$\frac{1}{p}=\frac{1}{a^2}+\frac{1}{b^2} <=> a^2b^2=p(a^2+b^2)$. Then $p|a$ or $p|b$. WLOG $b=xp$, then we have: $x^2p^2-a^2x^2p+a^2=0$. $\Delta=a^4x^4-4a^2x^2=a^2x^2(a^2x^2-4)$, then $a^2x^2-4=k^2 <=> (ax+k)(ax-k)=4$. By taking cases we have $ax=2$, then by taking cases again we have $a=b=p=2$.

parmenides51 wrote: Find all triplets $(a,b,p)$ where $a,b$ are positive integers and $p$ is a prime number satisfying: $\frac{1}{p}=\frac{1}{a^2}+\frac{1}{b^2}$ Equivalent to $(p-a^2)(p-b^2)=p^2$ So, $p-a^2=\pm 1, \pm p,\pm p^2$ $+p^2, +p$ not possible since $p-a^2<p$ By symmetry, $+1$ is also not possible since $p-a^2=+1\implies p-b^2=+p^2$ So, $p-a^2=-1,-p,-p^2$ We have to check 2 cases- $p-a^2=p-b^2=-p$, $p-a^2=-1, p-b^2=-p^2$ The former gives $a=b=p=2$ The latter is not possible since $1=\nu_p(LHS)\neq \nu_p(RHS)=2$

The equation is written: $a^{2} b^{2}=pa^{2}+pb^{2}$. Let $gcd.(a,b)=d \Rightarrow a=dm, b=dn, gcd.(m,n)=1$, for $m,n$ positive integers. The equation is now written: $d^{4} m^{2} n^{2}=pd^{2} m^{2}+pd^{2} n^{2} \Rightarrow d^{2} m^{2} n^{2}=pm^{2}+pn^{2} \Rightarrow m^{2}|pn^{2}$, and since $gcd.(m,n)=1$, we take that $m^{2}|p \Rightarrow m^{2}=1,p \Rightarrow m^{2}=1 \Rightarrow m=1$. Similarly we take that $n=1$. We now have: $d^{2}=2p \Rightarrow 2|d^{2} \Rightarrow 2|d \Rightarrow d=2d_{1} \Rightarrow 4d_{1}^{2}=2p \Rightarrow 2d_{1}^{2}=p \Rightarrow$ $\Rightarrow 2|p \Rightarrow p=2 \Rightarrow d_{1}^{2}=1 \Rightarrow d_{1}=1 \Rightarrow d=2 \Rightarrow a=b=2$. Hence $(a,b,p)=(2,2,2)$.