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?


marked as duplicate by Bill Dubuque elementary-number-theory Aug 5 at 14:49

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    $\begingroup$ Hint: $GCD(a+b, a-b)$ is either $1$ or $2$. $\endgroup$ – SiXUlm Aug 5 at 14:42
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    $\begingroup$ Separate the proof into two pieces. First consider the case $m=1$. Second, assuming the claim holds for $m=1$, proof it holds for all $m$. $\endgroup$ – quarague Aug 5 at 14:42