# For $a , b \in \mathbb{Z}^+$ prove that if $\gcd(a , b) = 1$ then $\gcd{(a^n + b^n , a^n - b^n)} \le 2 ~~~\forall n \in \mathbb{N}$.

For $a , b \in \mathbb{Z}^+$ prove that if $\gcd(a , b) = 1$ then $\gcd{(a^n + b^n , a^n - b^n)} \le 2 ~~~\forall n \in \mathbb{N}$.

I tried proving the contrapositive $$\gcd{(a^n + b^n , a^n - b^n)} > 2 \implies \gcd{(a , b) > 1}$$ using Bezout's lemma but that didnt help much much. Using Bezout's lemma on original statement, all I could conclude is that $\gcd{(a , b)} \mid \gcd{(a^n + b^n , a^n - b^n)}$ which is trivial. I wasn't able to prove the base case $n = 1$ while trying induction.

Any help would be appreciated.

• The gcd divides both $2a^n$ and $2b^n$. – Lord Shark the Unknown Dec 20 '17 at 18:05

We first note that $\gcd(a^n,b^n)=1$.
Suppose $x | a^n +b^n$ and $x|a^n -b^n$ then $x|2a^n$ and $x|2b^n$. Thus $x| 2 \gcd(a^n,b^n) = 2$.