A-x, B+x, x is all perfect square (Diophantine equation)

Question

How many solutions are there to following diophantine equations? (or, asymptotically?)

For positive integers $A, B$, $A-x, B+x, x$ is all perfect square.

First of all, the number of solutions $(m,n)$ that $A=m^2+n^2$ is $r'_2(A)$, that one can look up for explicit formula in here. http://mathworld.wolfram.com/SumofSquaresFunction.html

Therefore above the number of solution above must be less or equal to $r'_2(A)$. Now, a heuristic argument comes in. One can suppose that chance of $B+x$ being a perfect square is about $O(\frac{1}{\sqrt(B)})$, therefore vaguely one can argue that above solution has upper bound of $O(\frac{r'_2(A)}{\sqrt(B)})$. But notice that this argument has so many flaws on so many levels. Would anyone like to consider this problem? Or is this problem already quite famously solved?

Answer 1

Suppose that you indeed mean

For positive integers $A,B$, $A−x,B+x,x$ is all perfect square

One trivial bound for number of solutions is $A$. Indeed, if $A < x$, then $A - x < 0$ and thus cannot be a perfect square.

Now, suppose that $x = k^2$, $A-x = m^2$, $B+x = k^2$. It follows that $B = n^2 - k^2 = (n+k)\cdot (n-k)$. Suppose that $B = r \cdot s$. It follows that $n = \frac{r+s}2$, $k = \frac{r-s}2$. That defines bijection between paris $(n, k)$ and $(r, s)$ and thus the number of solutions is at most half of the number of factors of $B$ (since we consider ordered pairs only, as $n > k$).

Answer 2

This system of equations:

$$\left\{\begin{aligned}&A^2+x^2=c^2\\&B^2-x^2=z^2\end{aligned}\right.$$

Solutions have the form:

$$x=4tkp^2s^2$$

$$A=2(t^2-k^2)p^2s^2$$

$$z=4k^2s^4-t^2p^4$$

$$c=2(t^2+k^2)p^2s^2$$

$$B=4k^2s^4+t^2p^4$$

$t,k,p,s$ - integers asked us.