Abstract
has not yet been completely solved. Lehmer ([7], [8]) proved that the least integer n with r(n) =0 must be a prime >214,928,639,999. This limit was further raised by Lehmer and others. In this paper we shall take a different approach to the problem. It was proved by Gupta [5] that (1.2) r(21) 0 O (mod 230), r(20) 0 0 (mod 230+1), which implies that r(20) never vanishes. Now let p be an odd prime and let 0 be determined by p = 20n -1, n odd. In Section 2 we shall prove that, for 0=1,. . ,9, (1.3) r(p)0 (mod 20), r(p) $ 0 (mod 20+1). This suggests the conjecture* that (1.3) holds for all 0 > 1 and this in turn would imply that Tr(p) never vanishes. In Section 3, we shall show that (1.3) for all 0 ? 1 implies (1.4) r(p4m+l) = 0 (mod 20), -r(p4mn+1) 0 0 (mod 20+1), (1.5) r(p40+S) =0 (mod 20), 7r(p4m+S) 0 0 (mod 20+1). In Section 4 we give a new proof of Lehmer's result [7] that if r(p) 5d0 for a prime p, then 'r(pn) 5-?0 for n >2. Furthermore, we shall prove that if (1.3) is valid then, for M odd, (1.6) r(pkM-1) 0 (mod 20+k-1), r(p2kM-1) $ 0 (mod 20+k). It is to be noted that throughout the paper p denotes an odd prime and 0 is determined by p = 2'n -1, n odd. 2. We now proceed to prove (1.3) for 0= 1, * , 9. It is known that [9] (2.1) Tr(M) M2cr7(M) (mod 22). Let M=2n-1 be a prime where n is odd. Then from (2.1) we have r(2n-1) 2n (mod 22) and, since n is odd, Tr(2n-1)-0 (mod 2), r(2n-1) 0 0 (mod 22). Also it is known that ([3], [4]) (2.2) r(M) o-e(M) (mod 23). * Professor Lehmer had suggested the form r(p) oan (p) (mod 2?) for (1.3). He made the conjecture independently but did not publish it and it was communicated privately to the author. 757
Published Version
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