Given $a, b$ coprime integers, show that any factor of $a^2 - 2b^2$ is of the form $c^2 - 2d^2$

Question

This is an exercise from Fermat's Last Theorem: A Genetic Introduction to Algebraic Number Theory.

I imagine the relevant ring here is $Z[\sqrt{2}]$, i.e., we can factor $a^2 - 2b^2$ into $(a + b\sqrt{2})(a - b\sqrt{2})$. This ring is also a Euclidean domain, hence a UFD also. I don't have too much experience, and I'm not really sure where to go from here. Any advice? I guess I want to show any integer factor of $a^2 - 2b^2$ can be written as $(c + d\sqrt{2})(c - d\sqrt{2})$ but I don't know how to get there with what I have.

Answer 1

As written, this is false! For example $3\mid 3^2 -2(3^2)$. However, it is true if $a^2-2b^2$ is squarefree.

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First, observe that if $m = a^2-2b^2$ and $n = c^2-2d^2$, then $$mn= (ac-2bd)^2-2(ad-bc)^2$$ is also of that form. In particular, it is sufficient to show that every prime factor is of that form.

You're right that we want to work in $\mathbb Z[\sqrt 2]$.

It's easy to check that $2$ is of the required form. If $p$ is an odd prime, then we can rephrase the question as follows:

If $p\mid a^2-2b^2$, then $p$ is the norm of some element of $\mathbb Z[\sqrt2]$.

For odd primes $p$, this is equivalent to asking for $p$ to split completely in $\mathbb Z[\sqrt 2]$. So rephrasing again, the question is:

If $p$ is an odd prime and there exist $a,b$ with $a^2-2b^2 \equiv 0 \pmod p$, then $p$ splits completely in $\mathbb Z[\sqrt 2]$.

Can you finish from here? You'll need to use the fact that $a^2-2b^2$ is squarefree.