.
For example, the root is -1 ± i√5, so if we consider Gaussian integers, but in base b, it's not clear.
Perhaps for some b, the number 6 is not a digit, but for b>6, it is.
I think I need to guess that "has a friend" means that the number is a triangular number, and perhaps the problem intends for b>9 and perhaps only the smallest, but it says "all", and sum, but infinite.
Perhaps for b>9, and b such that the triangular number is integer, which it is, but perhaps they want the sum of the bases, but it diverges.
Unless the problem has a constraint that I missed.
Let's check the date: "Today is July 25, 2025, Friday" but probably not relevant.
Perhaps "step3" means something, but likely not.
Another thought: perhaps "friend" refers to the fact that the number is divisible by the base or something, but not.
Let's try to see if for b=10, it has a friend, what could it be.
Perhaps "has a friend" means that there is a number that is close to it or something.
Perhaps it means that the number is a sum of two squares or something.
For example, b^2 +2b+6 = a^2 + b^2 for some a,b, but then it's always possible or not.
For b=10, 126 = 11^2 +5^2 =121+25=146>126, 11^2 +3^2=121+9=130>126, 10^2 +5^2=100+25=125<126, 10^2+6^2=100+36=136>126, 9^2+7^2=81+49=130>126, 8^2+8^2=64+64=128>126, 7^2+8^2=49+64=113<126, 7^2+9^2=49+81=130>126, so 126 - 113 =13, not sum of two squares, but 126 = 11^2 +5^2? 121+25=146>126, no, 5^2=25, 11^2=121, 121+25=146>126. 1^2+11.13^2 not integer. 5^2+10.8^2 not. 3^2+11.22^2 not. In integers, 126 = 1^2 + 11.22^2 not, 2^2 + 11.13^2 not, 3^2 + 11.00^2 =9+121=130>126, 4^2 + 10.8^2=16+116.64=132.64>126, 5^2 + 10.6^2=25+112.36=137.36>126, 6^2 + 10.2^2=36+104.04=140.04>126, 7^2 + 8.8^2=49+77.44=126.44>126, 8^2 + 7.6^2=64+57.76=121.76<126, 8^2 + 7.7^2=64+59.29=123.29, 8^2 + 7.8^2=64+60.84=124.84, 8^2 + 7.9^2=64+62.41=126.41>126, so not sum of two squares. But for other b, it might be.
For b=11, 149, 12^2=144, 149-144=5, not square. 11^2+8^2=121+64=185>149, 10^2+7^2=100+49=149, yes! 10^2 +7^2=100+49=149.
So for b=11, N=149=10^2+7^2, sum of two squares.
For b=12, 144+24+6=174, 13^2=169, 174-169=5, not square. 12^2+7^2=144+49=193>174, 11^2+8^2=121+64=185>174, 10^2+8^2=100+64=164<174, 10^2+9^2=100+81=181>174, 9^2+8^2=81+64=145<174, 9^2+9^2=81+81=162<174, 9^2+10^2=81+100=181>174, so 174 - 145 =29, not sum of two squares, and 13.2^2=174.24>174, so no.
b=13,201, 14^2=196, 201-196=5, not square. 13^2+8^2=169+64=233>201, 12^2+9^2=144+81=225>201, 11^2+10^2=121+100=221>201, 10^2+9^2=100+81=181<201, 10^2+10^2=200<201, 10^2+11^2=100+121=221>201, 11^2+8^2=121+64=185<201, 11^2+9^2=121+81=202>201, so 202-201=1, so 202 = 11^2 +9^2, but 202 > 201, so for 201, 14^2=196, 201-196=5, not a square, and no other, so not sum of two squares.
b=14,230, 15^2=225, 230-225=5, not square. 14^2+8^2=196+64=260>230, 13^2+9^2=169+81=250>230, 12^2+10^2=144+100=244>230, 11^2+10^2=121+100=221<230, 11^2+11^2=242>230, 12^2+9^2=144+81=225<230, 12^2+10^2=244>230, 13^2+7^2=169+49=218<230, 13^2+8^2=169+64=233>230, so 230-218=12, not a square, and 15.2^2=231.04>230, so no.
b=15,261, 16^2=256, 261-256=5, not square. 15^2+6^2=225+36=261, yes! 15^2 +6^2=225+36=261.
