If integers $s$, $x$, $y$, $z$ are such that $s=5xy+5yz+5xz=2x^2+2y^2+2z^2$, then $10s$ is a perfect square I'm having a problem with a question:

Consider integer $s$ and integers $x,y,z$ such that
$$s = 5xy+5yz+5xz \quad\text{and}\quad s= 2x^{2} + 2y^{2} + 2z^{2}$$ Prove that $10s$ is a perfect square.

I'm pretty sure it has something in common with
$$(x+y+z)^{2} = x^{2} + y^{2} + z^{2} + 2(xy+yz+xz)$$ e.g. if I somehow (prob. by writing $10s$ as $4s + 6s$ etc.) show $10s$ equals $4, 9$ or $16$ times $(x+y+z)^{2}$, then I'm done, yet I have no clue how to show this.
Another idea was to check modulo, and I found at least one of $x,y,z$ is divisible by $5$ and at least two of $x,y,z$ are even still don't know how to use it.

$(xy+yz+xz)=\frac 15 s$ and $x^2 + y^2 + z^2 = \frac 12s$
Now $(x^2 + y^2 + z^2) + 2(xy + yz+xy)=  (x+y+z)^2$
so $\frac 12 s + \frac 25 s = \frac 9{10}s=(x+y+z)^2$
So $10s = \frac {100}9(x+y+z)^2 = (\frac {10}3(x+y+z))^2$.
Dang.
So close!
But if we can prove $x+y+z$ is divisible by $3$ we will be okay.  Can we?
Well, it must.  We have $\frac 9{10}s = (x+y+z)^2$ so $9x = 10(x+y+z)^2$ is an integer.  So $9\mid 10(x+y+z)^2$ so $9|(x+y+z)^2$ and $3|x+y+z$.  Hurray!  we are done.
