Originally Posted by
Unbeatable0 The only solution is (0,0,0) (over both the integers and the naturals).
Sketch of proof:
Since we're dealing with squares here, we can assume the integers are non-negative.
From what Tarask did, we're left with showing that $\displaystyle 2x^2+5y^2=z^2$ has no integer non-negative solutions.
From $\displaystyle z^2\equiv 2x^2 \pmod{5}$ we conclude, by first checking what a square can be modulo 5,
that $\displaystyle 5\mid x$ and $\displaystyle 5\mid z$.
Put this in the equation, check that $\displaystyle 5\mid y$ and replace $\displaystyle y$ with $\displaystyle 5y'$.
Now we have the same equation as before, and thus we can conclude by Fermat's infinite descent that the only
solution is (0,0,0).