Let $m$ be a squarefree integer $\neq 1$ . All numbers of the quadratic field $\mathbb{Q}(\sqrt{m})$ may be written in the form

(1)

(2)

(3)

Naturally, $p$ and $q$ are integers always when $l = 1$ . We suppose now that $l > 1$ . The latter of the equations (3) says that $q$ can be integer only when $$(\gcd(j,\,l))^2 = \gcd(j^2,\,l^2) \mid k^2m$$ (see divisibility in rings). Because $\gcd(j,\,k,\,l) = 1$ , we have by Euclid's lemma that $\gcd(j,\,l) \mid m$ . Since $m$ is squarefree, we infer that

(4)

So, by the latter of the equations (3), $4 \mid j^2-k^2m$ , i.e.

(5)

(6)

We have now obtained the following result:

• When $m \not\equiv 1 \pmod{4}$ , the integers of the field $\mathbb{Q}(\sqrt{m})$ are $$a+b\sqrt{m}$$ where $a,\,b$ are arbitrary rational integers;
• when $m \equiv 1 \pmod{4}$ , in addition to the numbers $a+b\sqrt{m}$ , also the numbers $$\frac{j+k\sqrt{m}}{2},$$ with $j,\,k$ arbitrary odd integers, are integers of the field.

Then, it may be easily inferred the

Theorem. If we denote

then any integer of the quadratic field $\mathbb{Q}(\sqrt{m})$ may be expressed in the form $$a+b\omega,$$ where $a$ and $b$ are uniquely determined rational integers. Conversely, every number of this form is an integer of the field. One says that 1 and $\omega$ form an integral basis of the field.

Bibliography

1

K. V¨AISÄLÄ: Lukuteorian ja korkeamman algebran alkeet. Tiedekirjasto No. 17.    Kustannusosakeyhtiö Otava, Helsinki (1950).