With reference to Noam's result, equivalent to $a(a - x^2) = (a - 1) y^2$, it isn't hard to show that this is rational, and $a, x, y$ can be expressed in terms of two parameers.

From this one can churn out possible values of $x$ by the bushel; but deciding whether any given value of $x$ is possible is harder, and may be no more practical using this result than those already discussed.

Anyway, I'll sketch how the solution is obtained.

Firstly, the N&S condition for the above, considered as a quadratic in $a$, to have rational $a$ when $x, y$ are rational is $(x^2 + y^2)^2 - 4 y^2 = z^2$ for some rational $z$.

This implies some rational $p$ for which:

$\frac{y}{x^2 + y^2} = \frac{p}{p^2 + 1}$

$\frac{z}{x^2 + y^2} = \frac{p^2 - 1}{p^2 + 1}$

Reciprocating the first and completing squares expresses it in the form:

$x^2 + (y - \frac{p^2 + 1}{2 p})^2 = (\frac{p^2 + 1}{2 p})^2$

so there must be some rational $q$ for which:

$y = \frac{p^2 + 1}{2 p} + \frac{q^2 - 1}{2 q} x $

$\frac{p^2 + 1}{2 p} = \frac{q^2 + 1}{2 q} x$

From the last one can express $x = x(p, q}$, and plugging this into the preceding equation gives $y = y(p, q)$. Finally, we can obtain $z$ and then $a$ also both in terms of $p, q$.

NOTE: As the PC I am typing this on doesn't format equations, I am "flying blind". So if the result looks hideous due to any missing dollars and suchlike, perhaps someone could edit it.

With reference to Noam's result, equivalent to $a(a - x^2) = (a - 1) y^2$, it isn't hard to show that this is rational, and $a, x, y$ can be expressed in terms of two parameters.

From this one can churn out possible values of $x$ by the bushel. Deciding whether any given value of $x$ is possible is harder, and may be no more practical using this result than those already discussed.

Anyway, I'll sketch how the solution is obtained.

Firstly, the N&S condition for the above, considered as a quadratic in $a$, to have rational $a$ when $x, y$ are rational is $(x^2 + y^2)^2 - 4 y^2 = z^2$ for some rational $z$.

This implies some rational $p$ for which:

$$\frac{y}{x^2 + y^2} = \frac{p}{p^2 + 1},\qquad \frac{z}{x^2 + y^2} = \frac{p^2 - 1}{p^2 + 1}$$

Reciprocating the first and completing squares expresses it in the form:

$$x^2 + (y - \frac{p^2 + 1}{2 p})^2 = (\frac{p^2 + 1}{2 p})^2$$

so there must be some rational $q$ for which:

$$y = \frac{p^2 + 1}{2 p} + \frac{q^2 - 1}{2 q} x, \qquad \frac{p^2 + 1}{2 p} = \frac{q^2 + 1}{2 q} x$$

From the last one can express $x = x(p, q)$, and plugging this into the preceding equation gives $y = y(p, q)$. Finally, we can obtain $z$ and then $a$ also both in terms of $p, q$.

NOTE: As the PC I am typing this on doesn't format equations, I am "flying blind". So if the result looks hideous due to any missing dollars and suchlike, perhaps someone could edit it.