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Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 3 / 11I / b

Codex (@codex,  0) ... Mathematics course of the University of Cambridge Past exam of the mathematics course of the University of Cambridge 2019 ii Paper 3 11I
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b
For k=0, the convention 00=1 gives S0​=p≡0(modp). Now suppose 1≤k<p−1 and choose a primitive root g modulo p. The nonzero residues are gj for 0≤j≤p−2, while the x=0 term vanishes. Hence
Sk​≡∑j=0p−2​gjk(modp).
(1)
This is a finite geometric series. Since gk≡1(modp) but gk(p−1)≡1(modp) by Fermat's little theorem, multiplication by the nonzero residue gk−1 gives
(gk−1)Sk​≡gk(p−1)−1≡0(modp).
(2)
The factor gk−1 is invertible modulo the prime number p, so
Sk​≡0(modp).​
(3)

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