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Let $\sqrt 7=\sum_{i=0}^\infty a_i 7^{-i}, 0\le a_i \le 6$ be the expansion of $\sqrt 7$ in base $7$.

I am curious about the following question: Is there a $K\in \mathbb{N}$ such that for any $n\ge K$, there exists an $ i\in (n, 2n] $ for which $ a_i\ne 6$?

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    $\begingroup$ I believe (the negation of) this roughly should mean that one of the Pell equations $x^2-7y^2 = d$ for some $|d| < 11$ should have infinitely many solutions with $y$ being a power of $7$, which might be possible to rule out by the linear forms in logarithms or some more elementary means? $\endgroup$
    – Aleksei Kulikov
    Commented Aug 9 at 11:30
  • $\begingroup$ Oh wait, you do allow $i = n$. Then I think I can prove that it is true (and indeed holds for $K = 0$). $\endgroup$
    – Aleksei Kulikov
    Commented Aug 9 at 11:56
  • $\begingroup$ I louked for equations of the form $x^2=7^y+d$, it seems that for some values of $d$, the question of finiteness of solutions is still open. $\endgroup$
    – user534817
    Commented Aug 9 at 15:12
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    $\begingroup$ "the question of finiteness of solutions is still open" -- if you mean in the sense that nobody cared to do it I would believe you, but I'm pretty sure it is doable with modern number theory tools. $\endgroup$
    – Aleksei Kulikov
    Commented Aug 9 at 15:20

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Assume for the sake of contradiction that $a_i = 6, i \in [n, 2n]$. Consider the rational number $$r = \sum_{i = 0}^{n-1} a_i7^{-i} + \frac{1}{7^{n-1}}.$$ Clearly, $r > \sqrt{7}$. On the other hand, since the next digits in $\sqrt{7}$ up to $i = 2n$ are $6$, we have $r - \frac{1}{7^{2n}} < \sqrt{7}$. This means that $$0 < r-\sqrt{7} < \frac{1}{7^{2n}}.$$

Let's multiply this by $(r+\sqrt{7})$ (note that $r < 4$, say, so this number is less than $7$). We will get $$0 < r^2-7 < \frac{1}{7^{2n-1}}$$

If we multiply this by $7^{2n-2}$ then the number in the center will be an integer. So, we get an integer between $0$ and $\frac{1}{7}$, but such an integer does not exist -- contradiction.

In general if we know it for $i\in [n, 2n-c]$ for some constant $c$, then we will get a solution $(x, y)$ to the Pell-like equation with $y$ being a power of $7$, and I think modern number theory knows how to rule such things out.

Update: since it was mentioned that $a_n$ is not necessarily $6$, the argument above needs a modification. Doing the same thing as above we will eventually reach an equation of the form $$x^2 - 7^{2n+1} = k, x, n\in \mathbb{N}_0,$$ where $k$ is allowed to be $1, 2, 3, 4, 5, 6$. The values $k = 3, 5, 6$ can be removed by reducing modulo $7$, $k = 4$ can be removed by reducing modulo $4$ and $k = 1$ can be removed by reducing modulo $3$. This leaves us with the equation $x^2-7^{2n+1}=2$. This one can't be removed by simple modulo considerations since $x = 3, n = 0$ is a solution.

By considering it modulo $19$ (a factor of $7^3-1 = 342$) we can see that $n$ must be divisible by $3$. So, by denoting $7^{n/3} = y$ we get an equation $x^2-7y^6 = 2$. I think there are some magic words one can say here, like a genus of the curve is too big and deduce finiteness, but sadly I do not know them. Alternatively, by denoting $y^2=z$ we get an elliptic curve $x^2-7z^3 = 2$ and if it has only finitely many rational points we will also get a desired result, but again I don't know how one could compute them (and if it will turn out to be finite at all).

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  • $\begingroup$ Thank you very much. In fact, the value $𝑛$ is not allowed. Sorry for this tipo. The interval should be $[n+1,2n]$. $\endgroup$
    – user534817
    Commented Aug 9 at 14:42
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    $\begingroup$ The curve defined by the equation $x^2 - 7y^6 = 2$ is hyperelliptic of genus $2$, which means that it has finitely many rational points by Faltings' theorem. It's probably not hard to determine the set of rational points; this is within the purview of computational arithmetic geometry. $\endgroup$
    – Stanley Yao Xiao
    Commented Aug 9 at 16:13
  • $\begingroup$ @StanleyYaoXiao Faltings's theorem is what I had in mind, except I did not know how to compute the genus. As for " It's probably not hard to determine the set of rational points; this is within the purview of computational arithmetic geometry. " -- it is one thing to say such things and the other is to carry out the computation, in my opinion; as far as I know in general Faltings's theorem is ineffective currently. $\endgroup$
    – Aleksei Kulikov
    Commented Aug 9 at 19:53
  • $\begingroup$ @AlekseiKulikov you are right that we do not have a general effective Faltings' theorem in the height aspect. However, for genus 2 and hyperelliptic curves in general there are algorithms that determine the set of rational points. Most of these methods are a clever application of the Chabauty-Coleman method. $\endgroup$
    – Stanley Yao Xiao
    Commented Aug 9 at 20:15
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    $\begingroup$ Maybe I miss something, but don't you only need finitely many integral points on the elliptic curve $x^2-7y^3=2$? That follows from a theorem of Siegel. Note that if $x^2-7y^3=2$, then $(7x)^2 - (7y)^3 = 98$, so we get a point on the Mordell curve $U^2 = V^3 + 98$. Data of of Bennett and Ghadermarzi at personal.math.ubc.ca/~bennett/BeGa-data.html shows that the only integer points $(U,V)$ there are $(\pm 21, 7)$. Tracing this back corresponds to the "trivial" solution $3^2-2^1 = 7^1$. For theoretical justification see personal.math.ubc.ca/~bennett/BeGh-LMSJCM-2015.pdf $\endgroup$
    – so-called friend Don
    Commented Aug 9 at 22:22

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