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# 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)