2 Jeffrey
نویسنده
چکیده
We consider the minimal number of solutions to (n) = (n + k) for xed values of k. Previous work of A. Schinzel concerning this problem is described, as is the outcome of recent computational searches conducted to extend Schinzel's results. Let k be a xed positive integer. In this note, we consider the minimum number of solutions to the equation (n) = (n + k); (1) where n is a positive integer and denotes Euler's totient function. In 1956, W. Sierpi nski 4] showed that there is always at least one solution to (1) for each xed k. He constructs a solution as follows: let p be the smallest prime that does not divide k, and let n = (p ? 1)k. Then n is a solution to (1). In 1958, A. Schinzel 2] showed that there are at least two solutions to (1) for all k 8 10 47. His method of proof was to split the problem into two cases, one for odd values of k and the other for even values of k. Of the two, the case for k odd is the more interesting, and that is what we shall focus on rst. Lemma 1 given below is the key result for the case when k is odd. The notation rj s means that each prime factor of r is a prime factor of s. Lemma 1. Suppose that the sequence of primes 3 = p 1 < p 2 < < p m satisses the conditions (i) (p i ? 2)jp 1 p 2 p i?1 (2 i m), (ii) (p i ? 1)j 2p 1 p 2 p i?1 (2 i m). Suppose that k is odd and is not divisible by p 1 p 2 p m , and let p j be the smallest prime in the sequence that does not divide k. Then n = p j k=(p j ? 2) is a solution to (1). The details of the proof are left to the reader, or may be found in 2, Lemma 1]. It is easy to verify that the solution to (1) given in Lemma 1 is distinct from that given by Sierpi nski's result, so that Lemma 1 assures two solutions for all odd k < p 1 p 2 p m .
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تاریخ انتشار 2007