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of two nontrivial rational numbers ?

Given an integer $n$, suppose that it has two decompositions : $p^2 +q^2 = s^2 +t^2 \ $ ($p,q,s,t$ are all positive and $(s,t) \not \in \{(p,q),(q,p)\} \ $).

We want to write $n$ as a product of two nontrivial rational numbers.

Let us write $p=s+x$ and $q=t-y$ where $x \in \mathbb{Z}$ such that $x\ge -s$ and $y \in \mathbb{Z}$ such that $y \le t$.

Rewriting the equalities we get : $n= s^2 +x^2 +2sx + t^2 +y^2 -2ty$.

Then $0= x^2+y^2 +2sx -2ty$.

In the same way we find that : $n= p^2+x^2- 2px + q^2 +y^2 +2qy$.

Then $0 = x^2 +y^2 -2px +2qy$.

Hence $2sx - 2ty = -2px +2qy \Leftrightarrow sx-ty=-px+qy\Leftrightarrow x(s+p)=y(t+q) \Leftrightarrow x= y\dfrac{t+q}{s+p}$.

If we substitute in $n$,we obtain : $n= p^2 +y^2\left(\dfrac{t+q}{s+p}\right)^2 - 2py\dfrac{t+q}{s+p}+q^2+y^2 + 2qy$.

But I'm stuck, it seems that I'm turning around...

Thanks in advance !

Maman
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  • actually your $n = p^2 + q^2 = r^2 + s^2$ leads to factoring $n,$ which cannot be prime. Let me find references, th name that comes to ming is Brillhart, the original technique due to Euler. Give me a minute – Will Jagy Dec 17 '21 at 01:07
  • First one is John Brillhart, A Note on Euler's Factoring Problem. The Montlhy of the MAA, volume 116 December 2009 – Will Jagy Dec 17 '21 at 01:13
  • http://zakuski.utsa.edu/~jagy/Brillhart_Euler_factoring_2009.pdf

    http://zakuski.utsa.edu/~jagy/Brillhart_Euler_factoring_2016.pdf

    – Will Jagy Dec 17 '21 at 01:18
  • This question was closed as I was composing a careful answer. I put the answer at the linked question, of which this one is judged to be duplicate – Will Jagy Dec 17 '21 at 03:20

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