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)

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