So for b=11 and b=15, it is sum of two squares.
For b=10, 126, is it sum of two squares? 11^2+5^2=121+25=146>126, 10^2+5^2=100+25=125<126, 10^2+6^2=100+36=136>126, 9^2+7^2=81+49=130>126, 8^2+7^2=64+49=113<126, 8^2+8^2=128>126, 7^2+8^2=49+64=113, same, 6^2+10^2=36+100=136>126, 5^2+10^2=25+100=125, 4^2+11^2=16+121=137>126, 3^2+11^2=9+121=130>126, 2^2+11^2=4+121=125<126, 1^2+11^2=1+121=122<126, 0^2+11.22^2 not, so 126 - 122 =4, so 11^2 +2^2=121+4=125<126, 11^2 +3^2=121+9=130>126, so no integer sum of two squares.
b=16,294, 17^2=289, 294-289=5, not square. 16^2+8^2=256+64=320>294, 15^2+9^2=225+81=306>294, 14^2+10^2=196+100=296>294, 14^2+9^2=196+81=277<294, 14^2+10^2=296>294, 13^2+11^2=169+121=290<294, 13^2+12^2=169+144=313>294, 12^2+11^2=144+121=265<294, 12^2+12^2=288<294, 12^2+13^2=144+169=313>294, 11^2+13^2=121+169=290, same as above, 10^2+14^2=100+196=296>294, so 294-288=6, not a square, and 17.1^2=292.41, 17.2^2=295.84>294, so no.
b=17,329, 18^2=324, 329-324=5, not square. 17^2+8^2=289+64=353>329, 16^2+9^2=256+81=337>329, 15^2+11^2=225+121=346>329, 14^2+12^2=196+144=340>329, 13^2+12^2=169+144=313<329, 13^2+13^2=338>329, 14^2+11^2=196+121=317<329, 14^2+12^2=340>329, 15^2+10^2=225+100=325<329, 15^2+11^2=225+121=346>329, 16^2+7^2=256+49=305<329, 16^2+8^2=256+64=320<329, 16^2+9^2=256+81=337>329, so 329-320=9, so 16^2 +8^2=256+64=320, and 329 - 320 =9, not help, but 18^2 +5^2=324+25=349>329, or 17^2 +10^2=289+100=389>329, so no. 329 - 313 =16, and 16 is 4^2, but 13^2 +12^2=169+144=313, then 329 = 313 +16, but not sum of two squares directly. 18^2 =324, 329-324=5, not a square. 1^2+18.13^2 not, so not sum of two squares.
b=18,366, 19^2=361, 366-361=5, not square. 18^2+6^2=324+36=360<366, 18^2+7^2=324+49=373>366, 17^2+9^2=289+81=370>366, 16^2+10^2=256+100=356<366, 16^2+11^2=256+121=377>366, 15^2+12^2=225+144=369>366, 14^2+13^2=196+169=365<366, 14^2+14^2=392>366, 13^2+14^2=365, same, so 366-365=1, so 14^2 +13^2 =196+169=365, then 366 = 365 +1 = 14^2 +13^2 +1^2, not two squares. 19^2 +5^2=361+25=386>366, so not.
b=19,405, 20^2=400, 405-400=5, not square. 19^2+8^2=361+64=425>405, 18^2+9^2=324+81=405, yes! 18^2 +9^2=324+81=405.
So b=19 also works.
So for b=11,15,19, it is sum of two squares.
And earlier b=35, 1301, is it sum of two squares? 36^2=1296, 1301-1296=5, not square. 35^2+6^2=1225+36=1261<1301, 35^2+7^2=1225+49=1274<1301, 35^2+8^2=1225+64=1289<1301, 35^2+9^2=1225+81=1306>1301, 34^2+11^2=1156+121=1277<1301, 34^2+12^2=1156+144=1300<1301, 34^2+13^2=1156+169=1325>1301, 33^2+14^2=1089+196=1285<1301, 33^2+15^2=1089+225=1314>1301, 32^2+16^2=1024+256=1280<1301, 32^2+17^2=1024+289=1313>1301, 31^2+18^2=961+324=1285<1301, 31^2+19^2=961+361=1322>1301, 30^2+19^2=900+361=1261<1301, 30^2+20^2=900+400=1300<1301, 30^2+21^2=900+441=1341>1301, 29^2+22^2=841+484=1325>1301, so 1300 = 30^2 +20^2 =900+400=1300, then 1301 - 1300 =1, so 1301 = 30^2 +20^2 +1^2, not two squares. 36^2=1296, 1301-1296=5, not a square, and no other, so not sum of two squares for b=35.
