Let n and p be positive integers greater than 1, with p being a prime. Show that if n divides p-1 and p divides n^3-1, then 4p-3 is a perfect square.

Let p = n^2+n+1
Then p-1 is divisible by n
p-1 = n^2 + n = n(n+1)
also n^3-1 is divisible by p
n^3-1 = (n-1)(n^2+n+1) = (n-1)p

So 4p-3 = 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 2015-02-28 11:26:35