$5 \mid k(k^2 +1)(k-1)^2 \implies k\not\equiv 4$ mod $5$

Question

Note that $k$ is the valency of a graph. In the textbook Algebraic Graph Theory (Godsil & Royle) it mentions (at the tail-end of an exercise) that $5 \mid k(k^2 +1)(k-1)^2 \implies k\not\equiv 4$ mod $5$. The main part of the exercise is to show that $5 \mid k(k^2 +1)(k-1)^2$, which I can show, but I don't see how that leads me to $k\not\equiv 4$ mod $5$. Any hints/help would be greatly appreciated.

Answer 1

If $k \equiv 4 \pmod{5}$, then \begin{align*} k-1 &\equiv 3 \pmod{5}\\ (k-1)^2 &\equiv 4 \pmod{5}\\ \end{align*} Likewise $k^2+1 \equiv 2 \pmod{5}$ Thus $$k(k-1)^2(k^2+1) \equiv 4 \cdot 4 \cdot 2 \equiv 2 \pmod{5}.$$

This contradicts the fact that the given expression is divisible by $5$.

Answer 2

We have $5|k(k^2+1)(k-1)^2\implies k(k^2+1)(k-1)^2=0 (mod 5)$

This is satisfied if $k$ is congruent to 0 or 1, or if $k^2+1=0(mod 5)$.

Of the only possible values, 0,1,2,3, and 4, $k^2+1=0( mod 5)$ only if $k=2$ or $k=3$.

So the equation is satisfied if $k=0,1,2,$ or $3$.

So $k<>4 (mod 5)$.