For the ones that are sum of two squares: b=11,15,19.
b=20,400+40+6=446, 21^2=441, 446-441=5, not square. 20^2+4^2=400+16=416<446, 20^2+5^2=400+25=425<446, 20^2+6^2=400+36=436<446, 20^2+7^2=400+49=449>446, 19^2+8^2=361+64=425<446, 19^2+9^2=361+81=442<446, 19^2+10^2=361+100=461>446, 18^2+11^2=324+121=445<446, 18^2+12^2=324+144=468>446, 17^2+13^2=289+169=458>446, so 446-445=1, so 18^2 +11^2=324+121=445, then 446 = 445 +1, not two squares. 21^2 +5^2=441+25=466>446, so not.
b=21,441+42+6=489, 22^2=484, 489-484=5, not square. 21^2+6^2=441+36=477<489, 21^2+7^2=441+49=490>489, 20^2+8^2=400+64=464<489, 20^2+9^2=400+81=481<489, 20^2+10^2=400+100=500>489, 19^2+10^2=361+100=461<489, 19^2+11^2=361+121=482<489, 19^2+12^2=361+144=505>489, 18^2+12^2=324+144=468<489, 18^2+13^2=324+169=493>489, 17^2+14^2=289+196=485<489, 17^2+15^2=289+225=514>489, 16^2+15^2=256+225=481<489, 16^2+16^2=512>489, so 489-485=4, so 17^2 +14^2=289+196=485, then 489 = 485 +4 = 17^2 +14^2 +2^2, not two squares. 22^2 +5^2=484+25=509>489, so not.
b=22,484+44+6=534, 23^2=529, 534-529=5, not square. 22^2+7^2=484+49=533<534, 22^2+8^2=484+64=548>534, 21^2+9^2=441+81=522<534, 21^2+10^2=441+100=541>534, 20^2+11^2=400+121=521<534, 20^2+12^2=400+144=544>534, 19^2+12^2=361+144=505<534, 19^2+13^2=361+169=530<534, 19^2+14^2=361+196=557>534, 18^2+14^2=324+196=520<534, 18^2+15^2=324+225=549>534, 17^2+16^2=289+256=545>534, so 534-530=4, so 19^2 +13^2=361+169=530, then 534 = 530 +4 = 19^2 +13^2 +2^2, not two squares. 23^2 +5^2=529+25=554>534, so not.
b=23,529+46+6=581, 24^2=576, 581-576=5, not square. 23^2+8^2=529+64=593>581, 22^2+9^2=484+81=565<581, 22^2+10^2=484+100=584>581, 21^2+11^2=441+121=562<581, 21^2+12^2=441+144=585>581, 20^2+12^2=400+144=544<581, 20^2+13^2=400+169=569<581, 20^2+14^2=400+196=596>581, 19^2+14^2=361+196=557<581, 19^2+15^2=361+225=586>581, 18^2+16^2=324+256=580<581, 18^2+17^2=324+289=613>581, 17^2+16^2=289+256=545<581, etc. 581-580=1, so 18^2 +16^2=324+256=580, then 581 = 580 +1, not two squares. 24^2 +5^2=576+25=601>581, so not.
b=24,576+48+6=630, 25^2=625, 630-625=5, not square. 24^2+5^2=576+25=601<630, 24^2+6^2=576+36=612<630, 24^2+7^2=576+49=625<630, 24^2+8^2=576+64=640>630, 23^2+9^2=529+81=610<630, 23^2+10^2=529+100=629<630, 23^2+11^2=529+121=650>630, 22^2+11^2=484+121=605<630, 22^2+12^2=484+144=628<630, 22^2+13^2=484+169=653>630, 21^2+13^2=441+169=610<630, 21^2+14^2=441+196=637>630, 20^2+15^2=400+225=625<630, 20^2+16^2=400+256=656>630, 19^2+16^2=361+256=617<630, 19^2+17^2=361+289=650>630, 18^2+17^2=324+289=613<630, 18^2+18^2=648>630, so 630-629=1, so 23^2 +10^2=529+100=629, then 630 = 629 +1, not two squares. 25^2 +5^2=625+25=650>630, so not.
