I'm pretty sure this is a very basic thing, but my background is in physics and I have never previously done any number theory.

We have as a theorem that for an algebraic number field $K$, $\alpha \in K$ is an algebraic integer if and only if its minimal polynomial in $\mathbb{Q}$ has coefficients in $\mathbb{Z}$. The minimum polynomial for $\alpha = a + b \sqrt{d}$ in the field $\mathbb{Q}[\sqrt{d}]$ is then:

$$ (X - (a + b \sqrt{d}))(X - (a - b\sqrt{d})) = X^{2} - 2aX + (a^{2} - b^{2}d) $$

Therefore $\alpha$ is an algebraic integer $\iff 2a \in \mathbb{Z}, a^{2}-b^{2}d \in \mathbb{Z}$.

Fine, but why is this claim true?

$$ d = 2,3 \; \bmod \; 4 \implies \mathscr{O}_{\mathbb{Q}[\sqrt{d}]} = \mathbb{Z}[\sqrt{d}] $$ $$ d= 1\; \bmod \; 4 \implies \mathscr{O}_{\mathbb{Q}[\sqrt{d}]} = \mathbb{Z} \bigg[\frac{1+\sqrt{d}}{2}\bigg]$$

Is this somehow related to the elementary theorem about when the sum of two squares is an integer?

  • 1
    $\begingroup$ You mean $\alpha$ an algebraic integer $\iff \ldots$. $\endgroup$ May 23, 2019 at 0:03
  • $\begingroup$ Yes, corrected thanks $\endgroup$ May 23, 2019 at 0:05
  • 2
    $\begingroup$ Note $a^2\equiv0$ or $1\pmod4$ and same with $b^2$; cf. this question $\endgroup$ May 23, 2019 at 0:06

2 Answers 2


You got $n=2a \in\Bbb Z$ and $a^2-db^2\in \Bbb Z.$ These imply $m=2b\in\mathbb Z$ and $n^2-dm^2\in\Bbb Z.$ $n^2$ and $m^2\equiv 0$ or $1\pmod4$. If $n^2\equiv dm^2\pmod4$ and $d\equiv2 $ or $3\pmod4$, the only solution is $n^2\equiv m^2\equiv0\pmod4$ so $n\equiv m\equiv 0\pmod2$ so $a, b \in\Bbb Z$. If $d\equiv1\pmod4$ there is also the solution $n^2\equiv m^2\equiv1\pmod4$, in which case $a$ and $b$ are half odd integers.

  • $\begingroup$ Note: the fact that $0^2\equiv2^2\equiv0 $ and $1^2\equiv3^2\equiv1\pmod4$ also means that if $p\equiv3\pmod4$ then $p$ cannot be expressed as a sum of two squares $\endgroup$ May 23, 2019 at 14:55

Yes, it is related to that result. If $a$ and $b$ are both rational numbers (not necessarily integers), and $d$ is an integer, then $a^2 - db^2$ is sure to be rational, but it may or may not be an integer.

If $a$ and $b$ are both halves of odd integers (call them $\alpha$ and $\beta$), and $d \equiv 1 \pmod 4$, it will happen that $$a^2 = \left(\frac{\alpha}{2}\right)^2 = \frac{\alpha^2}{4}$$ with $\alpha^2 \equiv 1 \pmod 4$, and likewise $$b^2 = \left(\frac{\beta}{2}\right)^2 = \frac{\beta^2}{4}$$ with $\beta^2 \equiv 1 \pmod 4$. Since $d \equiv 1 \pmod 4$ as well, we then have $$a^2 - db^2 = \frac{\alpha^2}{4} - \frac{d \beta^2}{4},$$ from which it obviously follows that $\alpha^2 - d \beta^2$ is a multiple of 4.

It often helps to work these things out with concrete examples. Since you mention a physics background, I'm going to assume you're well-versed in the arithmetic of complex numbers.

Try $d = -3$. Given the number $$\frac{-11}{2} + \frac{7 \sqrt{-3}}{2},$$ we readily see that the trace (the $2a$) is $-11$ and the norm (the $a^2 - db^2$) is 67. The relevant polynomial is then $x^2 + 11x + 67$.

Now try $d = -5$ and the number $$\frac{-11}{2} + \frac{7 \sqrt{-5}}{2}.$$ The trace is indeed $-11$ for this number as well, but the norm is... $$\frac{121}{4} + \frac{5 \times 49}{4} = \frac{183}{2},$$ and the polynomial is $2x^2 + 22x + 183$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.