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 :(
 A: If $5|(x^2+y^2+z^2)$ we can say that in $\Bbb Z_5$, $x^2+y^2+z^2= 0$ and in a) we saw the squares are $0,1,-1$ in $\Bbb Z_5$. So we can only have this when we have at least one $0$ and the other two $1$ and $-1$ or all three terms are $0$ (hence divisible by $5$). We cannot compensate two $1$'s or two $-1$'s and three of $1,-1$ is impossible anyway (we get an odd number).
A: a) is fine.
For b), you have proved in a) that any square is congruent to $0, 1$ or $-1\bmod 5$. Compute all possible sums of three of these (with possible repetitions). 
A: For part $(a)$, it suffices to enumerate among all residue classes modulo $5$. So we get:
$$
0^{2} \equiv 0 \pmod{5} \\
1^{2} \equiv 1 \pmod{5} \\
2^{2} \equiv 4 \pmod{5} \\
3^{2} \equiv 4 \pmod{5} \\
4^{2} \equiv 1 \pmod{5} \\
$$
Hence, $a^{2} \not \equiv 2,3 \pmod{5}$, for any integer $a$, since any integer can be written as $5k + t : t \in \{0,1,2,3,4\}$. 
For part $(b)$, suppose that exactly two of $x,y,z$ are multiples of $5$. Without loss generality, assume that $x,y$ are multiples of $5$ and $z$ is not. Then $z = 5k + t : t \ne 0$ and $x = 5x_{1}$ and $y = 5y_{1}$ for $x_{1},y_{1} \in \mathbb{Z}$. Then:
$$
x^{2} + y^{2} + z^{2} \equiv t^{2} \not \equiv 0 \pmod{5}
$$
so $5$ cannot divide $x^{2} + y^{2} + z^{2}$. Suppose that none of $x,y,z$ are multiples of $5$. Then:
$$
x^{2} + y^{2} + z^{2} \equiv a^{2} + b^{2} + c^{2} \not \equiv 0 \pmod{5}
$$
Since $a,b,c \in \{1,4\}$.
We proved the claim using the contrapositive.
A: 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$.)
