Integer solutions of $800000007 = x^2+y^2+z^2$

Question

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.

Answer 1

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 $N$.

Answer 2

Try checking modulo 8. Note that every odd square is congruent with 1 modulo 8.