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

Question

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.

score 1 · Answer 1 · answered Mar 08 '21 at 22:46

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 $d=e=f$.

score 1 · Answer 2 · answered Mar 09 '21 at 09:17

1

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

answered Mar 09 '21 at 09:17

Bill Dubuque

272,048

We used $\rm \color{#0a0}{FD} = $ Freshman's Dream (Binomial Theorem) for gcds and notation $,(x,y) :=\gcd(x,y)\ \ $ – Bill Dubuque Mar 09 '21 at 09:17
Update: I just found a dupe, so see my more general answer there ('ll leave this since it may be simpler to comprehend). – Bill Dubuque Mar 09 '21 at 09:25

score 0 · Accepted Answer · answered Mar 08 '21 at 22:37

0

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 ...

answered Mar 08 '21 at 22:37

Henry

157,058

That $k^2d^2|b^2$? – ddswsd Mar 08 '21 at 22:40
@ddswsd Yes. And so? – Henry Mar 08 '21 at 22:42
That $gcd(b,c)=kd$ and $d|gcd(b,c)$? I honestly don't know. @Henry – ddswsd Mar 09 '21 at 00:58
so $kd|b$ and $\gcd(a,b)\ge kd$ so $k=1$ and $\gcd(a,c)=d$. Similarly $\gcd(b,c)=d$ – Henry Mar 09 '21 at 01:00
I don't why I can't get this. Should that say $gcd(a,c)\ge kd$ since we know that $gcd(a,b)=d$? And I don't understand how $k=1$. Or is what you wrote how we show that $k=1$? @Henry – ddswsd Mar 09 '21 at 01:07
What I wrote was aimed at showing $k=1$ since $d= \gcd(a,c) \ge kd$ and $k$ and $d$ are positive integers – Henry Mar 09 '21 at 01:15
omg it just clicked, I understand this now. Thank you! – ddswsd Mar 09 '21 at 01:32

score 0 · Answer 4 · answered Mar 08 '21 at 22:43

0

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.

answered Mar 08 '21 at 22:43

J.G.

115,835

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

4 Answers4