If $gcd(a,b)=d$ and $a^2+b^2=c^2$, show that $gcd(a,c)=gcd(b,c)=d$.

Question

If $gcd(a,b)=d$ and $a^2+b^2=c^2$, show that $gcd(a,c)=gcd(b,c)=d$.

This is what I have so far. We know that $d|a$ and $d|b$ which implies that $d|c$, but that is not enough to show that $gcd(a,c)=gcd(b,c)=d$, only that they are at least $d$. Also $a^2=c^2-b^2$ and $b^2=c^2-a^2$. Any hints or solutions are greatly appreciated.

Answer 1

Hint:

• You have found $d|c$ so $d|\gcd(a,c)$ and $\gcd(a,c) = kd$ for some positive integer $k$.

• Then $k^2d^2 | a^2$ and $k^2d^2 | c^2$ so $k^2d^2 | (c^2-a^2)$.

• This implies ... and ...

Answer 2

If $f$ divides at least two of $a,\,b,\,c$, $f^2$ divides at least two of $a^2,\,b^2,\,c^2$ and hence all of them, so $f$ divides all of $a,\,b,\,c$. Hence $\gcd(a,\,b),\,\gcd(b,\,c),\,\gcd(c,\,a)$ all divide each other, and are equal.

Answer 3

Let $d = (a,b)$, $e = (a,c)$, $f=(b,c)$.

We have $d\mid a$ and $d\mid b$, then $d^2 \mid a^2+b^2 = c^2$, and then $d \mid c$.

Since $d \mid a$ and $d \mid c$ then $d \mid (a,c) = e$. So $d \mid e$

In practically the same way you can prove $d \mid f$, $e \mid d$, $e \mid f$, $f \mid d$ and $f \mid e$.

So $d=e=f$.

Answer 4

$\color{#c00}{(a,c)^2} \overset{\rm \color{#0a0}{FD}}= (a^2,c^2) = (a^2,a^2\!+\!b^2) = (a^2,b^2) \overset{\rm \color{#0a0}{FD}}= \color{#c00}{(a,b)^2}\,$ so $\,\color{#c00}{(a,c) = (a,b)}.\,$ $\overbrace{\text{Same for $(b,c)$}}^{\text{ by $\,(a\ b)\,$ symmetry}}$.