SAUDI ARABIAN MATHEMATICAL COMPETITIONS

Let $p$ be a prime satisfying the condition of the problem. By assumption, there is a positive integer $A$ such that $3^{p-1}-1=p{A}^2$.

It is clear that $p=2$ is a solution. Now, we consider $p>2$. Put $p-1=2k$, one has $(3^k-1)(3^k+1)=p{A}^2$. Since $(3^k-1,3^k+1)=2$, it follows that there are positive integers $B$ and $C$ such that either $$3^k-1=2p{B}^2,3^k+1=2C^2$$ or $$3^k-1=B^2,3^k+1=2p{C}^2.$$ But, the first case cannot hold since $2C^2=3^k+1\equiv 1(\mathrm{mod}3)$, we get $C^2\equiv 2(\mathrm{mod}3)$ which is impossible.

For the second case, if $k$ is odd then $4\mid 3^k+1=2p{C}^2$, hence $2\mid C$ (since $p$ is odd). This turns out to be that $3^k+1=2p{C}^2$ is divisible by $8$ which contradicts to $3^k+1\equiv 4(\mathrm{mod}8)$ (since $k$ is odd). Thus, $k$ must be even. Put $k=2m$, then $$2B^2=3^k-1=(3^m-1)(3^m+1).$$ Again, since $(3^m-1,3^m+1)=2$, there are positive integers $D,E$ such that either $$3^m-1=E^2,3^m+1=2D^2$$ or $$3^m-1=2E^2,3^m+1=D^2.$$ As above, the equality $3^m+1=2D^2$ leads to a contradiction. Hence, $3^m+1=D^2$, that is $$3^m=D^2-1=(D-1)(D+1).$$ Therefore, there are non-negative integers $t>s$ such that $D-1=3^s$ and $D+1=3^t$. This gives, $$2=(D+1)-(D-1)=3^t-3^s=3^s(3^{t-s}-1).$$ This happens if and only if $3^s=1$ and $3^{t-s}-1=2$, i.e. $s=0$ and $t=1$, we find that $p=5$ (satisfied).

In conclusion, $p=2;5$. $\square$