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let $a,b,c,d$ be positive integers satisfying $a^2+b^2=c^2+d^2$, $a\ge b$, $c\ge d$, $a>c$

prove that:

$a^2+b^2$ is divisible by $\dfrac{ac+bd}{\gcd(ac+bd,ab+cd)}$

I know very little about the equation $a^2+b^2=c^2+d^2=n$, like, does it have any special properties, or a formula for $n$?.

Some things i have

$ab+cd=\dfrac{(c+d+a-b)(c+d+b-a)}{2}=\dfrac{(a+b+c-d)(a+b+d-c)}{2}$

and $(a^2+b^2)^2=(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-cb)^2$

Please give your comments on the problem, thanks a lot.

Bill Dubuque
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bestty
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    I don't know if it helps, but there's a known parametrization of your equation. See here : https://math.stackexchange.com/questions/153603/diophantine-equation-a2b2-c2d2 – HackR Sep 03 '24 at 15:48
  • See also https://math.stackexchange.com/questions/719700/if-a-prime-can-be-expressed-as-sum-of-square-of-two-integers-then-prove-that-th – Gerry Myerson Oct 21 '24 at 01:41

1 Answers1

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This is part of an old result of Euler which shows that $\,n\,$ is composite if it has two essentially distinct representations as sums of squares (the proof constructs a proper factor of $\,n\,$ via a quick gcd computation). This part is easy algebra via $\,\rm\color{darkorange}G$ = gcd law and $\rm\color{#c00}{given\ equality}$

$$\begin{align} (\color{#90f}{cd+ab})\:\!\color{c00}{(a^2+b^2)}\ \,&\!\!= (a^2+b^2)cd+\overbrace{(\color{#c00}{c^2+d^2})}^{\!\!\!\!\!\!\!\textstyle =\, \color{#c00}{a^2+b^2}}ab\\[.4em] &\!\!= (\color{#0a0}{ac+bd})\:\!(ad+bc)\,\ \ {\rm by\ algebra}\\[.4em] {\rm thus}\ \ \ \ \ \color{#0a0}{ac+bd}&\mid (\color{#90f}{ab+cd})\:\!\color{c00}{(a^2+b^2)}\quad {\rm by\ above}\\[.4em] \overset{\rm\color{darkorange}G}\Longrightarrow\ \ \ \ ac+bd&\mid \color{#0a0}g_{\:\!\color{#90f}{0}}\,\color{c00}{(a^2+b^2)},\,\ \color{#0a0}g_{\:\!\color{#90f}{0}} = (\color{#0a0}{ac+bd},\color{#90f}{ab+cd})\\[.5em] \Longrightarrow\ \ \ \frac{ac+bd}{\color{#0a0}g_{\:\!\color{#90f}{0}}}\!&\mid a^2+b^2 \end{align}$$

Bill Dubuque
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