Let $d\in\mathbb{Z}$ be an integer which is not a square (it does not have to be squarefree, though).

Question. Assume that $\vert d\vert\geq 3$ to avoid special cases. Is is true that $\pi=-1+\sqrt{d}$ is irreducible in $R=\mathbb{Z}[\sqrt{d}]$ ?

Remark. I can show this it is true in the following cases:

  • $d<0$

  • $d=1\pm p$, where $p$ is prime.

Apart from these cases, I have no idea whether $\pi$ might be irreducible in full generality or not when $d>0$. (I tried to use a CAS to produce examples and counterexamples, but I am really bad at programming so I didn't get very far.)

Addendum. I tried to determine first the values of $d$ for which $\pi$ is a prime element. This happens to be the case exactly when $d=1\pm p$, $p$ prime, so this does not give any new insight.

Update (February 21,2024). In his answer, Keith Conrad gives examples of integers $d\not\equiv 1 \ [4]$ for which $-1+\sqrt{d}$ is not irreducible.

When $d\equiv 1 \ [4]$, there are also such examples. For example, $(-1+\sqrt{41})=(7+\sqrt{41})(-6+\sqrt{41})$ is a non trivial factorization (the first factor has norm $8$ and the second one has norm $-5$).

Another example is $-1+\sqrt{57}=(7+\sqrt{57})(8-\sqrt{57})$.

  • $\begingroup$ I don't know about the general question, but I tried the first example you can't prove ($d=5$) and found $-1+\sqrt{5} = (7-3\sqrt{5})(2+\sqrt{5})$. Literally just by guessing - sorry that I don't have a more informative method! $\endgroup$ Feb 12 at 12:59
  • 2
    $\begingroup$ @preferred_anon the number $2+\sqrt{5}$ is a unit, with inverse $\sqrt{5}-2$, so that factorization does not show $-1+\sqrt{5}$ is reducible. In fact, $-1+\sqrt{5}$ is irreducible in $\mathbf Z[\sqrt{5}]$. $\endgroup$
    – KCd
    Feb 12 at 13:01

1 Answer 1


Suppose $\mathbf Z[\sqrt{d}]$ has unique factorization, so it is the ring of integers in $\mathbf Q(\sqrt{d})$ (because UFDs are integrally closed), which implies $d$ is squarefree.

In a UFD, prime and irreducible elements are the same thing. An element $\alpha$ is prime exactly when the ideal $(\alpha)$ is a prime ideal, which makes $(\alpha)$ a maximal ideal: the residue ring $\mathbf Z[\sqrt{d}]/(\alpha)$ is finite and a finite integral domain is a field. The size of $\mathbf Z[\sqrt{d}]/(\alpha)$ is $|{\rm N}(\alpha)|$ and a finite field has prime-power order, so when $\mathbf Z[\sqrt{d}]$ has unique factorization, a necessary condition that an element $\alpha$ in this ring be irreducible is that the absolute value of its norm is a prime power.

Since $|{\rm N}(-1+\sqrt{d})| = |1 - d| = |d-1|$, when $\mathbf Z[\sqrt{d}]$ has unique factorization and $|d-1|$ is not a prime power, $-1+\sqrt{d}$ is not irreducible.

Example. When $d = 7, 11, 19, 22, 23, 31, 43, 46$, and $47$, $\mathbf Z[\sqrt{d}]$ is the ring of integers in $\mathbf Q(\sqrt{d})$ and has class number $1$, so it is a PID and thus has unique factorization. In these cases, $|d-1| = d-1$ is not a prime power.

In the first case, $d = 7$, an explicit factorization of $-1+\sqrt{7}$ into two non-units is $(\sqrt{7}+3)(2\sqrt{7}-5)$, where the factors are not units since they have norm $2$ and $-3$, respectively.

In the second case, $-1+\sqrt{11} = (3+\sqrt{11})(7-2\sqrt{11})$, where the factors have norm $-2$ and $5$.

  • $\begingroup$ Thank you. Do you know what happens when $d\equiv 1 \ [4]$ ? (so $\mathbb{Z}[\sqrt{d}]$ is not a UFD anymore) $\endgroup$
    – GreginGre
    Feb 12 at 14:21
  • $\begingroup$ After reading over my post, have you tried looking at $d \equiv 1 \bmod 4$ yourself? $\endgroup$
    – KCd
    Feb 12 at 14:57
  • $\begingroup$ Well, if $d\equiv 1 [4]$, $\mathbb{Z}[\sqrt{d}]$ is not a UFD, and irreducible and prime elements are not the same. I know when $-1+\sqrt{d}$ is prime, but that's it...If there is something i can use in your answer in this case, i missed it , i'm sorry. Maybe I could work out the potential irreducible factors of $-1+\sqrt{d} $ in $\mathbb{Z}[\frac{-1+\sqrt{d}}{2}]$ and see if any of them lie in $\mathbb{Z}[\sqrt{d}]$. Is that what you have in mind ? $\endgroup$
    – GreginGre
    Feb 12 at 15:13
  • $\begingroup$ For what is worth, i think i can show that $2$ is always irreducible in $\mathbb{Z}[\frac{-1+\sqrt{d}}{2}]$, and that $\frac{-1+\sqrt{d}}{2}$ is prime if and only if $d=1+4p$, where $p$ is prime. In particular, this should be enough to show that if $d=1+4p$, then $-1+\sqrt{d}$ is irreducible in $\mathbb{Z}[\sqrt{d}]$ as soon as$\mathbb{Z}[\frac{-1+\sqrt{d}}{2}]$ is a UFD . However, i cannot find any counterexample yet if $\frac{d-1}{4}$ is composite, as solving a generalized Pell equation is not a thing I am comfortable with. $\endgroup$
    – GreginGre
    Feb 12 at 15:44

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