<s>[INST] The set of differences of square rationals At first, we observe that $A:=\{ p^2-m^2 : p,m\in \Bbb{Z}\}=\mathbb{Z}\setminus (4\mathbb{Z}+2)$ (because an integer $a$ can be written
as the form $a=p^2-m^2$ if and only if $a\neq 4k+2$, for every integer $k$) and so $A^c=4\mathbb{Z}+2$. Hence, $B:=\left\{ p^2-m^2 : p,m\in \Bbb{Z}\setminus \{0\}\right\}=A\setminus \left\{\pm k^2: k\in \Bbb{Z} \mbox{ and } k^2\neq s^2-t^2, \mbox{ for all } s,t\in \Bbb{Z}\right \}$ 
Now, put 
$$
D:=\left\{ \frac{p^2}{q^2}-\frac{m^2}{n^2}: p,q,m,n\in \Bbb{Z}\setminus \{0\} \text{ and} \gcd(p,q)=
\gcd(m,n) =1\right\}
$$
We would like to determine this set and its complement $D^c$ in $\Bbb{Q}$ exactly.
It is clear that $B\subseteq D$. Is it true that $D\neq \Bbb{Q}$?
 [/INST] We have $4m = (m+1)^2-(m-1)^2$ for all $m \in \mathbb Z$. Therefore,
$$
\frac{a}{b}
=
\frac{4ab}{4b^2}
=
\frac{(ab+1)^2-(ab-1)^2}{(2b)^2}
=
\left(\frac{a b + 1}{2 b}\right)^2 - \left(\frac{a b - 1}{2 b}\right)^2
$$
When $ab=\pm 1$, one of the terms is zero. In this case, $\frac{a}{b}=\pm1$ and we can use
$$
1 = \left(\frac{5}{3}\right)^2 - \left(\frac{4}{3}\right)^2,
\quad
-1 = \left(\frac{4}{3}\right)^2 - \left(\frac{5}{3}\right)^2
$$
Thus, all rationals are the difference of two nonzero rational squares.
</s>