b=25,625+50+6=681, 26^2=676, 681-676=5, not square. 25^2+6^2=625+36=661<681, 25^2+7^2=625+49=674<681, 25^2+8^2=625+64=689>681, 24^2+9^2=576+81=657<681, 24^2+10^2=576+100=676<681, 24^2+11^2=576+121=697>681, 23^2+11^2=529+121=650<681, 23^2+12^2=529+144=673<681, 23^2+13^2=529+169=698>681, 22^2+13^2=484+169=653<681, 22^2+14^2=484+196=680<681, 22^2+15^2=484+225=709>681, 21^2+15^2=441+225=666<681, 21^2+16^2=441+256=697>681, 20^2+16^2=400+256=656<681, 20^2+17^2=400+289=689>681, 19^2+18^2=361+324=685>681, so 681-680=1, so 22^2 +14^2=484+196=680, then 681 = 680 +1, not two squares. 26^2 +5^2=676+25=701>681, so not.
b=26,676+52+6=734, 27^2=729, 734-729=5, not square. 26^2+7^2=676+49=725<734, 26^2+8^2=676+64=740>734, 25^2+10^2=625+100=725<734, 25^2+11^2=625+121=746>734, 24^2+11^2=576+121=697<734, 24^2+12^2=576+144=720<734, 24^2+13^2=576+169=745>734, 23^2+13^2=529+169=698<734, 23^2+14^2=529+196=725<734, 23^2+15^2=529+225=754>734, 22^2+15^2=484+225=709<734, 22^2+16^2=484+256=740>734, 21^2+16^2=441+256=697<734, 21^2+17^2=441+289=730<734, 21^2+18^2=441+324=765>734, 20^2+18^2=400+324=724<734, 20^2+19^2=400+361=761>734, 19^2+18^2=361+324=685<734, etc. 734-730=4, so 21^2 +17^2=441+289=730, then 734 = 730 +4 = 21^2 +17^2 +2^2, not two squares. 27^2 +5^2=729+25=754>734, so not.
b=27,729+54+6=789, 28^2=784, 789-784=5, not square. 27^2+6^2=729+36=765<789, 27^2+7^2=729+49=778<789, 27^2+8^2=729+64=793>789, 26^2+9^2=676+81=757<789, 26^2+10^2=676+100=776<789, 26^2+11^2=676+121=797>789, 25^2+11^2=625+121=746<789, 25^2+12^2=625+144=769<789, 25^2+13^2=625+169=794>789, 24^2+12^2=576+144=720<789, 24^2+13^2=576+169=745<789, 24^2+14^2=576+196=772<789, 24^2+15^2=576+225=801>789, 23^2+14^2=529+196=725<789, 23^2+15^2=529+225=754<789, 23^2+16^2=529+256=785<789, 23^2+17^2=529+289=818>789, 22^2+16^2=484+256=740<789, 22^2+17^2=484+289=773<789, 22^2+18^2=484+324=808>789, 21^2+18^2=441+324=765<789, 21^2+19^2=441+361=802>789, 20^2+19^2=400+361=761<789, 20^2+20^2=800>789, so 789-785=4, so 23^2 +16^2=529+256=785, then 789 = 785 +4, not two squares. 28^2 +5^2=784+25=809>789, so not.
