We all love primes!

Given : $\dfrac{a^2(b-a)}{(b+a)} = p^2$

$\implies (b-a) = \dfrac{2ap^2}{a^2 - p^2}$

Case 1: $\gcd(a,p) = 1$

We have,

$(b-a) = \dfrac{2ap^2}{(a+p)(a-p)}$

$\gcd(a+p,p^2) | \gcd(a+p,p) = 1 \ \implies \gcd(a+p,p^2) =1$

Similarly,

$\gcd(a-p,p^2) = \gcd(a+p,a) = \gcd(a-p,a) = 1$

$\implies \gcd(a^2-p^2,ap^2) = 1$

Since $\dfrac{2ap^2}{a^2-p^2} \in \mathbb{Z^{+}}$ , we have,

$a^2-p^2|2$

$\implies a^2 - p^2 = 1 \ \text{or} \ 2 \implies \mathrm{No \ Solution}$

Case 2: $\gcd(a,p) = p$

Let $a = \lambda p \ ; \ \gcd(\lambda , p) = 1$

$\implies b - \lambda p = \dfrac{2\lambda p}{\lambda ^2 -1}$

Since, $\dfrac{2\lambda p}{\lambda ^2 -1} \in \mathbb{Z^+}$ and $\gcd (\lambda , \lambda ^2 - 1) = 1$ , we have,

$\lambda ^2 - 1 | (2p)$

$(\lambda+1)(\lambda-1)|(2p)$

Note that $\lambda +1 > \lambda -1$

$\implies \lambda + 1 = p \ ; \ \lambda -1 =2 \implies \mathrm{No \ Solution}$

or,

$\lambda + 1 = 2p \ ; \ \lambda -1 =1 \implies \mathrm{No \ Solution}$

or,

$\lambda + 1 = p \ ; \ \lambda -1 =1 \implies \lambda =2 \ ; \ p=3$

Substituting back, we have,

$a =6 \ ; \ b=10 \ ; \ p=3$

$\implies a+b+c =\boxed{19}$

Rearranging the given equation we get: $a^2(b-a)=(b+a)p^2$ $\implies$ $a^2b-a^3=p^2b+p^2a$ $\implies$ $a^2b-p^2b=a^3+p^2a$ $\implies$ $b(a^2-p^2)=a(a^2+p^2)$ (1)

Now if a prime $k$ divides $a^2-p^2$ and $a^2+p^2$ then it also divides $(a^2+p^2)-(a^2-p^2)=2p^2$ so $k=p$ or $k=2$

And if a prime $k$ divides $a^2-p^2$ but not $a^2+p^2$ then $k$ divides $a$ by equation 1. But then $k$ divides $a^2-p^2$ and $k$ divides $a$ so $k$ divides $p^2$ . Thus $k=p$ .

So the only prime factors of $a^2-p^2$ are $2$ and $p$

Now if 4 $\mid$ $a^2-p^2$ and 4 $\mid$ $a^2+p^2$ then 4 $\mid$ $2p^2$ so $p=2$ . Then the only prime factor of $a^2-p^2=a^2-4$ is 2. So both $(a-2)$ and $(a+2)$ are powers of 2 which means $a=6$ . Substituting $p=2$ and $a=6$ in the original equation gives $b=7.5$ which is not an integer.

Now if 4 $\mid$ $a^2-p^2$ and 4 $\not\mid$ $a^2+p^2$ then 2 $\mid$ $a$ (by equation 1) so 4 $\mid$ $a^2$ and thus 4 $\mid$ $p^2$ $\implies$ 2 $\mid$ $p$ $\implies$ $2=p$ which we showed has no solution. Then 4 $\not\mid$ $a^2-p^2$

Now if $p^4$ $\mid$ $a^2-p^2$ then $p^2$ $\mid$ $a$ (as $(a^2+p^2)-(a^2-p^2)=2p^2$ and $p\neq2$ so it follows from equation 1) and thus $p^4$ $\mid$ $a^2$ . But then $p^4$ $\mid$ $a^2-p^2$ and $p^4$ $\mid$ $a^2$ so $p^4$ $\mid$ $p^2$ which is not possible. So $p^4$ $\not\mid$ $a^2-p^2$

Thus $a^2-p^2=2^kp^n$ where $k=0,1$ $n=0,1,2,3$

• $a^2-p^2=2$ $\implies$ $(a+p)(a-p)=2$ but $a+p>2$ so no solution
• $a^2-p^2=2p$ $\implies$ $(a+p)(a-p)=2p$ $\implies$ $(a+p)=p$ and $(a-p)=2$ $\implies$ no solution

or $(a+p)=2p$ and $(a-p)=1$ $\implies$ no solution

• $a^2-p^2=2p^2$ $\implies$ $(a+p)(a-p)=2p^2$ $\implies$ $(a+p)=2p$ and $(a-p)=p$ $\implies$ no solution

or $(a+p)=2p^2$ and $(a-p)=1$ $\implies$ no integer solution

or $(a+p)=p^2$ and $(a-p)=2$ $\implies$ no integer solution

• $a^2-p^2=2p^3$ $\implies$ $(a+p)(a-p)=2p^3$ $\implies$ $(a+p)=2p^3$ and $(a-p)=1$ $\implies$ no integer solution

or $(a+p)=2p^2$ and $(a-p)=p$ $\implies$ no positive integer solutions

or $(a+p)=p^3$ and $(a-p)=2$ $\implies$ no integer solution

or $(a+p)=p^2$ and $(a-p)=2p$ $\implies$ solution for $p$ is not prime

• $a^2-p^2=1$ $\implies$ $(a+p)(a-p)=1$ so no solution
• $a^2-p^2=p$ $\implies$ $(a+p)(a-p)=p$ $\implies$ $(a+p)=p$ and $(a-p)=1$ $\implies$ no solution
• $a^2-p^2=p^2$ $\implies$ $(a+p)(a-p)=p^2$ $\implies$ $(a+p)=p^2$ and $(a-p)=1$ $\implies$ no integer solution
• $a^2-p^2=p^3$ $\implies$ $(a+p)(a-p)=p^3$ $\implies$ $(a+p)=p^3$ and $(a-p)=1$ $\implies$ no positive integer solutions

or $(a+p)=p^2$ and $(a-p)=p$ which yields $p=3 , a=6$ .

So $p=3 , a=6$ are the only possible values for $a$ and $p$ .

Substituting these values in the original expression we find $b=10$ so $(a,b,p)=(6,10,3)$ is the only solution and $a+b+p=6+10+3=19$