Show that $\sum\limits_{k = 1}^p {{k^p} \equiv 0(\bmod p)}$.

Given that $\sum\limits_{k = 1}^p {{k^{p - 1}} \equiv n(\bmod p)}$ where $0 \leqslant n \leqslant p - 1$, find the value of n.

using Fermat’s little theorem,

${k^p} \equiv k(\bmod p)$     (M1)

therefore,

$\sum\limits_{k = 1}^p {{k^p} \equiv } \sum\limits_{k = 1}^p {k(\bmod p)}$     M1

$\equiv \frac{{p(p + 1)}}{2}(\bmod p)$     A1

$\equiv 0(\bmod p)$     AG

since $\frac{{(p + 1)}}{2}$ is an integer (so that the right-hand side is a multiple of p)     R1

[4 marks]

using the alternative form of Fermat’s little theorem,

${k^{p - 1}} \equiv 1(\bmod p),{\text{ }}1 \leqslant k \leqslant p - 1$     A1

${k^{p - 1}} \equiv 0(\bmod p),{\text{ }}k = p$     A1

therefore,

$\sum\limits_{k = 1}^p {{k^{p - 1}} \equiv } \sum\limits_{k = 1}^{p - 1} {1{\text{ }}( + 0)(\bmod p)}$     M1

$\equiv p - 1(\bmod p)$     A1

(so n = p − 1)

Note: Allow first A1 even if qualification on k is not given.

[4 marks]

Only the top candidates were able to produce logically, well thought-out proofs. Too many candidates struggled with the summation notation and were not able to apply Fermat’s little theorem. There was poor logic i.e. looking at a particular example and poor algebra.

Only the top candidates were able to produce logically, well thought-out proofs. Too many candidates struggled with the summation notation and were not able to apply Fermat’s little theorem. There was poor logic i.e. looking at a particular example and poor algebra.