# Download e-book for iPad: Algebra[Lecture notes] by K. Mathiak, Steffen Supra

By K. Mathiak, Steffen Supra

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**Extra resources for Algebra[Lecture notes]**

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If the answer is negative, it will provide a characterization of prime numbers. What can one say, anyway, when it is not possible to solve the problem? Only that the existence of composite integers n, for which ϕ(n) divides n − 1, is unlikely, for various reasons: (a) any such number must be very large (if it exists at all); (b) any such number must have many prime factors (if it exists at all); (c) the number of such composite numbers, smaller than any given real number x, is bounded by a very small function of x.

Euler’s Theorem. If gcd(a, n) = 1, then aϕ(n) ≡ 1 (mod n). Proof. Let r = ϕ(n) and let b1 , . . , br be integers, pairwise incongruent modulo n, such that gcd(bi , n) = 1 for i = 1, . . , r. Then ab1 , . . , abr are again pairwise incongruent modulo n and gcd(abi , n) = 1 for i = 1, . . , r. Therefore, the sets {b1 mod n, . . , br mod n} and {ab1 mod n, . . , abr mod n} are equal. Now, r r bi ≡ ar i=1 r abi ≡ i=1 bi (mod n). i=1 Hence, r (ar − 1) bi ≡ 0 (mod n) and so ar ≡ 1 (mod n). i=1 Just like for Fermat’s little theorem, it follows also from Euler’s theorem that there exists the smallest positive exponent e such that ae ≡ 1 (mod n).

By experimenting numerically with sequences an − bn , it is observed that, apart from a few initial terms, t∗n is composite. In fact, Schinzel indicated the following theorem in 1962. II. Some Fundamental Theorems on Congruences 35 Let k(m) denote the square-free kernel of m, that is, m divided by its largest square factor. Let e= 1, if k(ab) ≡ 1 (mod 4), 2, if k(ab) ≡ 2 or 3 (mod 4). If n/ek(ab) is integral and odd, and if n > 1, then an − bn has at least two distinct primitive prime factors, with only a few exceptions (of which the largest possible is n = 20).

### Algebra[Lecture notes] by K. Mathiak, Steffen Supra

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