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