I came across this question while studying the book Challenge and Thrill of Pre-College Mathematics:
If $a$ and $b$ are relatively prime integers, then prove that $$\gcd((a+b)^m, (a-b)^m)\leq2^m$$
This is my attempt:
Let $\gcd((a+b),(a-b))=k$. Then $k|(a+b)$ and $k|(a-b)$. Thus $k|(a+b)+(a-b)$ and $k|(a+b)-(a-b)\Rightarrow k|2a$ and $k|2b$.
Now either $k|2$, $k|a$, or $k|b$. If the first is true then $k\leq2$, and if the latter two are true then $k|\gcd(a,b) \Rightarrow k|1$, in which case we can again say that $k=1\leq2$.
Now we know that $\gcd((a+b)^m,(a-b)^m)=k^m$, and since $k\leq2$ it implies $k^m\leq2^m$, which proves our proposition. QED.
Since I'm new to this subject, I'm finding this proof slightly shoddy. Is the logic correct? I'm concerned about the proof that $k\leq2$ the most.