By G. H. & Wright, E. M. Hardy

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**Example text**

A direct substitution verifies the conclusion that none of them satisfies the congruence; and hence that the congruence has no solution at all. On the other hand the congruence x5 − x ≡ 0 mod 5 has the solutions x = 1, 2, 3, 4, 5 as one readily verifies; that is, this congruence has five solutions—the maximum number possible in accordance with the results obtained above. EXERCISES 1. Show that (a + b)p ≡ ap + bp mod p where a and b are any integers and p is any prime. 2. From the preceding result prove that αp ≡ α mod p for every integer α.

P − 1 (A) rx ≡ a mod p. 2) and consider the congruence This has always one and just one solution x equal to a number s of the set (A). Two cases can arise: either for every r of the set (A) the corresponding s is different from r or for some r of the set (A) the corresponding s is equal to r. The former is the case when a is a quadratic non-residue modulo p; the latter is the case when a is a quadratic residue modulo p. We consider the two cases separately. In the first case the numbers of the set (A) go in pairs such that the product of the numbers in the pair is congruent to a modulo p.

3 · 2 · 1; 2p for the first vertex can be chosen in p ways, the second in p − 1 ways, . , the (p − 1)th in two ways, and the last in one way; and in counting up thus we have evidently counted each polygon 2p times, once for each vertex and for each direction from the vertex around the polygon. Of the total number of polygons 21 (p − 1) are regular (convex or stellated) so that a revolution through 360◦ p brings each of these into coincidence with its former position. The number of remaining p-gons must be divisible by p; for with each such p-gon we may associate the p − 1 p-gons which can be obtained from it by rotating it through ◦ successive angles of 360 p .