# Algebraic proof of non-trivial solution to the Pell's equation

Let $d$ be a square-free positive integer, and consider the pairs $(x, y) \in \mathbb{Z}^2$ that satisfy:

$$x^2 - dy^2 = 1$$

The existence of a non-trivial solution to this equation (i.e. distinct from $(\pm 1, 0)$) is equivalent to showing existence of a torsion-free unit in $\mathbb{Z}[\sqrt{d}]$. All proofs I've seen of this fact uses the Dirichlet Approximation Theorem, and then follows by consecutive applications of the Pigeonhole Principle. But the group of units in $\mathbb{Z}[\sqrt{d}]$ is a purely algebraic construction, so I wonder if it's possible to prove this fact without resorting to the approximation theorem. Is there a known "algebraic" proof of this?

• I got my pages copied out of Stillwell, he does periodicity of the "river" directly, page 98. I would call it about 30 pages you need to understand, particularly learn to draw the pictures yourself; it is all completely elementary, just detailed and quite visual. He does the pigeonhole argument first, you say you are familiar with that. – Will Jagy May 20 '17 at 0:29
• Perhaps constructing a unit directly using the method of continued fractions turns out to be both elementary and algebraic, cf. mathworld.wolfram.com/PellEquation.html – sharding4 May 20 '17 at 22:44

not sure what would satisfy you as "algebraic," however, Conway's topograph method, done with full detail (more than shown in his book) the way I have been drawing the diagram on this site, puts all sorts of numbers and labels on a drawing of $PSL_2 \mathbb Z.$ The construction of $u^2 - d v^2 = 1$ is incorporated into finding the matrix $$P = \left( \begin{array}{rr} u & dv \\ v & u \end{array} \right)$$ which generates the (oriented) automorphism group of $$G = \left( \begin{array}{rr} 1 & 0 \\ 0 & -d \end{array} \right),$$ that is $$P^T G P = G.$$

The example in Hatcher is 19, $$\left( \begin{array}{rr} 170 & 39 \\ 741 & 170 \end{array} \right) \left( \begin{array}{rr} 1 & 0 \\ 0 & -19 \end{array} \right) \left( \begin{array}{rr} 170 & 741 \\ 39 & 170 \end{array} \right) = \left( \begin{array}{rr} 1 & 0 \\ 0 & -19 \end{array} \right).$$ My scanner makes strange colors out of graph paper. That's life. The cycle of length six forms are the Gauss-Lagrange "reduced" forms equivalent to $x^2 - 19 y^2.$ I made a point of including the light blue numerical labels so that you could see where the reduced forms occur in the diagram. A form $\langle A,B,C \rangle$ is a pink value $A,$ a blue edge number $B$ (always positive, the little blue arrow indicates a sort of orientation) then a pink value $C,$ such that $AC < 0$ and $B > |A+C|.$ The fact that this description is equivalent to the usual definition of "reduced" is on page 37 of Franz's 2010 book, Theorem 1.36.

Let me put some links to my answers, then I will go look for publications, there is a new one by Hatcher, this was initiated in Conway, there was some helpful detail in Stillwell. See pages 70 and 71 in Hatcher for $d=19$ and the location of $P$ in the diagram. Oh, it appears that, after I had assumed he'd quit, Weissman is publishing his book, to be available in August 2017.

http://math.stackexchange.com/questions/228356/how-to-find-solutions-of-x2-3y2-2/228405#228405

http://math.stackexchange.com/questions/487051/why-cant-the-alpertron-solve-this-pell-like-equation/487063#487063

http://math.stackexchange.com/questions/512621/finding-all-solutions-of-the-pell-type-equation-x2-5y2-4/512649#512649

http://math.stackexchange.com/questions/680972/if-m-n-in-mathbb-z-2-satisfies-3m2m-4n2n-then-m-n-is-a-perfect-square/686351#686351

http://math.stackexchange.com/questions/742181/find-all-integer-solutions-for-the-equation-5x2-y2-4/756972#756972

http://math.stackexchange.com/questions/822503/positive-integer-n-such-that-2n1-3n1-are-both-perfect-squares/822517#822517

http://math.stackexchange.com/questions/1078450/maps-of-primitive-vectors-and-conways-river-has-anyone-built-this-in-sage/1078979#1078979

