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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 $D\neq \Bbb{Q}$?

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  • $\begingroup$ You don't need the gcd conditions in your definition of $D$. $\endgroup$ – TonyK Dec 13 '19 at 12:54
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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.

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If I understood you correctly, $D=\mathbb Q$ since $$ \left(\frac{a b + 1}{2 b}\right)^2 - \left(\frac{a b - 1}{2 b}\right)^2 = \frac{a}{b} $$

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  • $\begingroup$ Thanks for your answer. What about the cases $ab=\pm 1$? $\endgroup$ – M.H.Hooshmand Dec 13 '19 at 11:31
  • $\begingroup$ Of course, there are some mistakes in my explanations, now I correct them. $\endgroup$ – M.H.Hooshmand Dec 13 '19 at 11:34
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The non degenerate quadratic form $X^2-Y^2$ obviously represents $0$ in $\mathbf Q$, i.e. the equation $x^2-y^2=0$ admits solutions $(x,y)\neq (0,0)$. But a non degenerate quadratic form $q(X)= \sum a_{ij}X_iX_j$ in $n$ variables with rational coefficients which represents $0$ represents all rationals. Indeed, $q(tX+Y)= t^2Q(X)+tb(X,Y)+f(Y)$, where $b(X,Y)$ is the bilinear form associated to the quadratic form $q(X)$. If $q(x)=0$ for a certain $x\neq (0,...,0)$, the non degeneracy of $q$ implies the existence of $y\in {\mathbf Q}^n$ s.t. $b(x,y)\neq 0$, so that $q(tx+y)$ is a non constant linear function of $t$ which takes all values in $\mathbf Q$ when $t$ runs through $\mathbf Q$. Note that the main property remains valid over any field of characteristic $\neq 0$.

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