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

So we can find integers $\displaystyle x $ and $\displaystyle y $ such that $\displaystyle (5n+3)x + (7n+4)y = 1 $ for all $\displaystyle n $. But I don't think this would be a very efficient way of doing this because you have to check it for all $\displaystyle n $.

Maybe we could to the following: Show that $\displaystyle 7n+4 $ has a multiplicative inverse modulo $\displaystyle 5n+3 $? So show that there exists an integer $\displaystyle y $ such that $\displaystyle (7n+4)y \equiv 1 (\text{mod} \ (5n+3)) $?