I'm studying Pell's equations of the form $x^2-dy^2=1$ for $d$ a square-free natural number and $x,y$ integers. In particular, I want to show that such an equation always has a solution.

I see that this is equivalent to finding units with positive norm in $\mathbb{Z}[\sqrt{d}]$. I know that there is always a fundamental unit $\epsilon$ which generates the multiplicative group of units, and $\epsilon$ is unique if we take it as the smallest unit with $\epsilon>1$. Clearly any even power of $\epsilon$ will have positive norm since $\mathbb{Q}(\sqrt{d})$ is real. Hence any solution of Pell's equation would be some power of $\epsilon$.

But why does there always exist an $n\in\mathbb{N}$ such that $\epsilon^n\in\mathbb{Z}[\sqrt{d}]$? Clearly this is the case when $\epsilon\in\mathbb{Z}[\sqrt{d}]$, but of course if $d\equiv 1$ (mod $4$) then it may be the case that $\epsilon\in\mathbb{Z}[\frac{1+\sqrt{d}}{2}]$.

Computation of specific examples seems to indicate that the cube of any fundamental unit will be in $\mathbb{Z}[\sqrt{d}]$. I tried proving this directly, but the amount of calculation makes me think that there is a more elegant method. I see an argument in the final question of this document, but by the heuristic law of "conservation of difficulty", I think that once one uses Dirichlet's Diophantine approximation theorem to classify the units of real quadratic number rings, there must be a method that follows easily.

Looking forward to hearing your thoughts!

  • $\begingroup$ Also, I’m not sure what you’re calling “conservation of difficulty”? Does it mean that a simple enough problem must have a simple solution? Or that a deep result must have a difficult proof? $\endgroup$
    – Aphelli
    Feb 9, 2021 at 17:01
  • $\begingroup$ @Mindlack It's something that one of my instructors said years ago: different solutions to the same problem usually have the same amount of difficulty, even if the difficulty appears in an earlier form like the development of the theory. His example was that quadratic reciprocity has an "elementary" proof, but it's quite technical. On the other hand it has a more elegant proof using ramification theory, but it's built on a highly non-trivial theory. Anyway, it's mostly a heuristic to say that with Dirichlet's approximation theorem, I felt like this solution should be straightforward. $\endgroup$
    – Ducky
    Feb 9, 2021 at 17:05
  • $\begingroup$ I don't think that it is a good idea to try re-inventing the wheel. See chapter 1 only of this, and perhaps as an interesting supplement, this. $\endgroup$ Feb 9, 2021 at 17:05
  • 1
    $\begingroup$ Can't we simply apply the unit theorem directly to $\Bbb Z[\sqrt{d}]$, regardless of what $d$ is mod $4$? $\endgroup$
    – user873979
    Feb 9, 2021 at 17:12

2 Answers 2


If $d-1$ is divisible by $4$, an integral element not in $\mathbb{Z}[\sqrt{d}]$ is some $\omega=\frac{a+b\sqrt{d}}{2}$ with $a,b$ odd. Its norm is the integer $N=\frac{a^2-db^2}{4}$.

Then $\omega^2=-N+a\omega$, and $\omega^3=-N\omega+a\omega^2=-aN+(a^2-N)\omega$.

If $N$ is odd, then indeed $\omega^3 \in \mathbb{Z}[\sqrt{d}]$.

Note that for instance $\frac{1+\sqrt{5}}{2}$ is a unit in $\mathcal{O}_{\mathbb{Q}(\sqrt{5})}$ of norm $-1$, and its square isn’t in $\mathbb{Z}[\sqrt{5}]$.


Question $3$ of the document you link to is not the last question of the document, so don't describe it as the "final question".

When $d \equiv 1 \bmod 4$, the cube of each unit in $\mathbf Z[(1+\sqrt{d})/2]$ is in $\mathbf Z[\sqrt{d}]$. To check the coefficients of $1$ and $\sqrt{d}$ in the cube of a unit are both integers, it is somewhat important to use the fact that you're using a unit (the norm is $\pm 1$). The cube of a nonunit in $\mathbf Z[(1+\sqrt{d})/2]$ might not be in $\mathbf Z[\sqrt{d}]$, depending on the value of $d$. For example, $(1+\sqrt{17})/2$ is an integer in $\mathbf Q(\sqrt{17})$ and $((1+\sqrt{17})/2)^3 = (13 + 5\sqrt{17})/2 \not\in \mathbf Z[\sqrt{17}]$. Note that it is convenient to realize that the integers of $\mathbf Q(\sqrt{d})$ for squarefree $d \equiv 1 \bmod 4$ can be described as $\{(a+b\sqrt{d})/2 : a, b \in \mathbf Z, a \equiv b \bmod 2\}$.

Another suggestion: rather than bootstrap knowledge of the unit group of the full ring of integers to tell you the structure of the unit group of $\mathbf Z[\sqrt{d}]$ when $d \equiv 1 \bmod 4$, you could run through the proof of the unit theorem to check it works for every order (subring of finite index in the full ring of integers) of a number field. The proof of the unit theorem does not use prime ideal factorization, for example. There is nothing about the proof of the unit theorem that does not carry over verbatim to orders. In fact, accounts of the solvability of Pell's equation in elementary number theory books (not books on algebraic number theory) don't even impose conditions like "$d$ is squarefree". The standard description of the solvability of Pell's equation $x^2 - dy^2 = 1$ is that it has an integral solution $(x,y)$ with $y \not= 0$ as long as $d$ is not a perfect square. It is completely irrelevant that $d$ may or may not be squarefree in proofs of solvability of Pell's equation.

If you want to see abstract existence of an $n \geq 1$ such that $\varepsilon^n \in \mathcal O$ when $\mathcal O$ has finite index in the full ring of integers $\mathcal O_K$ of a number field $K$ and $\varepsilon \in \mathcal O_K^\times$, think about the integral powers of $\varepsilon$ lying in the finite set $\mathcal O_K/\mathcal O$.


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