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Let $k/\mathbb{Q}$ be a number field, i.e. a finite field extension to $\mathbb{Q}$.  We recall that the ring of integers in k, denoted $\mathcal{O}_k$, is the ring

$\mathcal{O}_k := \{\alpha \in k : f(\alpha)=0 \text{ for some monic } f(t) \in \mathbb{Z}[t]\}.$

For $k= \mathbb{Q}$, the ring of integers is just the integers $\mathbb{Z}$, in which case we recall the  Fundamental Theorem of Arithmetic: that every integer $n \neq 0$ may be written as a finite product

$n=\pm 1\cdot p_1 \cdots p_k,$

in which the $\{p_i\}$ are prime and uniquely determined (up to permutation). Domains for which this holds are known in general as unique factorization domains (UFDs). For $k = \mathbb{Q}(\sqrt{d})$ — with $d \neq 0,1$ square-free — the ring of integers $\mathcal{O}_k$ will in general not be a UFD.  In fact, for $d <0$, the integers $\mathcal{O}_k$ have unique factorization only in the 9 cases

$d=-1, -2, -3, -7, -11, -19, -43, -67, -163.$

Far less is known in the case $d >0$ (in which case k is known as a real quadratic field), although an unproven conjecture dating back to Gauss suggests that there should be infinitely many real quadratic fields.  More recently, some heuristics stemming from Cohen suggests that the ring of integers in $\mathbb{Q}(\sqrt{d})$ should be a UFD with probability $\approx 0.75446$ as $d \to \infty$ on the square-free integers.

Here, we’ll focus on a more tractable variant of this problem:

Question: What can be said about the number $\tau(n)$ of distinct real quadratic fields $k=\mathbb{Q}(\sqrt{d})$ with $d\leq n$ for which $\mathcal{O}_k$ is not a UFD?

For a weak answer to the question above, we devote the rest of this article to the establishment of the following bound:

Theorem: As $n \to \infty$, we have

$\tau(n) \gg \sqrt{n}\log n,$

in which the implied constant is made effective (e.g. greater than $1/11$).