Number of x such that $m-x^2$ and $n-x^2$ are both squares For given integers $m$,$n$, what bounds are known on the number of positive integers $x$ such that $m-x^2$ and $n-x^2$ are both perfect squares? In particular, is the number of such $x$ bounded above by a constant independent of $m,n$?
A search for $m,n \le 5000$ gives three pairs for which there are 4 such x: (1370,2210),(2210,3050), and (3485,3965).
 A: (revised) I suspected that there is no bound, but now I am not as sure . In an example $m,n$ with many such $x$ it will have to be the case that $m$ can be written in many ways as a sum of two squares (and the same for $n$.) Also, $m^2-n^2$ will have to be expressed in many ways as a difference of $2$ squares. We know that the number of ways to do these three things can be derived from the prime factorizations of $m,n$ and $n-m$. So this tells us where to look for possible examples. The number of positive solutions of $m=x^2+y^2$ is $2^{k-1}$ when $m$ is a product of $k$ distinct primes all congruent to $1\mod{4}$ (This gives $2^k$ possible $x$. Also, the general formula is not much harder).  Consider the $126$ products of $5$ out of the $9$ primes $5, 13, 17, 29, 37, 41, 53, 61, 73.$ It turns out that $m=2401165, n=4254445$ allow for 7 such $x.$
Going out to $101$ and using products of $4$ primes one finds that there are $5$ choices of $x$ in the case of the relatively prime integers $m=1662965=5\cdot37\cdot89\cdot101$ and $n=3068117=13\cdot53\cdot61\cdot73.$ Note that $n-m=2^5\cdot3^2\cdot7\cdot17\cdot41$ which allows $48$ ways to have $n-m=y^2-z^2.$
Here is why I wonder if we can be confident of getting even larger numbers: The points $(x,y)$ with $x^2+y^2=m$ are the points of $\mathbb{Z}^2$ which are on the circle of radius $m$ (and are in the first quadrant.) So we are looking for two circle of integer radii $m \lt n$, centered at the origin, and having a large number of pairs lattice points one on the larger circle and the other vertically below and on the smaller circle. To have a large number of lattice points on the circle of radius $m$ will require $m$ to have many factors and hence be large. As $m$ grows the number of lattice points can grow but the circumference grows more rapidly leading to a sparser distributin of points. $m=13\cdot17\cdot29$ allows for $8$ possible $x$ values along a quarter circle of circumference $m\pi/2 \approx 10067.$ Now $m=5\cdot13\cdot17\cdot29$ would allow for twice as many points, but along a quarter circle with $5$ times the circumference (and this is the most favorable example.)
