34th Hellenic Mathematical Olympiad

The given equation can be written as: $p{a}^2+b^2=a^2{b}^2$ (1) Since $p$ is prime, from (1) we have: $p|a$ or $p|b$. We suppose that $p|a$, and hence $a=p{a}_1$, $a_1\in {\mathbb{N}}^{\ast }$. Moreover, we have that $$\begin{array}{rl}\frac{1}{p}>\frac{1}{b^2}& \Rightarrow b^2>p\Rightarrow b^2\ge p+1\Rightarrow \frac{1}{a^2}=\frac{1}{p}-\frac{1}{b^2}\ge \frac{1}{p}-\frac{1}{p+1}=\frac{1}{p(p+1)}\\ & \Rightarrow \frac{1}{p^2{a}_1^2}\ge \frac{1}{p(p+1)}=\frac{1}{p^2+p}\ge \frac{1}{2p^2}\Rightarrow \frac{1}{a_1^2}\ge \frac{1}{2}\Rightarrow a_1^2\le 2\Rightarrow a_1=1\Rightarrow a=p.\end{array}$$ Then equation (1) becomes: $$\begin{array}{rl}p{p}^2+b^2& =p^2{b}^2\iff p^2+b^2=p{b}^2\iff p^2=(p-1)b^2\iff p^2-1=(p-1)b^2-1\\ & \iff (p-1)(p+1)=(p-1)b^2-1\Rightarrow p-1{\mid }_{p-1>0}\iff p-1=1\iff p=2.\end{array}$$ Hence $a=p=2$ and from equation (1) we find $b=2$. Similarly we work if $p|b$.

Second solution: The given equation can be written as: $p{a}^2+b^2=a^2{b}^2$ (1) Solving with respect to $a^2$ we find $$a^2=\frac{p{b}^2}{b^2-p}=\frac{p(b^2-p)+p^2}{b^2-p}=p+\frac{p^2}{b^2-p}(2)$$ Since $a^2$ is integer, $b^2-p\mid p^2$, and hence $$b^2-p=1\text{or}{b}^2-p=p\text{or}{b}^2-p=p^2$$ $b^2-p=1\iff p=b^2-1=(b-1)(b+1)$ and since $p$ prime, $b-1=1$, hence $b=2$ and $p=3$. Then $a^2=12$, absurd. $b^2-p=p$ gives $b^2=2p$. Then $b$ is even, say $b=2b_1$, and hence $4b_1^2=2p$. Hence $2\mid p$, and therefore $p=2$ and $b=2$, $a=2$. * In the third case (2) gives $a^2=p+1\iff p=a^2-1=(a-1)(a+1)$, and since $p$ prime, $a-1=1$, that is $a=2$ and $p=3$, which gives $b^2=3$, absurd.