Could you state or direct me to results regarding the Diophantine equation $a^2+b^2=c^2+d^2$ over integers? Specifically, I am looking for a complete parametrization. In the case that a complete parametrization does not exist, I would be interested in seeing complete parametrizations for special cases.

For example, in the case that any one of the variables is zero, we are led to Pythagorean triples. Taking a variable to be some non-zero constant would be interesting. I'm looking for known results as I am not as familiar with the literature as many MO-ers out there.

  • $\begingroup$ What qualifies as a "complete" parameterization? This is somewhat unclear (to me, at least) even in the Pythagorean triple case. I know, of course, that $(p^2-q^2,2pq,p^2+q^2)$ is a parameterization for solutions $(a,b,c)$ of $a^2+b^2=c^2$, but is it "complete"? It doesn't include $(4,3,5)$, for example. $\endgroup$ May 9 '13 at 2:09
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    $\begingroup$ $a^2+b^2=c^2+d^2 \Longleftrightarrow a^2-c^2=d^2-b^2 \Longleftrightarrow (a-c)(a+c)=(d-b)(d+b)$. Now the primitive solutions of $rs=tu$ are exactly $(r,s,t,u) = (xx',yy',xy',x'y)$ with $\gcd(x,y)=\gcd(x',y')=1$ (and some positivity condition to avoid duplication with factors of $-1$). If $(a,b,c,d)$ is primitive then $(r,s,t,u)$ is primitive up to a factor of $2$ and satisfies $r\equiv s$ and $t\equiv u \bmod 2$. So just figure out what to do with $x,x',y,y' \bmod 2$ and you're done. $\endgroup$ May 9 '13 at 2:13

MacKay and Mahajan have a short easy-to-read article expanding on Noam's comment. (pdf link)

Edit (5/14/13): There is a whole bunch of random notes on relationships like this one and others that can be found here: https://sites.google.com/site/tpiezas/003

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    $\begingroup$ But the last puzzle was already solved by Fermat... $$ $$ ...Which also suggests an alternative approach to the equation $a^2+b^2=c^2+d^2$ starting from the factorization $(a+bi)(a-bi)=(c+di)(c-di)$ in ${\bf Z}[i]$. $\endgroup$ May 9 '13 at 4:18
  • $\begingroup$ Using the word "random" to describe the notes may be viewed as somewhat disparaging, as well as inaccurate. A term like "variegated" might be preferred, as I find more resemblance between Piezas's notes and a botanical garden than I do between his notes and some of the backyards I've seen lately. Gerhard "As Lovely As A Trie" Paseman, 2013.05.14 $\endgroup$ May 14 '13 at 16:48

This equation is quite symmetrical so formulas making too much can be written: So for the equation:



























number $a,b,p,s$ integers and sets us, and may be of any sign.


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