By arranging it $$ c=\frac{ab-a-b}{b} \in \mathbb{Z^+} \Rightarrow a=bk$$$$ k \ge 1 k\in \mathbb{Z^+}$$By arranging again, $$c+1=p^2 \Rightarrow k(b-1)=p^2 \Rightarrow k \in \{1, p, p^2\}$$We will prove either $a+b$ or $ab$ is perfect squre. It is obvious that $k$ is not a perfect square. If it was then, $ab=b^2k$ would be a perfect square. Then we have $k \in \{1, p\}$ $k=1$ Then we have $a=b \Rightarrow ab=a^2$. Contradiction. $k=p$ We have $a=bp$ and $a+b=b(a-(p^2-1))$ If we solve these $a=p^2+p$ $b=p+1$ $\Rightarrow a+b=p^2+2p+1=(p+1)^2$. Contradiction