Is $c=\sqrt{\frac{p^2+2a^2q^2-a^2}{a^2q^2}}$ possible, such that $p$ and $q$ are integers, and $a$ and $c$ are rationals? I was working on a problem, and I narrowed it down to the following:
$$c=\sqrt{\frac{p^2+2a^2q^2-a^2}{a^2q^2}}$$
such that $p$ and $q$ are integers, and $a$ and $c$ are rationals.
Does $c$ exist? So obviously, the numerator is not a perfect square, so it cannot be trivially simplified. Basically, my question is to remove the radical in any way to solve for $c.$
I have accepted an answer, however, it is not quite what I was looking for.
This question is very intricately related to another question of mine, that I have not been able to solve: (I don't know if promoting questions through others is legal)
Rational Solution Set to these equations and inequalities?
Please help me!

First of all, we need to require that $a$ and $q$ both be nonzero (look at the denominator of the right hand side). On the plus side, once we make this stipulation we have $2a^2q^2>a^2$ and so $\sqrt{p^2+2a^2q^2-a^2}$ - and hence $c$ itself - is at least a real number!

Now with that stipulation, note that the right hand side simplifies to $${\sqrt{p^2+2a^2q^2-a^2}\over \vert aq\vert}.$$ This is rational iff $p^2+2a^2q^2-a^2$ is the square of a rational number; sometimes it will be and sometimes it won't be. For example, taking $q=1$ we get $p^2+2a^2q^2-a^2=p^2+a^2$.  Sometimes this will be the square of a rational (e.g. $p=3, a=4$) and other times it won't be (e.g. $p=1, a=1$). So the answer to your question is a resounding maybe.
