By Michael Th. Rassias

The e-book offers a self-contained advent to classical Number conception. all of the proofs of the person theorems and the options of the routines are being offered step-by-step. a few historic comments also are provided. The ebook might be directed to complex undergraduate, starting graduate scholars in addition to to scholars who organize for mathematical competitions (ex. Mathematical Olympiads and Putnam Mathematical competition).

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**Additional info for Problem-Solving and Selected Topics in Number Theory: In the Spirit of the Mathematical Olympiads**

**Example text**

L. Cohen and H. J. J. te Riele in their joint paper [14] proved that if there exist odd perfect numbers n, then it must hold n > 10300 . (ii) Another well-known open problem related to perfect numbers is the following: Are there inﬁnitely many even perfect numbers? 4. In the previous two theorems we examined the case when the integer 2n − 1 was a prime number. Hence, it is worth mentioning that the integers of that form are called Mersenne numbers, after the mathematician Marin Mersenne who ﬁrst investigated their properties.

Xn . In that case, we obtain (an xni + an−1 xn−1 + · · · + a1 xi + a0 ) − (an xn0 + an−1 xn−1 + · · · + a1 x0 + a0 ) 0 i − xn−1 ) + · · · + a1 (xi − x0 ) = an (xni − xn0 ) + an−1 (xn−1 0 i = (xi − x0 )p(xi ), i = 1, 2, . . , n, where p(x) is a polynomial of n − 1 degree with integer coeﬃcients. Because of the fact that p | (an xni + an−1 xn−1 + · · · + a1 xi + a0 ) i and p | (an xn0 + an−1 xn−1 + · · · + a1 x0 + a0 ) 0 it is evident that p | (xi − x0 )p(xi ), i = 1, 2, . . , n. 46 4 Congruences The integers xi , x0 are distinct solutions and thus xi ≡ x0 (mod p).

Hence, σ1 (m) ≥ c(2n − 1) + c + 1 = 2n c + 1 > σ1 (m), since σ1 (m) = 2n c, which is a contradiction. Therefore, we obtain that c = 1 and consequently we get m = 2n − 1 and k = 2n−1 (2n − 1) since k = 2n−1 m. The only question which remains unanswered is whether 2n − 1 is a prime number. However, by (1) we have (2n − 1)σ1 (2n − 1) = 2n (2n − 1) or σ1 (2n − 1) = 2n = (2n − 1) + 1. Thus, the only divisors of 2n − 1 are the number itself and 1. Thus, clearly 2n − 1 is a prime number. 1 Related open problems (i) We observe that Euler’s theorem strictly refers to even perfect numbers.