<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] 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.
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