b=28,784+56+6=846, 29^2=841, 846-841=5, not square. 28^2+5^2=784+25=809<846, 28^2+6^2=784+36=820<846, 28^2+7^2=784+49=833<846, 28^2+8^2=784+64=848>846, 27^2+9^2=729+81=810<846, 27^2+10^2=729+100=829<846, 27^2+11^2=729+121=850>846, 26^2+11^2=676+121=797<846, 26^2+12^2=676+144=820<846, 26^2+13^2=676+169=845<846, 26^2+14^2=676+196=872>846, 25^2+13^2=625+169=794<846, 25^2+14^2=625+196=821<846, 25^2+15^2=625+225=850>846, 24^2+14^2=576+196=772<846, 24^2+15^2=576+225=801<846, 24^2+16^2=576+256=832<846, 24^2+17^2=576+289=865>846, 23^2+16^2=529+256=785<846, 23^2+17^2=529+289=818<846, 23^2+18^2=529+324=853>846, 22^2+18^2=484+324=808<846, 22^2+19^2=484+361=845<846, 22^2+20^2=484+400=884>846, 21^2+19^2=441+361=802<846, 21^2+20^2=441+400=841<846, 21^2+21^2=882>846, 20^2+20^2=800<846, 20^2+21^2=400+441=841<846, 20^2+22^2=400+484=884>846, 19^2+22^2=361+484=845<846, 19^2+23^2=361+529=890>846, 18^2+23^2=324+529=853>846, so many, but 846-845=1, and we have 22^2 +19^2=484+361=845, or 20^2 +21^2=400+441=841, 26^2 +13^2=676+169=845, same, so 846 = 845 +1 = 22^2 +19^2 +1^2 or 26^2 +13^2 +1^2, not two squares. 29^2 +5^2=841+25=866>846, so not.
b=29,841+58+6=905, 30^2=900, 905-900=5, not square. 29^2+8^2=841+64=905, yes! 29^2 +8^2=841+64=905.
So b=29 also works.
So b=11,15,19,29, and likely more.
But still infinitely many, since the triangular number case had infinitely many, and this may too.
For b=35, it was not, but b=45 from earlier, N=45^2 +2*45+6=2025+90+6=2121, is it sum of two squares? 46^2=2116, 2121-2116=5, not square. 44^2+13^2=1936+169=2105<2121, 44^2+14^2=1936+196=2132>2121, 43^2+16^2=1849+256=2105<2121, 43^2+17^2=1849+289=2138>2121, 42^2+18^2=1764+324=2088<2121, 42^2+19^2=1764+361=2125>2121, 41^2+20^2=1681+400=2081<2121, 41^2+21^2=1681+441=2122>2121, 40^2+21^2=1600+441=2041<2121, 40^2+22^2=1600+484=2084<2121, 40^2+23^2=1600+529=2129>2121, 39^2+22^2=1521+484=2005<2121, 39^2+23^2=1521+529=2050<2121, 39^2+24^2=1521+576=2097<2121, 39^2+25^2=1521+625=2146>2121, 38^2+25^2=1444+625=2069<2121, 38^2+26^2=1444+676=2120<2121, 38^2+27^2=1444+729=2173>2121, 37^2+26^2=1369+676=2045<2121, 37^2+27^2=1369+729=2098<2121, 37^2+28^2=1369+784=2153>2121, 36^2+27^2=1296+729=2025<2121, 36^2+28^2=1296+784=2080<2121, 36^2+29^2=1296+841=2137>2121, 35^2+29^2=1225+841=2066<2121, 35^2+30^2=1225+900=2125>2121, 34^2+30^2=1156+900=2056<2121, 34^2+31^2=1156+961=2117<2121, 34^2+32^2=1156+1024=2180>2121, 33^2+31^2=1089+961=2050<2121, 33^2+32^2=1089+1024=2113<2121, 33^2+33^2=2156>2121, 32^2+32^2=1024+1024=2048<2121, 32^2+33^2=1024+1089=2113<2121, 32^2+34^2=1024+1156=2180>121, 31^2+34^2=961+1156=2117<2121, 31^2+35^2=961+1225=2186>2121, 30^2+35^2=900+1225=2125>2121, so 2121-2117=4, and 31^2+34^2=961+1156=2117, so 2121 = 31^2 +34^2 +2^2, not two squares. 46^2 +5^2=2116+25=2141>2121, so not.
But for b=11,15,19,29, it works, and likely more, so sum infinite.
Perhaps "has a friend" means that it is a perfect square, and we need to find when, but never for b>1.
Perhaps for b=0, but not.
Another idea: perhaps "in the base b" means that the expression "x^2 +2x+6" is evaluated in base b, but with x also in that base, but x is a number, so it's complicated.
Perhaps x is an integer, and we need that for some x, the value is integer, which it is, or perhaps that it is b or something.
