RMO 3

Given $pq=r+1$

$r$ must be an odd prime because if $r$ is even, then $r=2$ and $r+1=3\Rightarrow r=3$ which not possible since $p,q$ are primes. Hence $r$ is odd.

This means that $pq$ is even and hence one of $p,q$ must be $2$ .

Suppose $p$ is $2$ . This means that $2q=r+1$

Squaring both sides

$4{ q }^{ 2 }={ r }^{ 2 }+1+2r\\ \Rightarrow 4{ q }^{ 2 }=2({ p }^{ 2 }+{ q }^{ 2 })+2r\\ \Rightarrow 2{ q }^{ 2 }=2{ p }^{ 2 }+2r\\ \Rightarrow 2{ q }^{ 2 }=8+2r\\ \Rightarrow { q }^{ 2 }=4+r\\ \Rightarrow { q }^{ 2 }-4=r\\ \Rightarrow (q+2)(q-2)=r$

Since $r$ is a prime, one of $(q+2),(q-2)$ must be equal to one. If $q+2=1 \Rightarrow q=-1$ which is not possible. Hence $q-2=1 \Rightarrow q=3 \Rightarrow r=5$

Hence we find the solution $(p,q,r)=(2,3,5)$ . Similarly taking $q=2$ we can get $(p,q,r)=(3,2,5)$ .

Hence there are two solutions.