# Perfect powers in the solutions of a certain Pell equation

The fundamental solution of the Pell equation $$x^{2}-3y^{2}=1$$ is $2+\sqrt{3}$.

It seems to me that if $x_{n}+y_{n}\sqrt{3}$, $x_{n}, y_{n} \in \mathbb{N}$, is a solution of the said Pell equation and $x_{n}$ is a power of $7$, then $n=2$.

Simple congruence arguments allow us to conclude that such an $n$ can't be either an odd natural number or divisible by $4$.

Do you see a nice way to show that $n$ cannot be congruent to $2$ modulo $4$ unless $n=2$?

In general, what are the results with which somebody interested in determining all the solutions $x_{n}+y_{n}\sqrt{d}$ of the Pell equation $x^{2}-dy^{2}=1$, subject to the additional constraint that either $x_{n}$ or $y_{n}$ be a non trivial perfect power, must be armoured?

To be 100% honest, I posed this question in http://math.stackexchange.com earlier today, but I am afraid that it might actually belong here.

Thank you very much for your knowledgeable replies.

• Here you have the link to the said MSE thread: math.stackexchange.com/questions/2034161/… – Jamai-Con Nov 28 '16 at 22:53
• There are techniques for solving this sort of thing in Cassels' book on local fields. In general this sort of question might be hard, but sometimes results like this are accessible. For example $x_6=1351$ and you might ask whether this is the largest power of 1351 which occurs in the sequence, but proving this might be difficult or maybe even out of reach. Cassels' arguments about powers in sequences defined by recurrence relations sometimes rely on coincidences. This question may or may not be accessible; furthermore if it is accessible the only approach might be computational. – znt Nov 29 '16 at 8:49
• @znt 1351=7*193 is not a prime, but a znt prime (a generalization of Grothendieck prime). – Fan Zheng Dec 12 '16 at 22:41

The standard appproach is via Baker's method of linear forms in logarithms. We have $x_n+\sqrt{3}y_n=(2+\sqrt{3})^n$, thus $2x_n=(2+\sqrt{3})^n+(2-\sqrt{3})^n$. Now assume that $x_n=7^m$, and consider $\Lambda=\log 2+ m\log 7-n\log(2+\sqrt{3})$. Then $\Lambda\neq 0$, since $(2-\sqrt{3}^n\neq 0$, but $\Lambda$ is very small. More precisely we have $$0<\Lambda = \log\frac{2\cdot 7^m}{2\cdot 7^m-(2-\sqrt{3})^n} < \frac{0.27^n}{7^m}$$ Baker proved a lower bound for arbitrary integral linear combinations of logarithms of algebraic numbers, see e.g. https://www.birs.ca/workshops/2012/12ss131/files/bugeaud_LFL.pdf . We get $$\log|\Lambda|> -C\log\max(n+1,m+1),$$ where $C$ depends on the number of summands in the linear form (in our case 3), the size and and algebraic structure of the summands, but is completely explicit (see (1.2) in the paper by Bugeaud). In our case we get $C<10^{16}$. Comparing these bounds we get an upper bound for $n$. Checking the remaining range is non-trivial, but often possible using methods from diophantine approximation.

If you consider the Ljunggren-Nagell theorem "elementary" (see On a result attributed to W. Ljunggren and T. Nagell), then there is an elementary proof in your case. You ask about $$7^{2m} = 1 + 3y^2.$$ Clearly $y$ is even, say $y = 2u$. Then the equation can be rewritten in the form $$\frac{7^m-1}{6} \cdot (7^m+1) = 2u^2.$$ Since $(7^m+1)-(7^m-1)=2$, the gcd of the two left-hand factors divides $2$. By unique factorization, each of the two left-hand factors is either a square or twice a square.

If $m$ is odd, the first left-hand factor is odd. So $$\frac{7^m-1}{6}= \square, \quad \text{which comparing with the last display forces}\quad 7^{m}+1 = 2\square.$$ If $m$ is even, both $(7^m-1)/6$ and $7^m+1$ are even. Moreover, $7^m+1 \equiv 2\pmod{4}$ --- hence, cannot be a square, so must be twice a square. Looking back at the displayed equation from the last paragraph, we see that again, $(7^m-1)/6=\square$. By Ljunggren's theorem, the only solutions to $(7^m-1)/6=\square$ are $m=1$ and $m=4$. But only $m=1$ is compatible with the condition $7^m+1=2\square$.

We know $x_n + y_n\sqrt{3} = (2+\sqrt{3})^n$. If $n=4k+2$ for some $k\ge 0$, then $$x_n + y_n\sqrt{3} = (2+\sqrt{3})^n = (7+4\sqrt{3})^{2k+1},$$ which implies that $$x_n = \sum_{i=0}^{k}C(2k+1, 2i) 7^{2k+1-2i}(4\sqrt{3})^{2i} = \sum_{i=0}^{k}C(2k+1, 2i) 7^{2k+1-2i}48^i$$

We need to consider how many factors of $7$ each $C(2k+1, 2i)$ contains. I don't have time now and will claim that: if $7^\alpha || 2k+1$, we have $7^{\alpha + 1} \mid 7^{2k-2i}C(2k+1, 2i)$ for $0\le i<k$. The proof uses the factors of a prime in a factorial expressed in the sum of floor functions.

If the claim holds, then $7^{\alpha + 2}$ divides all terms but the last one.

Claim: $m>n>0$, $p$ is a prime number, $\alpha >0$ such that $p^\alpha || m$ (that is, $p^\alpha \mid m$ and $p^{\alpha+1} \nmid m$). If $p^\beta || C(m, n)$, then $\beta \ge \alpha - s$, where $s\ge 0$ and $p^s \le n$ and $p^{s+1} < n$.

Proof of claim: $$\beta =\sum_{i=1}^{\infty} ([\frac{m}{p^i}] - [\frac{n}{p^i}] - [\frac{m-n}{p^i}])$$

Each term in the summation contributes $0$ or $1$. If $n=1$, the first $\alpha$ terms contribute $1$ each; for general $n$, except the first $s$ terms, the contribution of each term is at least as good as in $C(m, 1)$.

• This is morally the 7-adic proof in Cassel's spirit. It is neater to prove that for $n=2\pmod 4$, $v_7(x_n)=v_7(n)+1$. Then the result follows from the fact that $7^k$ grows much slower than $(2+\sqrt3)^{7^k}$. – Fan Zheng Dec 12 '16 at 22:44
• Thanks for comments. Glad to know that my proof has some correctness and thanks for more resource. – S. Y Jan 26 '17 at 20:45