$n = a^2 + b^2 = c^2 + d^2$. What are the properties of a, b, c and d?

If n is a positive integer that can be represented as the sum of two odd squares in two different ways:
$$n = a^2 + b^2 = c^2 + d^2$$
where $a$, $b$, $c$ and $d$ are discrete odd positive integers, what properties can be deduced about $a$, $b$, $c$ and $d$? There’s lots of information online about what properties of $n$ are, and its prime factors, but I can’t find, or deduce myself, anything about $a$, $b$, $c$ and $d$. Are there any relationships between them?

Solutions Collecting From Web of "$n = a^2 + b^2 = c^2 + d^2$. What are the properties of a, b, c and d?"

What you want is

A Note on Euler’s Factoring Problem
By: John Brillhart

This note consists of a brief introduction to Euler’s factoring problem and his results, as well as a complete and elegant solution to the problem given by Lucas and Matthews about a century later.

from the December 2009 Monthly, pages 928-931, see
http://www.maa.org/pubs/monthly_dec09_toc.html

Alright, I pasted in the pdf but the result was not entirely legible. Also there is some question of legality as the article is pretty recent. I can email the pdf to individuals who send me a request.

However, I can typeset Theorem 2:

Let $N>1$ be an odd integer expressed in two different ways as
$$N = m a^2 + n b^2 = m c^2 + n d^2,$$
where $a,b,c,d,m,n \in \mathbb Z^+, \; b < d,$ and
$\gcd(ma,nb) = \gcd(mc,nd) =1.$ Then
$$N = \gcd(N, ad-bc) \cdot \; \frac{N}{\gcd(N, ad-bc)}$$
where the factors are nontrivial.

Your case would be $m=n=1.$ Note that then if the theorem cannot be used because $\gcd(a,b) = g >1,$ then we have some factoring anyway, as $g^2 | N.$

Going to the Gaussian integers (the complex numbers of the form $x+yi$, where $x,y$ are ordinary integers, and $i=\sqrt{-1}$), we get $$(a+bi)(a-bi)=(c+di)(c-di)$$ Then there exist Gaussian integers $r,s$ such that $$a+bi=rs,\quad a-bi=\overline r\overline s,\quad c+di=r\overline s,\quad c-di=\overline rs$$ Note that $\overline r$ is the complex conjugate of $r$; if $r=x+yi$, then $\overline r=x-yi$. From this we get $$a=(rs+\overline r\overline s)/2,\quad b=(rs-\overline r\overline s)/2i,\quad c=(r\overline s+\overline rs)/2,\quad d=(r\overline s-\overline rs)/2i$$ You could possibly take this a step further by writing $r=x+yi$, $s=w+zi$ and multiplying everything out and combining like terms; I think that’s as close as you’ll get to relating $a,b,c,d$.