 Solution

Put $m=(p-1)/2$, so the requested modulus is $pm$. The factorial $(p-1)!$ is divisible by $m$, and so is $p-1=2m$. Also, Wilson's theorem gives $(p-1)!\equiv -1\equiv p-1(\mathrm{mod}p)$. Because $m$ and $p$ are coprime integers, their product divides the difference. Thus the Wilson factorial congruence modulo a triangular number isThe displayed residue lies in the standard range even at $p=3$.