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For what odd primes p is x = (2^(p - 1) - 1)/p an integer square?
Let be be some odd prime, and x = (2^(p - 1) - 1)/p
For what p is x an integer square, i.e. solve for p :
For what p is x an integer square, i.e. solve for p :
a^2 = (2^(p - 1) -1)/p
pa^2 = 2^(p - 1) -1
pa^2 = 2^(p - 1) -1
The trivial solutions are: p = 3 and p = 7
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