1
$\begingroup$

In Ian Stewart's book on Algebraic Number Theory, he gives the following proof that the solutions to $y^2 + 2 = x^3$ is $x=3, y = \pm 5$:

enter image description here

I'm confused about the step "since $c^2 + 2d^2 | 4y^2$ and $c^2 + 2d^2 | 8$ then $c^2 + 2d^2 | 4$". How is this implied?

Also, he lists several sets of valid solutions, like $c = \pm 2, d = 0$, but somehow none of these correspond to any "proper factors" of the two: why?

  • 0
    Hint for the first question:$y$ is an odd number.2019-05-15

1 Answers 1

4

On your first question: At the beginning, it was stated that $y$ is odd. Thus the greatest common divisor of $4y^2$ and $8$ is $4$. Since $c^2+2d^2$ divides both, it divides their greatest common divisor.

On your second question: The term "proper factor" here refers to a factor that isn't a unit, i.e. a factor that's part of the prime factorization of $y+\sqrt{-2}$, which is only unique up to units. The first option yields two units $\pm1$, which divide $y+\sqrt{-2}$ but aren't proper factors. The other options yield $\pm\sqrt{-2}$ and $\pm2$, none of which divide $y+\sqrt{-2}$.