# Proof Verification: Prove that $\gcd((a+b)^m, (a-b)^m)\leq2^m$ for relatively prime $a$ and $b$.

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.

• Just out of curiosity, how do you add the yellow box around questions? – Naman Kumar Jan 18 at 12:40

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 $$k=2$$.
• 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? – Naman Kumar Jan 18 at 12:24
• Say $10 \mid 2\cdot 5\cdot 2$ where $a=5$, $b=2$ – Aqua Jan 18 at 12:25