Diophantine equation

sorry, i can't use latex....T.T

We may factorize the left-hand side using Sophie Germain to get $(x^2 + 2x +2)(x^2 - 2x + 2) = py^4$ . To exploit this factorization let's compute $\gcd(x^2 + 2x + 2, x^2 - 2x + 2):$

If $d | x^2 + 2x + 2, x^2 - 2x + 2 \implies d | 4x, 2x^2 + 4 \implies d | 4x^2, 2x^2 + 4 \implies d | -8 \implies d | 8$ . So $\gcd(x^2 + 2x + 2, x^2 - 2x + 2) | 8$ . We may now quickly examine two cases.

Case 1: $y$ is even : Then $y = 2y_0$ and thus $x^4 + 4 = 16 p y_0 ^4 \implies x \equiv 0 \mod 2 \implies x = 2x_0 \implies 16x_0^4 + 4 = 16py_0^4 \implies 4 \equiv 0 \mod 16$ . This is a contradiction so $y$ must always be odd.

Case 2: $p = 2$ . If that is the case then $x^4 + 4 = 2y^4 \implies x \equiv 0 \mod 2 \implies x = 2x_0 \implies 16x_0^4 + 4 = 2y^4 \implies 8x_0^4 + 2 = y^4 \implies y \equiv 0 \mod 2$ . This implies Case 1 which we know leads to a contradiction.

We now have that on the right-hand side of $(x^2 + 2x +2)(x^2 - 2x + 2) = py^4$ there are no even factors. This means that for each prime power on the right-hand side, all of it has to go to either of the two factors on the left. They cannot share factors (otherwise their gcd would not divide 8). Let's suppose that $y = p^{4k} y_0^4$ . Then $(x^2 + 2x +2)(x^2 - 2x + 2) = p^{4k+1}y_0^4$ .

Case 1: $x^2 + 2x + 2 = p^{4k+1} m^4, x^2 - 2x + 2 = n^4$ . Focusing on $x^2 - 2x + 2 = n^4$ , this implies that $x^2 - 2x + 2 - n^4 = 0$ . This equation must have an integral discriminant so let's compute it: $D = 2^2 - 4(1)(2 - n^4) = 4 - 8 + 4n^4 = (2n^2)^2 - 4$ . Using the substitution $a = 2n^2$ , this would imply a solution to the equation $a^2 - 4 = b^2 \implies a^2 - b^2 = 4$ . To solve this simply examine the inequality $a^2 - b^2 \leq (b+1)^2 - b^2 = 2b + 1$ . And $2b + 1 \leq 4 \iff b \leq 1$ . Quickly checking, the only solution is $a = 2, b = 0$ . So $2n^2 = 2 \implies n^2 = 1$ . Plugging this into the original polynomial equation we get $x^2 - 2x + 2 - 1 = 0 \implies (x-1)^2 = 0 \implies x = 1$ . Plugging this into the other polynomial equation, $5 = p^{4k + 1} m^4 \implies p=5, k=0, m= \pm 1$ . Thus we have found the solution $p=5, x = 1, y= \pm 1$ .

Case 2: $x^2 + 2x + 2 = n^4, x^2 - 2x + 2 = p^{4k + 1}m^4$ . It is easy to see that as the discriminants are the same, the previous case will repeat itself, only that now $x = -1$ .