I have come across two equations with variables $x,y$ \begin{align*} (x+ay)^2+ 4 x y\\ (x-y)^2-4 c x y \end{align*} where $a,c\in \mathbb{Z}_+$ are some constants.
I would like to determine the values of x,y that generate perfect squares.
I know what to do in the 1 variables situation, but I don't think this is enough: First check to see if it is a perfect square and if it is not go through procedure that has been outlined in previous posts which I will outline real quick again in the next paragraph.
So for the first equation, if we view $y$ as a constant and set equal to $d^2$, can rewrite as (x+(a+2)y-d)(x+(a+2)y+d)=4(a+1)y^2. Then one can rewrite $4(a+1)y^2= b* \frac{(4(a+1)y^2}{b}$ for some divisor $b$ and solve: $$x= \frac{b+\frac{(4(a+1)y^2}{b}}{2}-(a+2),\quad d= \frac{b-\frac{(4(a+1)y^2}{b}}{2} .$$
Questions:
Do I need to do a similar procedure where I treat $x$ as a constant and repeat above for $y$ or would those be already included in the solution set above?
Even then, it just doesn't seem like this would be all the solutions though, as the outline above would implode if one wrote $y=f(x)$, and I can't quite seem to figure out what step I am missing?
For the system of equations .
$$\left\{\begin{aligned}&(x+ay)^2+4xy=z^2\\&(x-y)^2-4cxy=v^2\end{aligned}\right.$$
The solution can be written in this form.
$$x=(c^2+3c+2)(c^2+(2a+3)c+a^2+3a+2)$$
$$y=-(2c+a+3)(c^2+c-a-1)$$
$$z=c^4+2c^3-(6a+5)c^2-(4a^2+18a+14)c-a^3-6a^2-13a-8$$
$$v=3c^4+(4a+14)c^3+(a^2+12a+22)c^2+(a^2+8a+11)c-(a+1)^2$$