I think I need to assume that "has a friend" means that the number is a triangular number, and perhaps the problem is for b<10 or something, but it says b>9.
Perhaps "integer bases b>9" and "all", but perhaps only one or two.
From the triangular number case, for b>9, we have b=11,45,74, etc.
Perhaps the sum is for the bases where it is triangular, and perhaps they want the sum of the bases, but it's infinite.
Unless the problem is to find when it is a perfect square, and there are no such b, sum 0, but unlikely.
Perhaps for the polynomial to have a root in the base, but not.
Let's try to search for "polynomial has a friend in base b" online, but since it's a specific problem, hard.
Perhaps "friend" means that it is divisible by b, but then b|6, no.
Or by 1, always.
Another thought: in some contexts, a number is said to be friendly if it is a sum of distinct divisors or something, but for 126, sum of divisors 1+2+3+6+7+9+14+18+21+42+63+126=1+2=3, +3=6, +6=12, +7=19, +9=28, +14=42, +18=60, +21=81, +42=123, +63=186, +126=312 >126, so not perfect number.
Perhaps for the number to be 0 in some base, but not.
I recall that in some problems, "has a friend" means that it is a palindrome, but as said, not for this number.
For this number to be a palindrome in its base b representation, but for b>6, it's "126", which is not palindrome since 1≠6.
For b<7, but b>9, not possible.
For b=6, 126_6 = 136 +26 +6=36+12+6=54, which is "126" in base 6, but 1,2,6 are digits, 6 is not a single digit in base 6, since digits must be 0-5, so not valid.
For b=7, 149 +27 +6=49+14+6=69, digits 1,2,6, 6>5, not valid digit.
So for b>6, it is three digits, 1,2,6, not palindrome.
So not.
Perhaps for other values of the polynomial, but it's given as x^2+2x+6, so likely the number it represents when x is the base, but then it's the same.
Perhaps x is not the base, but the base is b, and x is an integer, and we need that for some x, the value is "friend" in base b.
But still vague.
Perhaps "friend" means that it is equal to the base or something.
I think I need to assume that "has a friend" means that the number is a triangular number, and perhaps the problem has a typo, and it's for b<10 or b>9 but only finite, or perhaps only when it is also a sum of two squares or something.
From earlier, for b=11, N=149, which is prime, and sum of two squares, and also for b=15, N=261=929, and 261=15^2 +6^2, and for b=19, N=405=581=59^2, and 18^2+9^2, for b=29, N=905=5181, and 29^2+8^2.
But for triangular numbers, we have b=11,45, etc.
For b=45, N=45^2 +245+6=2025+90+6=2121, is it triangular? If N= k(k+1)/2, then k(k+1)=22121=4242, k^2+k-4242=0, discriminant 1+44242=1+16968=16969, sqrt? 130^2=16900, 131^2=17161>16969, 130.2^2=16952.04, 130.3^2=16978.09>16969, 130.25^2=16972.5625, 130.24^2=16972.0576? 130^2=16900, 0.242130=62.4, 0.24^2=0.0576, so 16900+62.4+0.0576=16962.4576, 130.3^2=130^2 +2130*0.3 +0.3^2=16900+78+0.09=16978.09, 16969 - 16962.4576=6.5424, not integer, so not triangular.
For b=11, N=121+22+6=149, is it triangular? 2N=298, k(k+1)=298, k^2+k-298=0, d=1+1192=1193, not square, so not triangular.
So different.
Perhaps for this problem, "has a friend" means that it is a sum of two squares.
Then for b>9, we have b=11,15,19,29, and likely more.
For example, b=35, not, b=45, not, b=74, N=74^2 +2*74+6=5476+148+6=5630, is it sum of two squares? 75^2=5625, 5630-5625=5, not square. 74^2+8^2=5476+64=5540<5630, 74^2+9^2=5476+81=5557<5630, 74^2+10^2=5476+100=5576<5630, 74^2+11^2=5476+121=5597<5630, 74^2+12^2=5476+144=5620<5630, 74^2+13^2=5476+169=5645>5630, so 5630-5620=10, not a square, and 75^2 +5^2=5625+25=5650>5630, so not.
b=441, large, but likely yes.
But still infinite.
Perhaps the problem is to find when the polynomial has an integer root, but it doesn't.
Or when it has rational roots, but