# Prove that if $\forall n\in\Bbb N\quad a_{n+1}^2-a_{n+1}=a_n\in\Bbb Q$, then the sequence is constant

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.

• 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?
– D S
Commented Apr 12 at 17:14
• @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. Commented Apr 12 at 17:53
• @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. Commented Apr 26 at 18:00

• 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.

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 $$i and $$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.