Solve for $x, y, z$, and $t$

Question

Given the equations: $$6x^2+y^2=z^2\\x^2+ 6y^2=t^2$$ for all $x,y,z,t \in\mathbb{N}$. Solve for $x, y, z$, and $t$.

Answer 1

The only solution is zero. Adding the two equations we get

$$7(x^2+y^2)=z^2+t^2$$ thus $$z^2+t^2\equiv 0 (mod 7)$$ and if $zt \neq 0$ we have that $-1$ is a quadratic residue modulo $7$, as $7\equiv 3 (mod 4)$ this is not the case so $xt=0$ and it follows that all are zero.

Answer 2

Adding, $7(x^2+y^2)=z^2+t^2$, whence $7 \mid z^2+t^2 \Rightarrow 7 \mid z, t$. Thus $49 \mid z^2+t^2 \Rightarrow 7 \mid x^2+y^2 \Rightarrow 7 \mid x, y$. Then $(\frac{x}{7}, \frac{y}{7}, \frac{z}{7}, \frac{t}{7})$ is a solution as well. We continue in this way to get $7^n \mid x, y, z, t$ for all positive integers $n$, so $x=y=z=t=0$.

Answer 3

For the system of equations:

$\left\{\begin{aligned}&x^2+ay^2=z^2\\&ax^2+y^2=t^2\end{aligned}\right.$

If the ratio can be represented as: $a=p^2+3$

Then:

$x=p-1$

$y=p+1$

$z=p^2+p+2$

$t=p^2-p+2$