Integer solutions of $800000007 = x^2+y^2+z^2$ Prove that the equation, $800000007 = x^2+y^2+z^2$ has no solutions in integers.(That is $8$ followed by $7$ zeroes, with a $7$ at the end).
I tried checking modulo $3$, $5$, $7$, and $10$, but couldn't reach any  conclusions.

Hint: have a look at modulo $8$.
details: 
If $x^2+y^2+z^2 = A $
then $A\neq 7\mod 8$:
$$
x^2\in \{0,1,4\}\mod 8\\
x^2+y^2+z^2 \in \{0,1,4\}+\{0,1,4\}+\{0,1,4\} = 
\{0,1,2,3,4,5,6\}\mod 8.
$$
NB: the general equation $x^2+y^2+z^2 = A $ has integer solutions iff $A$ has not the form
$$
4^N(8k+7).
$$
Let us prove that if $A$ has this form there is no solution:
Assume it is not true.
Let $N$ be the smallest integer for which the equation has a solution of the form
$4^N(8k+7)$. $N>0$ because of the preceding proof. Then,
$$
x^2+y^2+z^2 =0\mod 4.
$$
As the squares are $0,1$ then $2$ divides $x,y,z$. 
Then
$$
\left(\frac x2\right)
^2+\left(\frac y2\right)^2+
\left(\frac z2\right)^2 = 4^{N-1}(8k+7);
$$ 
this is impossible from the definition of $N$.
