I'm stuck for a few days in the following problem:
If $GCD(a,b)=1$, then $GCD((a+b)^m , (a-b)^m )$ is at most $2^m$.
Can you give me a hint?
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1Hint: $GCD(a+b, a-b)$ is either $1$ or $2$. – SiXUlm Aug 05 '19 at 14:42
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1Separate the proof into two pieces. First consider the case $m=1$. Second, assuming the claim holds for $m=1$, proof it holds for all $m$. – quarague Aug 05 '19 at 14:42