$p=b\sqrt{\frac{a-8b}{a+8b}}\Leftrightarrow a=b\cdot \dfrac{8p^2+8b^2}{b^2-p^2}\in \mathbb{N}$ Case 1: $(b,p)=1$.Let $d=(p^2+b^2,b^2-p^2)$,then $d\mid 2(b^2,p^2)=2$ so $d=1,2$ If $2\not |b^2-p^2$ (i.e $d=1$) then we must have $b^2-p^2|b^2+p^2$,since $d=1$ we must have $b^2-p^2=1$,contradiction. So $2|b^2-p^2$ and $d=2$.Ser $b^2-p^2=2t,b^2+p^2=2e,(e,t)=1$ and $t\mid 8$ so $b^2-p^2|16\Leftrightarrow b^2-p^2=2,4,8,16$ $b^2-p^2=2\overset {b+p>b-p}{\Leftrightarrow}b-p=1,b+p=2$,contradiction $b^2-p^2=4\overset {b+p>b-p}{\Leftrightarrow} b-p=1,b+p=4$,contradiction $b^2-p^2=8\overset {b+p>b-p,b-p\equiv b+pmod 2}{\Leftrightarrow} b-p=2,b+p=4\Leftrightarrow b=3,p=1\not \in \mathbb{P}$,contradiction So $b^2-p^2=16\Leftrightarrow b-p=2,b+p=8\Leftrightarrow\boxed{ b=5,p=3\,\,\, and \,\,\,a=85}$ Case 2:$b=kp,k\in \mathbb{N}$ $a=\dfrac{8kp(p^2+k^2p^2)}{k^2p^2-p^2}\in \mathbb{N}\Leftrightarrow \dfrac{8kp(k^2+1)}{k^2-1}\in \mathbb{N}$ Since $(k,k^2-1)=1$ we must have $\dfrac{8p(k^2+1)}{k^2-1}\in \mathbb{N}$ If $2|k$ then $(k^2+1,k^2-1)=1$ and $(8,k^2-1)=1$ so $k^2-1=p\Leftrightarrow (k-1)(k+1)=p\Leftrightarrow k=2,p=3$ So $\boxed {p=3,b=2\cdot3=6,a=80}$ If $2\not |k $ then $(k^2+1,k^2-1)=2$.Let $k^2+1=2t,k^2-1=2e,(e,t)=1 $. We must have $\dfrac{8p\cdot2t}{2e}\in \mathbb{N}\Leftrightarrow k^2-1|16p$ So we have that $k^2-1=1,2,4,8,16,p,2p,4p,8p,16p$ Obviously $k^2-1\neq 1,2,4,16$ so $k^2-1=8,p,2p,4p,8p,16p$ If $p=2$ then $k^2-1=8$ so $k=3$. So $\boxed {(a,b,p)=(60,6,2)}$ If $p\neq 2$ since $8|k^2-1$ we must have $k^2-1=8p,16p$ If $k^2-1=8p$ since $(k-1,k+1)|2$ easly we have $p=3,k=5$ so $\boxed{(a,b,p)=(130,15,3)}$ If $k^2-1=16p$ easly we have $\boxed{(a,b,p)=(369,45,5),(175,21,3)}$ So $(a,b,p)=(85,5,3),(80,6,3),(60,6,2),(130,15,3),(369,45,5),(175,21,3)$

And $\forall p \in \mathbb{P},\;a=30p,b=3p$