Problem 5

With $a=p$, the equation becomes $p^p=b!+p$. For $p=2$, checking $2\le b\le3$ gives only $(a,b,p)=(2,2,2)$ (since $2^2=4=2!+2$). For $p=3$, checking $3\le b\le5$ gives only $(a,b,p)=(3,4,3)$ (since $3^3=27=4!+3$).