Since $(a-b)\in \mathbb{P}$ it is obvious that $(a,b)=1$, otherwise we would have $a-b=a'd-b'd=d(a-b)$ which cannot be a prime for $d>1$ Since $ab$ is a perfect square and $(a,b)=1$ we have $a=x^{2} \ b=y^{2}$ So $x^{2}-y^{2}=(x-y)(x+y)=p\in \mathbb{P}$ Since $x>y$ we have $x-y=1 \ x+y=p$ So $\{x,y\}=\{\frac{p+1}{2},\frac{p-1}{2}\}$ for all $p\in \mathbb{P}$ such that $p\neq 2$, so $\{a,b\}=\{\frac{(p+1)^{2})}{4},\frac{(p-1)^{2})}{4}\}$ Regards Piotr Rutkowski p.s. It really was on Baltic Way?

Let $\gcd(a, b)=d$, $a=du$ and $b=dv$, where $d, u, v\in\mathbb N$ Then $d\cdot\gcd(u, v)=\gcd(du, dv)=\gcd(a, b)=d\implies\gcd(u,v)=1$ $ab$ is a perfect square $\implies d^2uv$ is a perfect square $\implies uv$ is a perfect square. But $\gcd(u,v)=1\implies u$ and $v$ are perect squares. So let $u=x^2$ and $v=y^2$, where $x, y\in\mathbb N$ Suppose that $d>1$ then $d|a-b$ But $a-b$ is a prime so $d=a-b=d(u-v)\implies u-v=1$ Then $(x-y)(x+y)=x^2-y^2=u-v=1$ Then $x-y>0$ so $2=1\cdot 2\leq (x-y)(x+y)=1$, which is false. So $d=1\implies\gcd(a,b)=1$ So $a=x^2$ and $b=y^2$ Because $a-b$ is a prime number let $a-b=p$ where $p$ is a prime number. So $(x-y)(x+y)=a-b=p$ $x+y>x-y>0\implies x+y=p$ and $x-y=1$ So $x=\frac{p+1}{2}$ and $y=\frac{p-1}{2}$ So $a=\left(\frac{p+1}{2}\right)^2$ and $b=\left(\frac{p-1}{2}\right)^2$, where $p$ is a prime number.