Problems & Puzzles: Problems Problem 1. David Silverman Question
There is not much to say other than if you find another case, or if you argue a good reason to show that there is not going to be any other case, we would like to know about it… Solution Jud McCranie sends his results and comments on this problem : "On the other hand, I tried the Silverman Problem last night too. That is something where I don't expect another solution. I tested it up to m=14,000,000. The reason I don't expect any more solutions of it is that if you look at the factorization of the numerator and denominator, there always seem to be plenty of primes in the denominator that aren't canceled out by primes in the numerator" (10/July/98) And adds (at 23/July/98) "I checked the Silverman problem up to m=105,097,565 and found
no other solutions. It is unlikely that there are any other solutions. Consider the factorization of the numerator and
denominator of the product. At m=4 the product is 12, but at m=5 the product is multiplied by 12/10, introducing a factor of
5 in the denominator that isn't cancelled in the numerator. At m=8, p=19, p+1=2*2*5, and the numerator has a factor of 5 to
cancel out with the denominator, and the product is once again an integer. At m=9, p=23, a factor of 11 is introduced into the
denominator which isn't canceled out until m=14, p=43, p+1=2*2*11. However, by this time, another factor of 5 entered
Does anybody wants to verify and to give a formal proof of the McCranie observation that we have underlined?... *** Frank Rubin wrote (12/12/2002):
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