http://math.stackexchange.com/questions/1091310/infinitely-many-systems-of-23-consecutive-integers/1093382#1093382

http://math.stackexchange.com/questions/1132799/finding-integers-of-the-form-3x2-xy-5y2-where-x-and-y-are-integers

http://math.stackexchange.com/questions/1221178/small-integral-representation-as-x2-2y2-in-pells-equation/1221280#1221280

http://math.stackexchange.com/questions/1404023/solving-the-equation-x2-7y2-3-over-integers/1404126#1404126

http://math.stackexchange.com/questions/1599211/solutions-to-diophantine-equations/1600010#1600010

http://math.stackexchange.com/questions/1667323/how-to-prove-that-the-roots-of-this-equation-are-integers/1667380#1667380

http://math.stackexchange.com/questions/1719280/does-the-pell-like-equation-x2-dy2-k-have-a-simple-recursion-like-x2-dy2

http://math.stackexchange.com/questions/1772594/find-all-natural-numbers-n-such-that-21n2-20-is-a-perfect-square/1773319#1773319

• Thanks, I am taking a look right now at the Stillwell's and Conway's references! – Henrique Augusto Souza May 19 '17 at 18:17

Given the Dirichlet Unit Theorem which states that the group of units in the ring of integers of a number field is finitely generated and has rank equal to $r = r_1 + r_2 − 1$, where $r_1$ is the number of real embeddings and $r_2$ half the number of complex embeddings of the field, you're guaranteed the existence of a fundamental unit in the ring of integers of a field $\mathbb{Q}[\sqrt{d}]$ as a real quadratic field has $r_1=2$. It's simple enough to translate back and forth between units and solutions of Pell's Equation as you've noted.

So to answer your question we would need to look at proof's of Dirichlet's Unit Theorem, and, in fact, most proofs apply Minkowski's Theorem to a logarithmic embedding of the units into a real vector space. Such proofs wouldn't qualify as "algebraic" proofs, but it is possible to give a more purely algebraic proof of Dirichlet's Unit Theorem by introducing the concept of ideles. Such an approach can be seen in Weil's Basic Number Theory or in Cassels and Frohlich Algebraic Number Theory where the unit theorem follows from the compactness of a certain idele group called $\mathbb{J}_K^1/K^\times$.

• Indeed, it follows easily from the Dirichlet's Unit Theorem. What I'm trying to do is precisely to prove a special case of this theorem, for real quadratic fields - and do it using more elementary arguments, yet in the context of algebraic integers - not resorting to approximations. I've already managed to prove that the rank is at most 1, but I'm still failing to show that it isn't zero. See my other question: math.stackexchange.com/questions/2287205/… – Henrique Augusto Souza May 19 '17 at 18:20
• I would strongly suspect that you will not find any solution which is both more purely algebraic and as or more elementary than the solution you know. – sharding4 May 19 '17 at 18:33
• This seems to be the case... I was really trying to avoid the Dirichlet Approximation theorem, but if I need to resort to it in order to prove existence of non-trivial elements, then showing that it is one-generated is the simplest part. I'll look up the Unit Theorem proof again and see if the argument for showing that the image of the logarithmic embedding has the full rank of the trace-zero subgroup can be simplified in the special case of real quadratic fields. – Henrique Augusto Souza May 19 '17 at 18:41
• I was mistaken. Even the idele-theoretic proofs of Dirichlet's Unit Theorem ultimately depend on a logarithmic embedding into a real euclidean space. Seems impossible to avoid. – sharding4 May 19 '17 at 19:38

Assume that $d = p \equiv 1 \bmod 4$ is a prime number. The field of $p$-th roots of unity contains the real quadratic number field $k = {\mathbb Q}(\sqrt{p})$. The norm of the cyclotomic unit $1 + \zeta$ is a unit in $k$, and you will have an algebraic proof of the solvability of the Pell equation if you can show that this unit is not $\pm 1$. This must be east to prove using the sine function (this proof is due to Dirichlet).

If you want to avoid transcendental function, you will have to invoke congruences, but I haven't checked how far this will get you. Things will get more technical in general; for primes $p \equiv 3 \bmod 4$ you already have to look at the field of $4p$-th roots of unity.