The question is asking us to prove that $x^2+y^2+z^2=999$ has no integer solutions.
Attempt at a solution:
So I've noticed that since 999 is odd, either one of the variables or all three of the variables must be odd.
If I assume that only one variable is odd, I can label the variables like this:
$$x=2k_1+1$$
$$y=2k_2$$
$$z=2k_3$$
By substituting, and doing some algebra, I can conclude that $k_1^2+k_2^2+k_3^2+k_1=249.5$, which is not possible since all $k_i\in\Bbb Z$.
If all three are odd, I can rename the variables like this:
$$x=2k_1+1$$
$$y=2k_2+1$$
$$z=2k_3+1$$
Eventually I conclude that $k_1^2+k_2^2+k_3^2+k_1+k_2+k_3 = 249$, but I don't know where to go from there.
An alternative I've considered is brute-forcing it, but I'd rather avoid that if I can. Any assistance here would be greatly appreciated.
Best Answer
Using congruences . . .
Odd squares are always $1 \pmod 8$, hence also $1 \pmod 4$.
Even squares are always $0 \pmod 4$, hence either $0 \text{ or } 4 \pmod 8$.
Since $x^2 + y^2 + z^2$ is odd, either $x,y,z$ are all odd, or exactly one of $x,y,z$ is odd.
If $x,y,z$ are all odd, then $x^2 + y^2 + z^2 \equiv 3 \pmod 8$, contradiction, since $999 \equiv 7 \pmod 8$.
If exactly one of $x,y,z$ is odd, then $x^2 + y^2 + z^2 \equiv 1 \pmod 4$, contradiction, since $999 \equiv 3 \pmod 4$.