<s>[INST] Show that $a^2$ cannot be congruent to 2 or 3 mod 5 for any integer a)  Show that $a^2$ cannot be congruent to $2$ or $3  \bmod 5$ for any integer.
$a^2≡2,3 \pmod5$ 
$a≡0,±1,±2\pmod5 )⟹a^2≡0,1,4$.
But $2,3$ are not congruent to $0,1,4\pmod5$.
I am not sure if I did it right, please check it for me
(b)  Show that if $5\mid x^2+y^2+z^2$ then either exactly one of, or each of  $x, y, z$ are a multiple of $5$.
I really can not figure it out :(
 [/INST] You did a) just find.  As $(-2,-1,0,1,2)$ is a complete residue class $\mod 5$ it must be that $a\equiv 0,\pm 1, \pm 2\pmod 5$ and therefore $a^2\equiv 0, 1,4\equiv -1$ none are congruent to $2$ or to $3\equiv -1$.
As for $b$ we have $x^2,y^2, z^2 \equiv 0, 1,-1\mod 5$ so we mus have some how that $x^2 + y^2 + z^2 \equiv (0|1|-1) + (0|1|-1) + (0|1|-1)\equiv 0$.
And it should be clear of all the combintations:
$0 + 0 + 0 \equiv 0$ (Hit)
$0 + 0 + 1\equiv 1$,$0+0-1\equiv -1; 0+1+1\equiv 2$ (Misses)
$0 + 1\equiv -1\equiv 0$ (Hit)
$0 -1 -1\equiv -2; 1+1+1\equiv 3; 1+1-1\equiv 1; 1-1-1\equiv -1; -1-1-1\equiv -3$ (Misses)
The only way $x^2 + y^2 + z^3 \equiv 0\pmod 5$ is if all $x^2,y^2, z^2 \equiv 0$ in which cae $x,y, z\equiv 0$ and are all multiples of $5$ 
or if one of them is a multiple of $5$ and the other two .... are not.  (Note, we never said that the other two could be anything; we only said they can't be multiples of $5$.  One must be $x$ (or $y$) $ \equiv \pm 1$ so $x^2$( or $y^2$) $\equiv 1$ and the other must be $y$ (or $x$) $\equiv \pm 2$ so that $y^2$ (or $x^2$) $ \equiv 4\equiv -1$.)
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