Q. Find the integer solutions to \[ y^2 = x^3 – 2 \]

We want to find all integer solutions to the classic Mordell equation. We will solve this using factorization in the ring \( \mathbb{Z}[\sqrt{-2}] \), which acts as a Unique Factorization Domain.

\[ y^2 = x^3 – 2 \]

First, rewrite the equation as \( y^2 + 2 = x^3 \). We factor the left side over \( \mathbb{Z}[\sqrt{-2}] \) to get \( (y + \sqrt{-2})(y – \sqrt{-2}) = x^3 \).

We must determine the parity of \( y \). If \( y \) were an even integer, then \( y^2 \equiv 0 \pmod{4} \), meaning \( x^3 = y^2 + 2 \equiv 2 \pmod{4} \). However, no perfect integer cube can be congruent to \( 2 \pmod{4} \), as cubes are strictly \( 0, 1, \) or \( 3 \pmod{4} \). Therefore, \( y \) must be an odd integer. Consequently, \( x^3 \) is odd, making \( x \) odd as well.

Next, we check for common divisors of the factors \( (y + \sqrt{-2}) \) and \( (y – \sqrt{-2}) \). Any common divisor \( d \) must divide their geometric difference, which is \( 2\sqrt{-2} \). The norm of \( 2\sqrt{-2} \) is \( 8 \), while the norm of \( y \pm \sqrt{-2} \) is \( x^3 \), which is odd. Because they share no common prime factors, \( (y + \sqrt{-2}) \) and \( (y – \sqrt{-2}) \) are coprime.

Because their product is a perfect cube and they are coprime in a Unique Factorization Domain, each individual factor must itself be a perfect cube multiplied by a unit. The only units in \( \mathbb{Z}[\sqrt{-2}] \) are \( 1 \) and \( -1 \), both of which are themselves perfect cubes. Therefore, we can confidently write \( y + \sqrt{-2} = (a + b\sqrt{-2})^3 \) for some integers \( a \) and \( b \).

Expanding the right side using the binomial theorem gives \( a^3 + 3a^2 b \sqrt{-2} + 3a(b\sqrt{-2})^2 + (b\sqrt{-2})^3 \), which algebraically simplifies to \( (a^3 – 6 a b^2) + (3 a^2 b – 2 b^3)\sqrt{-2} \).

By equating the imaginary parts of both sides, we get \( 1 = 3 a^2 b – 2 b^3 \), which factors as \( 1 = b(3 a^2 – 2 b^2) \). Since \( a \) and \( b \) are integers, \( b \) must be a factor of \( 1 \). This leaves exactly two possibilities: \( b = 1 \) or \( b = -1 \).

If \( b = -1 \), the equation becomes \( 1 = -1(3 a^2 – 2) = 2 – 3 a^2 \), implying \( 3 a^2 = 1 \), which yields no integer solution for \( a \).

If \( b = 1 \), the equation becomes \( 1 = 1(3 a^2 – 2) \), meaning \( 3 a^2 = 3 \), which cleanly simplifies to \( a^2 = 1 \). This gives two valid roots: \( a = 1 \) or \( a = -1 \).

Now, we equate the real parts, yielding \( y = a^3 – 6 a b^2 = a(a^2 – 6 b^2) \). Substituting \( b = 1 \) and \( a = 1 \), we get \( y = 1(1 – 6) = -5 \). Substituting \( b = 1 \) and \( a = -1 \), we get \( y = -1(1 – 6) = 5 \). Thus, \( y = 5 \) or \( y = -5 \).

Substitute \( y = \pm 5 \) back into the original equation to find \( x \). This gives \( 25 = x^3 – 2 \), which implies \( x^3 = 27 \), leading directly to \( x = 3 \). The only valid integer solutions \( (x, y) \) are \( (3, 5) \) and \( (3, -5) \).