BULGARIAN NATIONAL MATHEMATICAL OLYMPIAD

Answer: $(p,a,m)=(2,1,1),(2,3,2),(2,7,3),(3,1,1),(3,7,2),(3,25,3),(5,1,2),(2,15,4),(5,101,3)$. Direct checks for $p=2,3,5$ give the above solutions. For $p\ge 7$ the Wilson theorem implies $a\equiv 1(\mathrm{mod}p)$. Moreover, obviously $p-1|a-1$. Therefore $a=kp(p-1)+1$ for some integer $k\ge 0$. If $k\ge 6$ then $a\ge 6p^2-6p+1>5p^2$, a contradiction. So we have $0\le k\le 5$. After division by $p-1$ we obtain $$(p-2)!+kp=p^{m-1}+p^{m-2}+\cdots +1.$$ Since $p\ge 7$, the $2$ and $\frac{p-1}{2}$ are different and appear in $(p-2)!$, i.e. $p-1|(p-2)!$. This gives $k\equiv m(\mathrm{mod}p-1)$. If $m\ge p$, then $$(p-2)!+kp=p^{m-1}+p^{m-2}+\cdots +1>p^{p-1}>(p-1)!,$$ which easily gives a contradiction. Thus $m\le p-1$. If $k=0$, then $m=p-1$ and $$(p-2)!=p^m-1\ge p^{p-1}-1>(p-1{)}^{p-1}>(p-1)!,$$ a contradiction. If $k=1$ or $2$, then $m=1$ or $2$, respectively, which is impossible. If $k=3$, then $m=3$, i.e. $(p-2)!=(p-1{)}^2$, which is impossible. For $k=4$ we get $m=4$ and $(p-2)!=(p-1)(p^2+2p-1)$, whence $$p-2|p^2+2p-1=p^2-4+2(p-2)+7,$$ i.e. $p=7$, which does not give solutions. For $k=5$ we obtain $m=5$ and $(p-2)!=(p-1)(p^3+2p^2+3p-1)$, whence $$p-2|p^3+2p^2+3p-1=p^3-8+2(p^2-4)+3(p-2)+21,$$ i.e. $p=7$, which does not give solutions.