Show that if a Diophantine equation has a solution then both $x$ and $y$ must be odd

Question

Show that if the Diophantine equation $y^2=x^3+ 2$ has a solution, then $x$ and $y$ must both be odd.

How do I take into account the condition that $y^2=x^3+ 2$ has a solution? How do I take this into account to prove the question?

Answer 1

Re-writing that as $y^2 - x^3 = 2$, the only possibilities are that $y$ and $x$ are both even, or $y$ and $x$ are both odd.

If they are even, then write $y=2k$ and $x=2t$. Then:

$$4k^2-8t^3 = 2 \\ 2k^2 - 4t^3 = 1$$

Which is impossible.

Answer 2

Suppose one of $x$ or $y$ is even.

• First, show that then both $x$ and $y$ are even.

• Since $y$ is even, what is $y^3$ (mod 4)?

• Since $x$ is even, what is $x^2$ (mod 4)?

• And what is $x^2+2$ (mod 4)?

Do you see why this is a problem?