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I came across this question while studying the book Challenge and Thrill of Pre-College Mathematics:

If $a$ and $b$ are relatively prime integers, then prove that $$\gcd((a+b)^m, (a-b)^m)\leq2^m$$

This is my attempt:

Let $\gcd((a+b),(a-b))=k$. Then $k|(a+b)$ and $k|(a-b)$. Thus $k|(a+b)+(a-b)$ and $k|(a+b)-(a-b)\Rightarrow k|2a$ and $k|2b$.

Now either $k|2$, $k|a$, or $k|b$. If the first is true then $k\leq2$, and if the latter two are true then $k|\gcd(a,b) \Rightarrow k|1$, in which case we can again say that $k=1\leq2$.

Now we know that $\gcd((a+b)^m,(a-b)^m)=k^m$, and since $k\leq2$ it implies $k^m\leq2^m$, which proves our proposition. QED.


Since I'm new to this subject, I'm finding this proof slightly shoddy. Is the logic correct? I'm concerned about the proof that $k\leq2$ the most.

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  • $\begingroup$ Just out of curiosity, how do you add the yellow box around questions? $\endgroup$ – Naman Kumar Jan 18 at 12:40
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This is not correct: Now either $k|2$, $k|a$, or $k|b$.

It is better if you take prime $p\mid k$ then $p|2$, $p|a$, or $p|b$ and thus $p=2$.

So the only prime which divdes $k$ is $2$, so $k=2^l$. Now if $l\geq 2$ then $4\mid a+b$ and $4\mid a-b$ so $4\mid 2a$ so $2\mid a$ and the same is true for $b$, that is $2\mid b$. Thus $l\leq 1$ and so $k=2$.

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  • $\begingroup$ Thank you for answering. I understood your proof, but I'm afraid I don't understand why it is necessary to take some prime $p \mid k$ instead of simply solving using $k$. Can you please elaborate? $\endgroup$ – Naman Kumar Jan 18 at 12:24
  • $\begingroup$ Say $10 \mid 2\cdot 5\cdot 2$ where $a=5$, $b=2$ $\endgroup$ – Aqua Jan 18 at 12:25
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    $\begingroup$ Oh, that makes a lot more sense now. Thank you. $\endgroup$ – Naman Kumar Jan 18 at 12:31

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