Solving an integer equation

Question

Is it true that if: $x$, $y$, $z$ and $t$ are integers such that:

$xz + 7yt = 0$ $and$ $yz + xt = 0$,

then $x = y = 0$ $or$ $z = t = 0$?

Why or why not?

Unless I have miscalculations somewhere back in the problem I'm trying to solve, the above should be true. But I'm not finding a way to prove it so. So I'm starting to suspect whether it's true or not.

I first thought that since $xz$ = $-7yt$, then $7$ divides $x$ or $7$ divides $z$, then I took the three subsequent cases and reached some ugly results.

Any help would be appreciated. Thanks.

Answer 1

Note that $$\eqalign{(x^2-7y^2)(z^2-7t^2) &=x^2z^2-7x^2t^2-7y^2z^2+49y^2t^2\cr &=x^2z^2+14xyzt+49y^2t^2-7x^2t^2-14xyzt-7y^2z^2\cr &=(xz+7yt)^2-7(xt+yz)^2\cr &=0\ .\cr}$$ Therefore either $x^2-7y^2=0$, and since $\sqrt7$ is irrational this implies $x=y=0$; or, similarly, $z=t=0$.

Answer 2

We have $xz=-7yt$ and $yz=-xt$. Multiply. We get $z^2xy=7xyt^2$. If neither $x$ nor $y$ is $0$, we get $z=t=0$, since $\sqrt{7}$ is irrational.

Chase down what happens if one of $x$ or $y$ is $0$. You will quickly find that if the other is not $0$ then $z=t=0$.