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)