$(a,a+p)=(a,p)$ so we have two cases: 1) if $(a,p)=p$ then $a=px^2$ and $b=py^2$ so we get $py^2-px^2=p$ which implies $a=0$ which is impossible 2) if $(a,p)=1$ then $a=x^2$,$b=y^2$ so since $y^2-x^2$ we get $a=\left(\frac{p-1}{2}\right)^2$ $b=\left(\frac{p+1}{2}\right)^2$ so $\frac{p^2+1}{2}=3 mod 5$ which implies $p=5$. Indeed we see that $(a,b)=(4,9)$ works

We have: $b-a=p$ so $b=a+p,$ for some prime $p.$ $2a+p \equiv 3 (\mod 10)$ $a(a+p)=k^{2},$ for some natural $k.$ So we have $a^{2}+ap=k^{2} \iff (2a+p)^{2}=4k^{2}+p^{2} \iff (2a+p-2k)(2a+p+2k)=p^{2},$ so $2a+p-2k=1$ and $2a+p+2k=p^{2},$ which implies $4a+2p=p^{2}+1,$ but $4a+2p \equiv 6 (\mod 10)$ so $p^{2}$ is divisible by 5, so the only solution is $p=5, a=4, b=9.$