# If $a$ and $b$ are coprime integers then prove that $\gcd{((a+b)^m, (a-b)^m)} \leq 2^m$. [duplicate]

The question I am working on is this one:

If $$a$$ and $$b$$ are coprime integers then prove that $$\gcd{((a+b)^m, (a-b)^m)} \leq 2^m$$.

Here is the progress that I have made:

$$\gcd{(a, b)} \mid a-b$$ and $$\gcd{(a, b)} \mid a+b$$.

Hence, $$\gcd{(a, b)} \mid \gcd{((a+b), (a-b))} \implies \gcd{((a+b), (a-b))} \geq \gcd{(a, b)} = 1$$.

Now, $$\gcd{((a+b), (a-b))} \mid 2a$$ and $$\gcd{((a+b), (a-b))} \mid 2b$$. From a similar line of reasoning as above, $$\gcd{((a+b), (a-b))} \geq \gcd{(2a,2b)} = 2\cdot\gcd{(a,b)} = 2$$.

Hence, I have proved $$1 \leq \gcd{((a+b), (a-b))} \leq 2$$.

Now, if I manage to prove that if $$\gcd{(a,b)}=d$$, then $$\gcd{(a^n,b^n)}=d^n$$, it would allow me to finish the above proof.

I tried to justify this by saying that if $$a$$ and $$b$$ have some prime factors in common, $$a^n$$ and $$b^n$$ have the same prime factors in common, but they will be exponentiated by $$n$$. However, I doubt I will be able to write this explanation in a exam or take the above for granted. I need help formalizing this.

If $$\gcd(a,b)=d$$ then there exist $$p,q$$ such that$$a=dp\\b=dq\\\gcd(p,q)=1$$therefore $$a^n=d^np^n\\b^n=d^nq^n\\\gcd(p^n,q^n)=1$$and we can write $$\gcd(a^n,b^n)=d^n$$
• I see. We use the fact that if $\gcd{(a,b)} = 1$, then $\gcd{(a^n,b^n)}$ will also be equal to $1$ to our advantage. Thanks for the answer! – eem Aug 17 '19 at 13:19