Junior Macedonian Mathematical Olympiad

After simplifying the equation we get $2r(p+q)=(p-q{)}^2$. Since $r$ is prime, it follows that $r$ is a divisor of $p-q$, so that $r^2$ is a divisor of the right-hand side of the last equality. This implies that $r$ is a divisor of $2(p+q)$. If $r>2$, then $r$ is a divisor of $(p+q)$, so that $r$ has to be a divisor of both $p$ and $q$, but since $p$ is prime, that is possible only if $p=r$ and $q=sr$. After simplification we get $2(1+s)=(s-1{)}^2$, from where $s^2-4s-1=0$. The last equality has no integer solutions, so in this case the equation has no solution.

If $r=2$, then $p$ and $q$ have the same parity, the case when $p=2$ is impossible together with the previous case $p=r$, so they have to be odd. Let $a\ne 2$ be a prime divisor of $p+q$, then $a$ must be a divisor of $p-q$ too, so that it must be a divisor of $p$ and $q$, which is possible only if $p=a$ and $q=sa$, in this case we get $4(1+s)=a(s-1{)}^2$, from where $a^2-(2a+4)s+(a-4)=0$, with solutions $\frac{a+2\pm \sqrt{a^2+4a+4-a^2+4a}}{a}=\frac{a+2\pm 2\sqrt{2a+1}}{a}$. If the number $\sqrt{2a+1}$ is an integer, then it is odd, so $2a+1=4b^2+4b+1$, from where $a=2b(b+1)$, from where it cannot be prime.

Therefore $p+q$ and $p-q$ must be powers of 2, i.e. $p-q=2^k$ and $p+q=2^{2k-2}$, from where $2p=2^k+2^{2k-2}$ and $2q=2^{2k-2}-2^k$, and since $p$ and $q$ are odd, $k=1$, but then $p+q=1$, which is impossible. It follows that the equation has no prime number solutions.