IMO 1977 LL NET26

Let p be a prime number greater than 5. Let V be the collection

IMO 1977 LL NET26

Origin: NET

Problem

Let p be a prime number greater than 5. Let V be the collection of all positive integers n that can be written in the form n = kp + 1 or n = kp −1 (k = 1, 2, . . .). A number n \inV is called indecomposable in V if it is impossible to find k, l \inV such that n = kl. Prove that there exists a number N \inV that can be factorized into indecomposable factors in V in more than one way.

Solution

The result is an immediate consequence (for G = {−1, 1}) of the following generalization. (1) Let G be a proper subgroup of Z∗ n (the multiplicative group of residue classes modulo n coprime to n), and let V be the union of elements

of G. A number m \inV is called indecomposable in V if there do not exist numbers p, q \inV , p, q ̸\in{−1, 1}, such that pq = m. There exists a number r \inV that can be expressed as a product of elements indecomposable in V in more than one way. First proof. We shall start by proving the following lemma. Lemma. There are infinitely many primes not in V that do not divide n. Proof. There is at least one such prime: In fact, any number other than \pm1 not in V must have a prime factor not in V , since V is closed under multiplication. If there were a finite number of such primes, say p1, p2, . . . , pk, then one of the numbers p1p2 \cdot \cdot \cdot pk +n, p2 1p2 \cdot \cdot \cdot pk +n is not in V and is coprime to n and p1, . . . , pk, which is a contradiction. [This lemma is actually a direct consequence of Dirichlet’s theorem.] Let us consider two such primes p, q that are congruent modulo n. Let pk be the least power of p that is in V . Then pk, qk, pk−1q, pqk−1 belong to V and are indecomposable in V . It follows that r = pk \cdot qk = pk−1q \cdot pqk−1 has the desired property. Second proof. Let p be any prime not in V that does not divide n, and let pk be the least power of p that is in V . Obviously pk is indecomposable in V . Then the number r = pk \cdot (pk−1 + n)(p + n) = p(pk−1 + n) \cdot pk−1(p + n) has at least two different factorizations into indecomposable factors.