If $GCD(a,b)=1$, then $GCD((a+b)^m , (a-b)^m )$ is at most $2^m$? [duplicate]

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 StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Aug 5 at 14:49

• Hint: $GCD(a+b, a-b)$ is either $1$ or $2$. – SiXUlm Aug 5 at 14:42
• 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 $m$. – quarague Aug 5 at 14:42