Primes

Solving the equation:

$p^{n} = \frac{(p-1)(p)}{2}$

$p^{n-1} = \frac{p-1}{2}$

$2p^{n-1} = p-1$ .......... (1)

Since $2p^{n-1} > p > p-1$ for all $n \geq 2$ , $n$ must be equal to $1$ (otherwise equation (1) cannot be true as the left hand side will always greater than the right hand side).

So, subtituting $n=1$ into equation (1), we have $p-1=2p^{1-1} = 2p^{0} = 2(1) = 2$

Hence, $p = 3$ . It is obvious that there is only one possible $p$ . So, the answer is 3.