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?

Question

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?

Answer 1

Checking a solution given by Winther in comments beneath the OP led me to the following:

In general, if $z=x+y$, then $a=x$.

This is seen by writing

$$\sqrt{2(\sqrt x+\sqrt z)(\sqrt y+\sqrt z)}=\sqrt x+\sqrt y+\sqrt z$$

squaring both sides and expanding to get

$$2(\sqrt{xy}+\sqrt{xz}+\sqrt{yz}+z)=x+y+z+2(\sqrt{xy}+\sqrt{xz}+\sqrt{yz})$$

and then cancelling left and right, leaving $z=x+y$.

Whether this gives all solutions to the OP's equation remains to be seen. (I'm not offering an opinion one way or the other.)

Answer 2

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.