Consider a positive integer n and a prime number p.
The following conditions are satisfied simultaneously:
- p divides n3-1
- n divides p-1
Does 4p-3 always correspond to a perfect square?
- If not, provide an example.
- If so, prove it.
n divides p-1 then means we must have p-1>=n, or a slightly weaker statement p>n.
So then p divides n^3-1. Factor the right side into (n-1)*(n^2+n+1). Since p>n then it cannot divide n-1, so it must divide n^2+n+1.
The simplest way for p to divide n^2+n+1 is by equality, so a class of solutions is the set of primes p of the form n^2+n+1. The OEIS lists these as sequence https://oeis.org/A002383
But if p=n^2+n+1 then 4p-3=4n^2+4n+1=(2n+1)^2, which is a perfect square.
What I am currently stuck on is if p<n^2+n+1. because p>n we must have p be the largest prime divisor of n^2+n+1. But so far I am unable to find any examples or exclude this case.
Some analytic progress
