Conditions for $x^2 + ax - a$ to be a perfect square. Consider $y = x^2 + ax - a$ for positive integral values of $a,x$.
I am ultimately looking for sufficient conditions for $y$ to be a perfect square, but necessary conditions or hints will also be helpful.
I have already tried a few techniques in modular arithmetic, and considered the values of $a,x$ for which $y$ lies between 2 consecutive squares. The latter only provides a bound for $x$ depending on the value of $a$ chosen:
$4x \leq (a + 1)^2$ for odd $a$ and $8x \leq a^2 + 4$ for even $a$.
 A: HINT.-Maybe do you want to try with the following.
$$x^2+ax-a=y^2\iff \left(x+\frac a2\right)^2+1=y^2+\left(\frac a2+1\right)^2$$
Now you have the general solution of the equation $$X^2+Y^2=Z^2+W^2$$ is given by
$$\begin{cases}2X=tM+sN\\2Y=tN-sM\\2Z=tM-sN\\2W=tN+sM\end{cases}$$ where $t,s,M,N$ are arbitrary integers.
A: Just a Hint
Suppose that
$\exists b :$
$y=b^2$ then
$-a=\frac{(x-b)(x+b)}{x-1}$.
which gives 
$b=-1 \;,\; a=1-x$ and
$x=1\; y=1$.
A: Not sure if this is what you are looking for:
$$x^2+ax-a=k^2 \Rightarrow x_{1,2}=\frac{-a \pm \sqrt{a^2-4a-4k^2}}{2}$$ 
You need this to be an integer. This happens if and only if $$a^2-4a-4k^2$$ is a perfect square ( note that this has the same parity as $a$). 
Therefore 
$$a^2-4a=4k^2+m^2$$
This implies that $a^2-4a$ is the sum of two squares.
Conversely, if $a^2-4a$ is the sum of two squares, since $a^2-4a \in \{0,1\} \pmod{4}$, one of the squares must be even.
Therefore 
$$ a^2-4a=4k^2+m^2$$
and setting 
$$x=\frac{-a \pm m}{2}$$ we get that $a$ and $m$ have same parity thus $x \in \mathbb Z$ and 
$$y=k^2$$
