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This proposition was proposed by my deskmate. And I gave a method to work out it.

So I want to communicate with masters of mathematics here.

This is my proof process:

"For $p\in \mathbb{Q}$, choose $k\in \mathbb{Q}$ (and $k \neq 0$)

construct $m, n$ like:

$m$ = $\frac{k^2+p}{2k}$

$n$ = $\frac{k^2-p}{2k}$

And $m^2-n^2$ = $(m+n)(m-n)$ = $p$."

Anton Vrdoljak
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Absolute Value
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  • Seems needlessly complex...and we only need one example, not a family. If we look at the system $m+n=p$, $m-n=1$ we get $m=\frac {p+1}2$ and $n=\frac {p-1}2$. That works. – lulu Oct 21 '23 at 16:09
  • Thank you very much，I used this complex method because my deskmate need a formula to study the universal sense.(by the way, my deskmate'name is chaochao) – Absolute Value Oct 21 '23 at 16:16
  • Well, your formula does give a family of solutions, though you ought to specify $k\neq 0$. Looking at the expressions, we easily get $m+n=k, m-n=\frac pk$. – lulu Oct 21 '23 at 16:18
  • Sorry very much.I am changing it now .And your equation is also where my inspiration comes from. – Absolute Value Oct 21 '23 at 16:38

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