The Theory of Numbers by Robert Daniel Carmichael
By Robert Daniel Carmichael
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Additional resources for The Theory of Numbers
L. That is, (1) can have only the n solutions already found. This completes the proof of the theorem. EXERCISES 1. Construct a congruence of the form a0 xn + a1 xn−1 + . . + an ≡ 0 mod m, a0 ≡ 0 mod m, having more than n solutions and thus show that the limitation to a prime modulus in the theorem of this section is essential. 2. Prove that x6 − 1 ≡ (x − 1)(x − 2)(x − 3)(x − 4)(x − 5)(x − 6) mod 7 for every integer x. 3. How many solutions has the congruence x5 ≡ 1 mod 11? the congruence x5 ≡ 2 mod 11?
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.
CHAPTER 3. ELEMENTARY PROPERTIES OF CONGRUENCES 30 But f (b) ≡ 0 mod p, (b − a)α ≡ 0 mod p. Hence, since p is a prime number, we must have f1 (b) ≡ 0 mod p. By an argument similar to that just used above, we may show that f1 (x) − f1 (b) may be written in the form f1 (x) − f1 (b) = (x − b)β f2 (x), where β is some positive integer. Then we have f (x) ≡ (x − a)α (x − b)β f2 (x) mod p. Now this process can be continued until either all the solutions of (1) are exhausted or the second member is a product of linear factors multiplied by the integer a0 .