Can $a=\left(\sqrt{2(\sqrt{y}+\sqrt{z})(\sqrt{x}+\sqrt{z})}-\sqrt{y}-\sqrt{z}\right)^2$ be an integer if $x$, $y$, and $z$ are not squares? Let $\gcd(x,y,z)=1$.Can we find 3 non-perfect squares $x,y,z\in \mathbb{Z},$  such that $a \in \mathbb{Z} \geq 2$
$$a=\left(\sqrt{2(\sqrt{y}+\sqrt{z})(\sqrt{x}+\sqrt{z})}-\sqrt{y}-\sqrt{z}\right)^2$$
I cannot seem to find any such triplets. Any hints on how to prove it?

Let $y=k^2$, $z=r^2$ and $x=c^2$(with $k,r,c$ integers) then the expression becomes: $$(\sqrt {2\cdot (k+r)(c+r)}-k-r)^2$$
Now for $a$ to be an integer $$2(k+r)(c+r)=n^2$$
with $n$ integer.
We put $$k+r=2^{2p+1}\cdot t^s$$
And $$c+r=t^v$$
(or vice versa,with $v$ and $s$ both odd or even).
If $r=1$ $$k=2^{2p+1}\cdot t^s-1$$
And $$c=t^v-1$$. This can be a solution.
