Show that $5n+3$ and $7n+4$ are relatively prime for all $n$.

Hint: $\gcd(5n+3,7n+4)=\gcd(35n+21,35n+20)$, why?

Once you can verify my question, all that remains is to show that consecutive integers are coprime.

If $p$ is a prime divisor of $5n+3$ and $7n+4$ then $$5n+3 \equiv 0 \pmod p$$ and $$7n+4 \equiv 0 \pmod p.$$ At least one of these is not satisfied when $p \in \{5,7\}$. Otherwise, $7$ and $5$ are invertible modulo $p$ and we can rearrange these equations as $$\frac{-4}{7} \equiv n \equiv \frac{-3}{5} \pmod p.$$ This implies $-20 \equiv -21 \pmod p$, giving a contradiction, since $p \geq 2$.

Since $7(5n+3) - 5(7n+4)=1$ the greatest common divisor is $5n+3$ and $7n+4$ is $1$ (by Bezout's identity).