<s>[INST] Proving non-trivial integer solutions don't exist for $z^2=10x^2-5y^2$ Given $z^2=10x^2-5y^2$. 
I am wanting to show this equation has integer solutions only $0=x=y=z$. 
I attempted by considering modulo $2$, however this didn't give me a contradiction. Any help or hints would be appreciated. 
Solution:
Consider modulo $5$. Then we have that $z^2 \equiv 0 \mod5$. This is impossible as:
$0^2=0 \mod5$
$1^2=1 \mod5$
$2^2=4 \mod5$
$3^2=4 \mod5$
$4^2=1 \mod5$
 [/INST] Outline of proof:


*

*If there is a non-trivial solution, there is a solution with $x$ odd.

*There are no solutions when $x$ odd.


We'll look entirely modulo $8.$

First note that if $x,y,z$ are solutions with $x=0$, then $y=0,z=0.$
So if there is any solution $(x,y,z)\neq (0,0,0),$ there must be one with $x>0.$ Then find the least $x>0$ such that there are integers $y,z$ such that $z^2=10x^2-5y^2.$
If $x$ is even, then $10x^2\equiv 0\pmod{8}.$  We then have $z^2+5y^2\equiv 0\pmod 8.$ 
But $$z^2\equiv 0,1,\text{ or } 4\pmod 8\\\text{and}\\5y^2\equiv 0,5,\text{ or } 4\pmod{8}.\tag{1}$$ So to get $z^2+5y^2\equiv 0\pmod 8,$ we must have $y,z$ both even. Then we get another solution $(x',y',z')=(x/2,y/2,z/2),$ contradicting that $(x,y,z)$ is a solution with minimal $x>0.$
So $x$ must be odd.

When $x$ is odd, we have $10x^2\equiv 2\pmod{8}.$
By (1) we get that $$z^2+5y^2\equiv 0,1,4,5,\text{ or }6\pmod{8}$$
So it is not possible for $z^2+5y^2\equiv 2\pmod{8}.$ But that means there is no $(x,y,z)\neq 0$ such that $z^2=10x^2-5y^2.$

More generally, if $a\equiv 1\pmod{8},$ $b\equiv 5\pmod 8$ and $c\equiv 2,3,7\pmod 8$ then there are no integer solutions $(x,y,z)$ with $x\neq 0$ to the equation:
$$az^2+by^2=cx^2.$$
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