I was reading a book on number theory and it was talking about the pell equation. It finds a trivial solution and then begins raising it to different powers which generated other solutions. The book did not prove that this would yield every integer solution so I was wondering if this was somehow an obvious conclusion despite my not seeing it. the particular form of the pell equation was


any help would be appreciated.

  • $\begingroup$ 'Pell" with a capital "P" $\endgroup$
    – Jean Marie
    Oct 30, 2019 at 22:48

2 Answers 2


There is a general result, which sees this case as a very special case. It is Dirichlet’s unit theorem (see https://en.m.wikipedia.org/wiki/Dirichlet%27s_unit_theorem ).

Here, a simpler proof may be attempted.

Let $U$ be the set of real numbers of the form $x+y\sqrt{2}$ such that $x$ and $y$ are integers and $x^2-2y^2=1$. Note that $U$ is stable under product and $1,-1 \in u$ and if $z \in U$, then $z^{-1} \in U$ too.

Let $z=x+y\sqrt{2} \in U$ be such that $z > 1$. Then $\overline{z}=x-y\sqrt{2}=z^{-1} > 0$ so $x > |y|\sqrt{2}$, hence $x > 0$. Now, since $z > 1$, $z > \overline{z}$ so $y > 0$.

As a consequence, let $z=x+y\sqrt{2}, z’=x’+y’\sqrt{2} \in U$ be greater than $1$, assume $x < x’$. Then, $2y^2=x^2-1 < x’^2-1=2y’^2$, therefore $y < y’$ and $z < z’$. Similarly, if $x > x’$, $z > z’$, so $z < z’ \Leftrightarrow x < x’$. In other words, $z \in U_{>1} \longmapsto x \in \mathbb{N}$ is increasing.

As a consequence, $U_{>0}$ is a subgroup of $\mathbb{R}_{>0}$ that has no sequence converging to $1$, so it is a $\{\alpha^n,\, n \in \mathbb{Z}\}$ for some $\alpha \in U$, $\alpha >0$.

Note that the proof holds for every Pell-Fermat equation $x^2-Dy^2=1$.


It isn't so obvious, and depends on the fact that $(1-\sqrt 2)(1+\sqrt 2)=-1$ so that $1+\sqrt 2$ and $1-\sqrt 2$ are units in $\mathbb Z(\sqrt 2)$.

Now suppose that $x^2-2y^2=1$ then $(x+y\sqrt 2)(x-y\sqrt 2)(1+\sqrt 2)(1-\sqrt 2)=-1$

We can pair the terms in different ways to get new solutions to $x^2-2y^2=\pm 1$

So if take the positive signs together we can derive $X=x+2y, Y=x+y$ and we find $X^2-2Y^2=x^2+4xy+4y^2-2x^2-4xy-2y^2=-x^2+2y^2=-1$

If we mixture signs we get $X=x-2y, Y=x-y$ and $X^2-2Y^2=x^2-4xy+4y^2-2x^2+4xy-2y^2=-x^2+2y^2=-1$

And obviously if we start with $x^2-2y^2=-1$ we get $X^2-2Y^2=1$ in each case.

Now it is clear that if $(x,y)$ is a solution to the $\pm1$ case so is $(|x|, |y|)$. If we steer clear of the trivial solution $x=\pm 1, y=0$ we can assume that $x$ and $y$ are both positive.

If $x=y$ then $x=y=1$. Otherwise $x^2=2y^2\pm 1$ and we have $2y\gt x\gt y$ and $0\lt 2y-x \lt x$. Whence if $x\gt 1$ we can "go down" to another solution with a smaller positive $x$. So we must end with $x=1$ and the solution $x=y=1$.

Since rules for finding new solutions have been derived from the original unit with $x=y=1$ by taking powers, all solutions are in fact of this form. The solutions with $+1$ correspond to the even powers.

[There are bits above where you should check the argument]

With quadratics there is at most one fundamental unit of this kind, so once you have found it you get all the solutions (for finding it see Pell Equations and Continued Fractions).


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