Let n and p be positive integers greater than 1, with p being a prime. Show that if n divides p1 and p divides n^31, then 4p3 is a perfect square.
Let p = n^2+n+1
Then p1 is divisible by n
p1 = n^2 + n = n(n+1)
also n^31 is divisible by p
n^31 = (n1)(n^2+n+1) = (n1)p
So 4p3 = 4(n^2+n+1)3 = 4n^2+4n+1+1 = (2n+1)^2
Note: p is not necessarily prime but it is odd.
I should call this a partial solution, because not all primes are of the form n^2+n+1. It could be the case that there is a solution involving a prime not of this form.

Posted by Jer
on 20150228 11:26:35 