Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.]

share|improve this question
    
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
2  
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

1 Answer 1

up vote 6 down vote accepted

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.

share|improve this answer
    
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? –  Hauke Reddmann Jun 24 '11 at 13:22

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.