Primes Propaganda

n^3+n^2+n+1 = p^6

n^3+n^2+n = p^6 - 1

n^2 x (n+1) = (p^3+1) (p^3-1)

Notice: From the LHS, if we input some values of n, we always obtain products having the last digit (Either 0, 2 or 6)

From the RHS, however, when we input some values of p, where p is odd prime, we always obtain digits other than (0, 2 or 6)

Hence, there is no solution.

Rewrite the equation as $(n+1)(n^2+1)=p^6$ Note that both of the factors on the LHS are greater than 1. Also, observe that the only prime divisor of both of the factors of LHS is $p$ . It means that $p|n+1$ and $p|n^2+1$ . So $p|n(n+1)-(n^2+1)=n-1$ and $p|n+1-(n-1)=2$ Therefore $p=2$ and $(n+1)(n^2+1)=64$ It is clear that $n<5$ , but none of the possibilities work. No solution.