# Unsolved Problem from AmMathMonthly

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 notations 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 >=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 Chebychev 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 split it into two subcases; 239^2+1=2*13^4 killed one of them.)

[Obvious note: I would give the exact MAA reference...if I could.]

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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. – Mark Bennet Jun 22 '11 at 12:54
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 – Gerald Edgar Jun 22 '11 at 13:38

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.