<s>[INST] Diophantine-like equations 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
$$\begin{align}
b^2+c^2+d^2-3a^2&=-8n(x-y)\\
a^2+c^2+d^2-3b^2&=8n(x+y)\\
a^2+b^2+d^2-3c^2&=-8n(y-x)\\
a^2+b^2+c^2-3d^2&=-8n(x+y)
\end{align}\tag2$$
 [/INST] 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. 
</s>