Find all triples $latex (p, x, y) $ such that $latex p^x = y^4 + 4 $, where $latex p $ is a prime and $latex x, y $ are natural numbers.
Hint 1: p cannot be 2. For if p is 2 then $latex p^x $ is even which implies $latex y^4 + 4 $ is even or y is even. Suppose y = 2y'. Then $latex 2^x = (2y')^4 + 4 = 4(4y'^4 + 1) = 4 \times odd $. Now x cannot be 1 or 2 as $latex 2^x $ must be greater than 4. Hence $latex x \ge 3 $. Let x = x' + 3. Thus
$latex 2^{(x' + 3)} = 4 \times odd $
$latex 2^3 \times 2^{x'} = 4 \times odd $
$latex 8 \times 2^{x'} = 4 \times odd $
$latex 2 \times 2^x = odd $ which is impossible.
Hint 2: Suppose p is odd. Then $latex p^x = (y^2 + 2y + 2 ) (y^2 - 2y +2) $. Hence there can be two cases:
Case 1: $latex y^2 + 2y + 2 = p^x ; y^2 -2y + 2 = 1 $ (as $latex 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: $latex 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 $latex 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 $latex p^x = y^4 + 4 $ since p divides left side, p must divide right side and hence p must divided $latex 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)