INMO-2008

We have $p^n \; = \; y^4 + 4 \; = \; (y^2 + 2)^2 - 4y^2 \; = \; (y^2 + 2y + 2)(y^2 - 2y + 2)$ and hence $y^2 + 2y + 2 = p^a$ . $y^2 - 2y + 2 = p^b$ for some $0 \le b \le a$ with $a+b = n$ . But then $y^2 - 2y + 2$ divides $y^2 + 2y + 2$ , and hence $y^2 - 2y + 2$ divides $4y$ . But $y^2 - 2y + 2 - 4y \; = \; (y - 3)^2 - 7 \; > \; 0 \hspace{2cm} y \ge 6$ so we deduce that $1 \le y \le 5$ . It is easy to check that $y^4 + 4$ is only a prime power for $1 \le y \le 5$ when $y=1$ . Thus we have $y=1$ , $p=5$ . $n=1$ , so that $y+p+n = \boxed{7}$ .