So I was solving a problem and encountered a specific system of equations that I don't know if a solutions exists for it or not.
$$\begin{align} 4ny&=d^2-a^2\\ -4nx+4ny&=d^2-b^2\\ 4nx&=d^2-c^2\\ \end{align}\tag1$$
where $0\leq x,y\leq n$ and $a,c,d\geq n$ and $0\leq b\leq n\sqrt{2}$ and $a,b,c,d,n\in \mathbb{N}$, I would appreciate it if any could suggest a hint as to how I should proceed. Or worded in a different way I want to solve the following system of equation with the same conditions on the variables
$$\begin{align} b^2+c^2+d^2-3a^2&=-8n(x-y)\\ a^2+c^2+d^2-3b^2&=8n(x+y)\\ a^2+b^2+d^2-3c^2&=-8n(y-x)\\ a^2+b^2+c^2-3d^2&=-8n(x+y) \end{align}\tag2$$
I think you made a typo somewhere as $(1)$ and $(2)$ do not have equivalent solutions. Let's assume $(2)$ has no typos. Then what you can do is treat it as four equations in four unknowns $a,b,c,d$.
$$\begin{align} b^2+c^2+d^2-3a^2&=-8n(x-y)\\ a^2+c^2+d^2-3b^2&=8n(x+y)\\ a^2+b^2+d^2-3c^2&=-8n(y-x)\\ a^2+b^2+c^2-3d^2&=-8n(x+y) \end{align}\tag2$$
I get,
$$\begin{align} a^2&=d^2-4n\,y\\ b^2&=d^2-4n(x+y)\\ c^2&=d^2-4n\,x\\ \end{align}\tag3$$
Then solve for this as three equations in three unknowns $x,y,d$ which yields,
$$\begin{align} x&=\frac{a^2-b^2}{4n}\\ y&=\frac{-b^2+c^2}{4n}\\ d&=\sqrt{a^2-b^2+c^2}\\ \end{align}\tag4$$
If you substitute $(4)$ into $(2)$ you will see that it will solve it. But not if into $(1)$, so there must be a typo somewhere.
Anyway, what I described is the general method you can use. And there are infinitely many integers $a,b,c$ such that $d$ is also an integer, though you have to choose those that satisfy the additional constraints that you gave.