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.


You can finish it this way.

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 $$\gcd(a^n,b^n)=d^n$$

  • $\begingroup$ 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! $\endgroup$ – eem Aug 17 '19 at 13:19
  • 1
    $\begingroup$ You got it..... $\endgroup$ – Mostafa Ayaz Aug 17 '19 at 13:20

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