On the Congruence 2 n - 2 ≡1 (mod n)

Abstract

There exist infinitely many positive integers n such that 2~2 = 1 (mod n). In the monograph (5) I proposed the following problem (problem 18, p. 138): Let a, k > 1 be fixed positive integers. Do there exist infinitely many composite n such that «| a*- 1? Put a = 2, k = in the above problem. Since by Fermat's theorem 2p~l = 1 (mod p) for odd primes p, if = 1 (mod n) and n > 2, n must be composite. R. Matuszewski and P. Rudnicki (with the aid of the computer K-202 in Warsaw) checked that below 4208 such integers do not exist. The following theorem holds: congruence 2m = 1 (mod2m - 1) it follows that (2m)V-V/m = 1 (mod2m - 1), 2213 s 1 (mod2m - 1) and 2~2 = 1 (modn) for n = 2m - 1. Thus 22 = 1 (mod n) for n = 2» - 1, where «0 = 4700063497. Suppose now that 22 = 1 (mod «), and n > 8. Let/? be a primitive factor of the number 2~2 - 1 (a prime factor of 2 - 1 is said to be primitive if it does not divide any of the numbers 2m - 1 for m = 1,2,...,n — 1. By a theorem of K. Zsigmondy (7) such a prime factor exists for any n > 6 and is of the form nt + 1). Now we shall show that nx = np is also a solution of the congruence 21_2 = 1 (mod «,). We have p = 2(n - 2)k + 1, where k is a positive integer and p > 2« - 3 > n and (/?,«)= 1. Thus