I need to prove that the fraction $\frac{21n+4}{14n+3}$ is irreducible for every natural number $n$. Irreducible means that the numerator and denominator have no common divisors other than 1; they're coprime.

So, I need to show that the greatest common divisor (GCD) of $21n+4$ and $14n+3$ is 1 for all natural numbers $n$.

Let me denote $a=21n+4$ and $b=14n+3$. I need to show that $\gcd (a,b)=1$.

I recall that for two expressions like this, I can use the Euclidean algorithm to find their GCD. The Euclidean algorithm says that $\gcd (a,b)=\gcd (b,a\mathrm{mod}b)$.

So, let me compute $a\mathrm{mod}b$. That is, divide $a$ by $b$ and find the remainder.

Set up the division: $\frac{21n+4}{14n+3}$.

I could think of it as finding the remainder when $21n+4$ is divided by $14n+3$.

Let me do polynomial division or just algebraic manipulation.

Assume that $21n+4=q\cdot (14n+3)+r$, and I need to find q and r.

Since both are linear in n, I can solve for n in terms of the divisor, but that might be messy.

Set $d=\gcd (21n+4,14n+3)$.

Since d divides both, it must divide any linear combination of them.

So, let me find a linear combination that eliminates n.

Consider $3\times (21n+4)-4\times (14n+3)$, because 3 and 4 are the coefficients.

Compute that:

$3(21n+4)=63n+12$

$4(14n+3)=56n+12$

Now, $3a-4b=(63n+12)-(56n+12)=63n+12-56n-12=7n$.

So, $3a-4b=7n$.

But n is a variable, so this still has n in it. Not constant.

d divides any linear combination, so d divides $3a-4b=7n$.

Similarly, d divides b = 14n + 3.

So, d divides 7n and 14n + 3.

Now, since d divides 7n and 14n + 3, I can use this.

Set d divides 7n, so 7n ≡ 0 mod d.

And d divides 14n + 3, so 14n + 3 ≡ 0 mod d.

But since 7n ≡ 0 mod d, then 14n ≡ 0 mod d, so 14n + 3 ≡ 3 ≡ 0 mod d, so d divides 3.

Is that right?

If 7n ≡ 0 mod d, and d divides 14n + 3.

But 14n = 2 * 7n, so if 7n ≡ 0 mod d, then 14n ≡ 0 mod d, so 14n + 3 ≡ 0 + 3 ≡ 3 mod d, and since d divides 14n + 3, it must be that d divides 3.

So, d | 3.

That means d is a divisor of 3, so d=1 or d=3.

But it could be 3, so I need to show that d cannot be 3.

In other words, I need to show that 3 does not divide both 21n+4 and 14n+3 for any n, or rather, for all n, they are not both divisible by 3.

d could be 3, but I need to show that it's impossible for d to be 3, meaning that for all n, gcd is 1, not 3.

But from above, d divides 3, so d=1 or 3, and if I can show that d is never 3, then it must be 1.

So, I need to show that 21n+4 and 14n+3 are never both divisible by 3 for any natural number n.

In other words, it is impossible for 3 to divide both.

So, let's check when 21n+4 is divisible by 3.

21n is always divisible by 3 because 21÷3=7, so 21n ≡ 0 mod 3.

Thus, 21n + 4 ≡ 0 + 4 ≡ 1 mod 3, since 4÷3 is 1 with remainder 1.

4 mod 3 is 1.

So 21n + 4 ≡ 1 mod 3, which means it is never divisible by 3, for any n.

21n is divisible by 3, 4 is 1 mod 3, so sum is 1 mod 3, not 0, so indeed, 21n+4 is never divisible by 3.

Similarly, for 14n+3: 14 mod 3 is 2, since 14÷3=4*3=12, remainder 2.

So 14n + 3 ≡ 2n + 0 mod 3? 3 is divisible by 3, so 3≡0 mod 3.

So 14n + 3 ≡ 2n + 0 ≡ 2n mod 3.

This is divisible by 3 only if 2n ≡ 0 mod 3.

Since 2 and 3 are coprime, this implies n ≡ 0 mod 3.

So if n is divisible by 3, then 14n+3 is divisible by 3.

But from above, 21n+4 is never divisible by 3, so it cannot be that both are divisible by 3.

Therefore, d cannot be 3, so since d divides 3 and d cannot be 3, it must be that d=1 for all n.

Is that it?

Let me summarize.

We have that any common divisor d of a and b must divide any linear combination, in particular, d divides 3a - 4b = 7n, but more importantly, since d divides a and b, and we have that d divides 3, as shown earlier.

From the linear combination, we have d divides 7n and d divides 14n+3.

Then, since d divides 7n, and d divides 14n+3, and 14n = 27n, so 14n+3 = 2(7n) + 3.

Since d divides 7n, let 7n = k d for some integer k.

Then 14n+3 = 2*(7n) + 3 = 2kd + 3.

Since d divides this, d divides 2kd + 3, and d divides 2kd, so d must divide 3.

Yes, so d|3.

Thus d=1 or 3.

Now, for d to be 3, it must divide both a and b.

But a = 21n+4 ≡ 1 mod 3, as 21n≡0, 4≡1, so a ≡1 mod 3, not 0, so never divisible by 3.

Therefore, d cannot be 3, so d=1.

Hence, the fraction is irreducible for all natural n.

But is that sufficient? Natural numbers include positive integers, and for all n, it should hold.

What if n is such that b=14n+3 is zero? But 14n+3=0 when n=-3/14, not natural number, so denominator never zero for natural n.

Also, for n=1, let's compute: a=211+4=25, b=141+3=17, gcd(25,17)=1, since 17 prime, 25 not multiple.

n=2, a=42+4=46, b=28+3=31, 31 prime, 46 not multiple, gcd=1.

n=3, a=63+4=67, b=42+3=45, 67 and 45, 67-45=22, gcd(45,22)= gcd(22,45-2*22=45-44=1), gcd(22,1)=1.

Similarly, if n divisible by 3, say n=3, a=67 not div by 3, b=45 div by 3, but a not, so gcd not 3.

n=6, a=216+4=126+4=130, b=146+3=84+3=87, gcd(130,87).

87 into 130: 1 time, 130-87=43, then gcd(87,43), 87-2*43=87-86=1, gcd(43,1)=1.

And 130 div by 2,5, etc, 87 div by 3, but no common factors.

So seems correct.

In the beginning, I had 3a - 4b = 7n, and d divides 7n.

But since d divides a and b, and a and b are integers, d divides 7n.

But n could be anything, so d divides 7n, but d also divides b=14n+3.

From d|7n and d|14n+3, since 14n+3 = 27n + 3, and d|7n, so d|27n, and d|27n +3, so d| (27n +3 - 2*7n) =3, same as before.

So no issue.

To be thorough, suppose d divides both, then d|3 as above, and since a not div by 3, d=1.

Another way: the fraction is irreducible if gcd(numerator, denominator)=1.

And we showed that any common divisor must divide 3, but since numerator not divisible by 3, no common divisor 3, so only 1.

But what about other divisors? Since d must divide 3, the only possible common divisors are 1 and 3, and 3 is impossible, so only 1.

Yes.

I think that's the proof.

So, to write it neatly.