We do not need to solve five constraints. Three will do.
The circle passes through $F_1$ and $F_2$ with the center directly between them. So $\angle F_1PF_2$ with $P$ on the circle measures $90°$, therefore
$P$ also lies in the ellipse whose major axis is given and must be the sum of the lengths from the foci to $P$:
The given area of $\triangle F_1PF_2$ which is a right triangle must be half the product of the legs:
Square both sides of (2), and on the squared left side plug in the familiar binomial formula for the square of a sum. Use (3) to eliminate the cross product term and now you can get $|PF_1|^2+|PF_2|^2$ for use in Equation (1).
Not unlucky at all (subtle hint at the answer), you should find that $|F_1F_2|^2$ is a perfect square and thus $|F_1F_2|$ is a whole number. In fact, although you don't need it, $|PF_1|$ and $|PF_2|$ wind up being whole numbers too.