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This question was posted, downvoted and closed today (2022 Thailand Olympiad problem) and 8 days ago ($f(x+1)^{2} - f(x+1) = f(x)$. What values of $f(1)$ allow $f(x)$ to be always rational if $x$ is natural number?). I want to re-ask it more precisely and with some more "context", propose an answer, and ask whether you have other solutions.

Prove that the only sequences $(a_n)$ of rational numbers such that $$a_{n+1}^2-a_{n+1}=a_n$$ are the two constant sequences $0$ and $2$.

Note that we don't assume $a_n\ge0$ a priori.

My first step (thanks to @cansomeonehelpmeout's comment on the previous post):

Letting $b_n=2a_n-1$, the problem is equivalent to: prove that the only sequences $(b_n)$ of rational numbers such that $$b_{n+1}^2=2b_n+3$$ are the two constant sequences $-1$ and $3$.

My next steps: I intend to prove that the $b_n$s must be integers, obviously $\ge-\frac32$, but also $\le3$, whence the conclusion.

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    $\begingroup$ I am not sure, but I think I read a site policy about asking the same question as a closed one. Are you certain you're not breaking any rules? $\endgroup$
    – D S
    Commented Apr 12 at 17:14
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    $\begingroup$ @DS Yes I am certain, because this is not the same question. As explained, it is really improved. It was closed (twice) for lack of "context" (essentially: lack of effort) and this lack is widely repaired here. I think I was not allowed to simply rewrite the closed posts and ask for their reopening, because I was not the author. $\endgroup$ Commented Apr 12 at 17:53
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    $\begingroup$ @AnneBauval I agree entirely with you. Quite a while ago, I added the missing context to save somebody else's question from imminent closure and then got a lot of flak for "putting my words into the author's mouth". If posting your own question instead is also against the rules, then this is a double bind. $\endgroup$ Commented Apr 26 at 18:00

2 Answers 2

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  • For each $n\in\Bbb N$, write $$b_n=\frac{p_n}{q_n}$$ with $p_n\in\Bbb Z$ and $q_n\in\Bbb N$ coprime. Since $p_{n+1}^2q_n=q_{n+1}^2(2p_n+3q_n)$ and $q_{n+1}^2$ is coprime to $p_{n+1}^2$, it divides $q_n$, so that (by induction) $$q_n^{2^{n-1}}\mid q_1.$$ As a consequence, all $q_n$s for $n$ large enough are equal to $1$, i.e. all $b_n$s for $n$ large enough are integers: $$\exists N\quad\forall n\ge N\quad b_n\in\Bbb Z.$$

  • Note moreover that $b_n\ge-\frac32$ (due to the recurrence relation) hence $$\forall n\ge N\quad b_n=-1\quad\text{or}\quad b_n\ge3$$ (since the values $b_n=0,1,2$ must be excluded, as for them, $2b_n+3$ is not a perfect square). And if some $b_n$ equals $-1$, so do all of them (before and after it).

  • Finally, if the sequence of integers $(b_n)_{n\ge N}$ is $\ge3$ then (due to the recurrence relation) it is non-increasing (i.e. $b_{n+1}\le b_n$) hence stationary, at a value $b\ge3$ such that $b^2=2b+3$, i.e. $b=3$. And if some $b_n$ equals $3$, so do all of them (before and after it).

Q.E.D.

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First it's easy to see that if $a_n=1$ for some $n$, then $a_{n+1}$ cannot be rational. If $a_n=0$ for some $n$, then $a_m=0$ for all $m\le n$, and $a_{n+1}=0$ or $a_{n+1}=1$ which has been ruled out. That is, if $a_n=0$ for some $n$, then $a_n=0$ for all $n$. Therefore we may assume $a_n\not=0, 1$ for all $n$.

If $a_n\in\mathbb Z$ for some $n$, then clearly $a_i\in\mathbb Z$ for all $i<n$ and $a_i\in\mathbb Z$ for all $i>n$ due to the solution of $x^2-x=a\in\mathbb Z$ is always integral over $\mathbb Z$ (or apply the rational root theorem for high school students who haven't seen the definition of integral elements). That is, if $a_n\in\mathbb Z$ for some $n$, then this holds for all $n$. In this case, $a_{n+1}-1$ is a nonzero integer.

$$|a_n|=|a_{n+1}(a_{n+1}-1)|\ge |a_{n+1}|$$

So $|a_n|$ eventually stabilizes, and if $|a_n|=|a_{n+1}|\not=0$, then $|a_{n+1}-1|=1$, $a_{n+1}=2$ or $a_{n+1}=0$ (ruled out by assumption). Hence $a_n=2$ for some $n$. It's easy to show from here that $a_n\equiv 2$ for all $n$.

If $a_n$ is not an integer for some $n$, then it's not an integer for all $n$. Note that if $a_{n+1}=\frac{a}{b}$ where $b\ge 2, \gcd(a,b)=1$, then $a_n = \frac{a(a-b)}{b^2}$ where $\gcd(a(a-b), b^2)=1$. Therefore by induction the denominator of $a_1$ has to be bigger than $b^{2^n}\ge 2^{2^n}$ for all $n$, which is absurd.

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