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Proof(by contraposition): the tagged duplicate is trying to use bezout's identity while this post is trying not to

Suppose $\gcd(a^2,b^2)>1$ then $a^2$ & $b^2$ have a common factor greater than 1. W.L.O.G., it could be that $a\mid b$ or $a=b^n$ for some $n\in \Bbb{Z}$ or $a=b^{\frac{m}{k}}$ for some $k,m\in \Bbb{Z}$.

Case 1: $a|b$ iff $|a|\leq |b|$. Assuming that $a\neq 1$ and $b\neq 1$, then $gcd(a,b)\neq 1$.

Case 2: $a=b^n$ iff $b=\sqrt[n]{a}$, then $\gcd(a,b)=\gcd(a,\sqrt[n]{a})=\sqrt[n]{a} \neq 1$ assuming $a\neq 1$.

Case 3: $a=b^{\frac nk}$ iff $b=\sqrt[k]{a^n}$, then $\gcd(a,b)=\gcd(a,\sqrt[k]{a^n}) \neq \sqrt[k]{a^n} \neq 1$ assuming $a \neq 1$.

Thus, for all cases, if $\gcd(a,b)=1$, then $\gcd(a^2,b^2)=1$. QED

I think what I wrote is correct, but I am not that confident it is.

TheLast Cipher
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1 Answers1

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Your proof should be simpler. Note that if a prime $p$ divides $a^2$ and $b^2$ then $p$ divides $a$ and $b$ (use Prove that if $p$ is a prime, $a$ is an integer, and $p$ divides $a^2$, then $p$ divides $a$.).

Robert Z
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  • Suppose $gcd(a^2,b^2)=d$ then $d|a^2$ and $d|b^2$. If $d$ is composite then there exist a prime $p$ that divides $d$ which implies $p|a^2$ and $p|b^2$. By Euclid's Lemma, if $p|a^2$ then $p|a$ or $p|a$ and same thing goes why $p|b$. Thus, $p|a$ and $p|b$. Thus, by contraposition, the above claim is true. QED...? – TheLast Cipher Oct 22 '17 at 10:43
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    Correct. Just say if $d>1$ then there is a prime $p$ which divides $d$. – Robert Z Oct 22 '17 at 10:45