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Here is a simply described but fiendishly diophanterrorizing problem I asked on AMM eons ago. Maybe you can shed some light upon it.

0.2 (base 4) = 0.2 (continued fraction)
0.24 (base 6) = 0.24 (continued fraction)
Find all examples of
0.$xyz$... (base B) = 0.$xyz$... (continued fraction).

First of all, both choices of notation define a rapidly closing interval nesting, and already on post-comma digit 2, you're down to one number by a simple > / < argument. But you may not use 0 for CF and $\ge B$ for base $B$, and thus almost any base $B$ will run into a dead end sooner or later. (It's fun to experiment with low $B$.)

Obvious Thing 1: 1-digit solution 0.$n$ for $B=n^2$.
Educated guess 2: There are only two solutions with two digits. (The second was listed in the MAA Answer Column; juggling with Chebyshev polynomials I had a sort of proof for that case, but it probably had more holes than a Menger sponge and so it wasn't printed there).
Wild guess 3: There is no solution with more than two digits, for the reasons above.

Can you at least prove case 2? (The MAA discussion splits it into two subcases; $239^2+1=2\times 13^4$ killed one of them.)

This is Monthly problem 10507. The problem appeared in February 1996 (volume 103, issue 2, page 173), and the discussion (and solution to other parts) in March 1998 (volume 105, issue 3, page 276). Thanks to Gerald Edgar for the pointer.

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  • $\begingroup$ It took a little time to work out the continued fraction notation here. Also there is an implicit assumption that B is a positive integer, otherwise the two digit problem reduces to a quadratic in B and there are many solutions. Likewise the three digit problem would give a cubic. $\endgroup$ Jun 22 '11 at 12:54
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    $\begingroup$ MathSciNet provides a reference... Reddmann, Hauke; Group, USA Problems; Problems and Solutions: Solutions: Numbers with the same Continued Fraction and Base b Expansions: 10507. Amer. Math. Monthly 105 (1998), no. 3, 276–277 $\endgroup$ Jun 22 '11 at 13:38
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Here is a solution for the case you ask. But first let me say that given the nature of the question it would probably get better answers at artofproblemsolving. What follows is a lot of very elementary number theory, and an appeal to a result of Ljunggren from 1942.

So we have $B\geq 2$ and $x,y\in \{0,1,\dots,B-1\}$ satisfying $$\frac{x}{B}+\frac{y}{B^2}=\frac{1}{x+\frac{1}{y}}$$ or in other words $B^2y=(xy+1)(Bx+y)$. Let $a=\gcd(x,y)$ and $x=am, y=an$. We have $$B^2n=(Bm+n)(a^2mn+1)$$ where $\gcd(m,n)=1$. Since $\gcd(n,a^2mn+1)=1$ we have that $n$ is a factor of $Bm+n$ therefore there is an integer $k$ so that $B=kn$. The equation simplifies to $$n^2k^2=(km+1)(a^2mn+1)$$ We see that $\gcd(km+1,k^2)=1$ so $km+1$ divides $n^2$, but also $\gcd(n^2,a^2mn+1)=1$ so $n^2$ divides $km+1$. We conclude that $km+1=n^2$ and $a^2mn+1=k^2$. In particular $k^2-1$ is divisible by $n$, and $n^2-1$ is divisible by $k$, so that $$\frac{k^2+n^2-1}{kn}=t\in \mathbb Z.$$ Now some Vieta jumping shows that $k,n$ are consecutive terms in the sequence $a_0=0,a_1=1$ and $a_{n+1}+a_{n-1}=ta_n$. Let $k=a_{p+1}$ and $n=a_p$, the equations reduce to $$m=nt-k=a_{p-1}, a^2m=kt-n=a_{p+2}.$$ Now, it is not hard to prove that our sequence is a strong divisibility sequence so that $$a^2=\frac{a_{p+2}}{a_{p-1}}$$ implies that $p+2$ is divisible by $p-1$ which only happens if $p\in \{2,4\}$. So in particular we either have $a^2=t^3-2t$ or $a^2=t^3-3t$. The second equation doesn't have non-trivial solutions because if $\gcd(t,t^2-3)=1$ then $t^2-3$ is a square which is not possible, and if $\gcd(t,t^2-3)=3$ then $3(t/3)^2-1$ is a square $-1\pmod{3}$ which is also a contradiction. For the first equation, similarly we conclude that $t=2r$ must be even and that $r(2r^2-1)$ is a perfect square, so $r=s^2$ is also a perfect square. This finally brings us to the equation $2s^4-1=l^2$, which has solutions only for $s=1$ and $s=13$ as was proved by W. Ljunggren in "Zur Theorie der Gleichung x^2+1=Dy^4" (Avh. Norske Vid. Akad. Oslo I. 5, 27pp.). This proves that the two solutions you had are the only ones.

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  • $\begingroup$ Yes, half of this line appeared in AMM. Essentially, my complete proof was the same as yours (if you replace "strong divisibility sequence" with "juggling Chebychev polynomials" and "proof" with "handwaving" :-). THX for making it rigorous (and to Gerald for digging up the reference, I have no experience with MathSciNet). P.S. Any chance at least the 3-digit case may be attacked by similar approaches? $\endgroup$ Jun 24 '11 at 13:22

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