No solutions to $x^2+y^2+z^2 = 7t^2$ A problem in my book asks me to show that there are no solutions to 
$$x^2+y^2+z^2 = 7t^2$$
in the integers apart from $(x,y,z,t)=(0,0,0,0)$.
The solution states that reducing modulo $4$ we see that $x,y,z,t$ must be even
and dividing through we get a smaller solution.
I don't understand how we can conclude that everything is even. If $(x,y,z,t)$ are all $1$ mod $4$ it still seems to hold.

Hmm.... Well, if $x,y,z$ are even then $t$ is, of course and if, $x,y,z$ are all odd then $t$ is. and then  $\mod 4$ we get $x^2 + y^2 + t^2 \equiv 3\mod 4$ and $7t^2 \equiv 3\pmod 4$ so that's not a problem.  (Yet.) 
Indeed It will always be the case that $(2m+1)^2 + (2n+1)^2 + (2k+1)^2 \equiv 3 \equiv 7(2j+1)^2$.
But we can rule out $1$ or $2$ odds.
If one of $x,y,z$ are odd and $t$ are odd but the rest are even we'd have $x^2 + y^2 + z^2\equiv 1 \pmod 4\not\equiv -1\equiv 7t^2$.
If two of $x,y,z$ are odd and $t$ is even we would have $x^2 + x^2 +z^2 \equiv 2\pmod 4$ and $7t^2 \equiv 0 \pmod 4$.
So $\mod 4$ yields either $x,y,z,t$ are all odd or all even.
$\mod 8$ we have $(8m\pm 1,3)^2 \equiv 1\pmod 8$ and so all odd would result in $x^2 + y^2 + z^2 \equiv 3 \not \equiv 7\equiv t^2$.
So that's it.  If $x^2 + y^2 + z^2 = 7t^2$ then all $x,y,z,t$ are even but wolog we can assume $x,y,z,t$ have no factor $d$ in common be because $x^2 + y^2 + z^2 = 7t^2 \iff (\frac xd)^2 + (\frac yd)^2 + (\frac zd)^2=7(\frac td)^2$.
