# INMO 2008 Problem 2

Find all triples $(p, x, y)$ such that $p^x = y^4 + 4$, where $p$ is a prime and $x, y$ are natural numbers.

Hint 1: p cannot be 2. For if p is 2 then $p^x$ is even which implies $y^4 + 4$ is even or y is even. Suppose y = 2y'. Then $2^x = (2y')^4 + 4 = 4(4y'^4 + 1) = 4 \times odd$. Now x cannot be 1 or 2 as $2^x$ must be greater than 4. Hence $x \ge 3$. Let x = x' + 3. Thus

$2^{(x' + 3)} = 4 \times odd$

$2^3 \times 2^{x'} = 4 \times odd$

$8 \times 2^{x'} = 4 \times odd$

$2 \times 2^x = odd$ which is impossible.

Hint 2: Suppose p is odd. Then $p^x = (y^2 + 2y + 2 ) (y^2 - 2y +2)$. Hence there can be two cases:

Case 1: $y^2 + 2y + 2 = p^x ; y^2 -2y + 2 = 1$ (as $y^2 - 2y + 2 < y^2 + 2y + 2$ ). Solving the second equation for y gives us solution y = 1 , x= 1 and p =5.

Case 2: $y^2 + 2y + 2 = p^m ; y^2 -2y + 2 = p^n$ where x = m +n and m > n . Subtracting the second equation from the first we get $4y = p^n ( p^ (m-n) + 1 )$. This implies p divides y as p does not divide 4 (as p is not equal to two). But considering the equation $p^x = y^4 + 4$ since p divides left side, p must divide right side and hence p must divided $y^4 + 4$. If p divides y then p will also divide 4 which cannot be.

Hence there are no other solution. Only solution is (5, 1, 1)

This site uses Akismet to reduce spam. Learn how your comment data is processed.

### Cheenta. Passion for Mathematics

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.