Find all triples $(x,y,z)$ of positive integers such that $5(x²+2y²+z²)=2(5xy-yz+4xz)$... Find all triples of $(x,y,z)$ of positive integers such that $5(x^2+2y^2+z^2)=2(5xy-yz+4xz)$, and at least one of $x,y,z$ is prime
My approach: Put $x,y,z=2$ one by one then we get value as $(x,y,z)=(14/3,2,10/3)$ and $(14/5,6/5,2)$. Now since RHS is a multiple of 2 LHS should also be $0 \pmod 2$, so at least one of $x,y,z$ must be equal to $2.$
So there is no integral solution.

Solving as a quadratic in $x$, we have
$$
x = \pm \frac{1}{5}\sqrt{-(5y-3z)^2} + y + \frac{4z}{5}
$$
Since $(5y-3z)^2$ is non-negative, the only possibility is $5y-3z = 0$ or
$$y = \frac{3z}{5}$$
This gives
$$
x = 0 + \frac{3z}{5} + \frac{4z}{5} = \frac{7z}{5}
$$
Now substitute $x$ and $y$ in terms of $z$ in the original equation.
