Prove that for all natural numbers $k$, $k^5-k$ is a multiple of $10$.

Question

Prove that for all natural numbers $k$, $k^5-k$ is a multiple of $10$.

I found several answers for this here, but none of which would have considered seperately the cases $\pmod{2}$ and $\pmod{5}.$

Since any integer $n \equiv 0,1 \pmod{2}$ we have that $k^5-k \equiv0 \pmod{2}$ thus divisible by $2$.

However, the case $k^5-k \equiv0 \pmod{5}$ seems to be a bit trickier. Any integer $n \equiv 0,1,2,3,4 \pmod{5}$, but is this of any help as it seemed to be in the case of $\pmod{2}$?

Verify that the congruence class of $k$ modulo 5 is always the same as the the congruence class of $k^5$ modulo 5. — Laars Helenius, Oct 25 '20 at 09:51

score 1 · Accepted Answer · answered Oct 25 '20 at 09:47

1

For the case $\mathrm{5}$ use Fermat's theorem.

answered Oct 25 '20 at 09:47

Alexandr Taymenev

126

$k^5 \equiv k \pmod{5} \Rightarrow k^5-k \equiv k-k \equiv 0 \pmod{5}$? – Oct 25 '20 at 09:49
$k^{\phi(5)} = k^4 \equiv 1 (mod 5)$ – Alexandr Taymenev Oct 25 '20 at 09:51
Doesn't Fermat's theorem state that $a^p \equiv a \pmod{p}$? – Oct 25 '20 at 09:53
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This is equivalent statements =) – Alexandr Taymenev Oct 25 '20 at 09:54

Prove that for all natural numbers $k$, $k^5-k$ is a multiple of $10$.

1 Answers1