<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] As Hoseyn Heydari's comment stated, you should consider checking modulo $5$ instead for your equation of
$$z^2 = 10x^2 - 5y^2 \tag{1}\label{eq1}$$
As the RHS is a multiple of $5$, this means $z = 5z_1$ for some integer $z_1$. Thus, \eqref{eq1} becomes
$$5z^2_1 = 2x^2 - y^2 \tag{2}\label{eq2}$$
Since the possible congruences of squared values modulo $5$ are $0, 1$ and $4$, then $2x^2$'s values modulo $5$ must be $0, 2$ or $3$ (as $2 \times 4 \equiv 3 \pmod 5$). Thus, since the LHS is a multiple of $5$ and the only match between the $2$ sets of $\{0,2,3\}$ and $\{0,1,4\}$ is $0$, this means $2x^2 - y^2$ can only be congruent to $0$ modulo $5$ if both $x$ and $y$ are as well. Let $x = 5x_1$ and $y = 5y_1$ to get
$$z^2_1 = 10x^2_1 - 5y^2_1 \tag{3}\label{eq3}$$
Note this is of the same form as \eqref{eq1}, but with each value being $5$ times smaller. As the procedure can be repeated indefinitely, this is only possible if each of the values were $0$ initially, giving that the sole solution is $x = y = z = 0$.
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