TITLE: Proving $\gcd((a+b)^m,(a-b)^m)\le2^m$ for coprime $a,b$ and $m \in \Bbb N$
QUESTION [5 upvotes]: If $a$ and $b$ are two coprime integers then prove that $\gcd((a+b)^m, (a-b)^m) \leq 2^m$
My attempt:
I took two cases first one in which a and b are both odd then very clearly the gcd will be 2^m but I'm stuck in the second case in which one of the integer is even and the other one is odd I am reaching no where I tried everything even binomial expansion but I'm stuck.

REPLY [8 votes]: Let $p$ be a prime that divides both $(a+b)^m$ and $(a-b)^m$. Then $p$ divides $a+b$ and $a-b$. So $p$ divides the sum and difference, that is, $2a$ and $2b$. Since $a$ and $b$ are relatively prime, we get $p=2$.
Now let $2^k$ be the highest power of $2$ that divides $a+b$, and $2^l$ be the highest power of $2$ that divides $a-b$. Let $e$ be the minimum of $k$ and $l$. 
Then $2^e$ divides $2a$, and $2^e$ divides $2b$. Since $a$ and $b$ are relatively prime, we have $e\le 1$.
If $a$ and $b$ have different parities, then $2$ does not divide $a+b$, so $e=0$.  
Suppose now that $a$ and $b$ are both odd. Then for one of them, the highest power that divides it is $2$, and for the other it is $\gt 2$. It follows that the gcd of $(a+b)^m$ and $(a-b)^m$ is $2^m$.
We conclude that in fact the gcd is either $1$ or $2^m$.