for each natural number n, the numbers $4^{4n + 2} + 3^{n + 2}$ and $3^{n + 2}$ are coprime.

Question

I am trying to check that, for each natural number n, the numbers $\ 4^{4n + 2} + 3^{n + 2}$ and $\ 3^{n + 2}$ are coprime.

This tells me that gcd(a,b)=1. Now I need to see if applies to all n

I am not sure if this the right way and how to do it $\ dn = gcd(4^{4n + 2} + 3^{n + 2}, 3^{n + 2})$

I would appreciate if anyone could help me to understand the right way to do it.

Thanks

Roberto

Answer 1

Hint:

Set $ a=4^{4n + 2}$ and $b= 3^{n + 2}$ and use $\gcd(a+b,b)=\gcd(a,b)$.