If $a_{0}\geq 0$ then $a_{n}\geq 0\forall n$. By $xy\leq (\frac{x+y}{2})^{2}$ we have $a_{n+1}\leq \frac{a_{n}^{2}}{4}$, therefore exists $n_{0}$ such that $0\leq a_{n}<1\forall n>n_{0}$. Now, for $n>n_{0}$ we have $a_{n+1}=0$, and we have done! If $a_{0}<0$, Who can post ?

say $a_{0}<0$ then since $0\leq a_{0}-[a_{0}]<1$ wehave if $a_{0}$ is negative other numbers are either $0$ or are negative so let's then conside the series of nonegative real no.s $-a_{0},-a_{1},-a_{2},\dots$ rest can be proved by N.T.TUAN's method

N.T.TUAN wrote: we have $a_{n+1}\leq \frac{a_{n}^{2}}{4}$, therefore exists $n_{0}$ such that $0\leq a_{n}<1\forall n>n_{0}$. It is not correct. We have $a_{n+1}=[a_{n}]\{a_{n}\}$, therefore if $0\le a_{n}\le 1$, then $a_{n+k}=0 \ \forall k\in N$. If $a_{n}>1$, then $a_{n+1}<[a_{n}]$, therefore $a_{[a_{0}]+k}=0 \ \forall k\in N$. If $a_{n}<0$, then $a_{n+1}\le 0$. If $-1<a_{0}<0$, then $a_{1}=-\{a_{0}\}=-a_{0}-1, a_{2}=-a_{1}-1=a_{0}$, therefore $a_{n+2}=a_{n}$.

Sorry, because I think it is $a_{n+1}\leq \frac{a_{n}}{4}$ I am wrong!

if $[a_{0}]\geq 0$,then if$a_{0}\geq 1$we know that $a_{n+1}<[a_{n}]$ thus $[a_{n+1}]<[a_{n}]$ so there must be an $n$St$[a_{n}]=0$ so $a_{n+1}=a_{n+2}=.......=0$ if $[a_{0}]<0$ we can easily prove that: if $[a_{m}]=s<-1$,then there must exsist a $k$,St $[a_{k}]<[a_{m}]$ thus we can suppose $[a_{m}]=-1$ then $a_{n}+a_{n+1}=-1$for any $n\geq m$

Quote: a_{n+1}=\left[a_{n}\right]\cdot \left\{a_{n}\right\} Prove it! I remember that we define: $[x]$ as the maximum integer n such that $n \le x$, i.e.: [4] = 4, [4.5] = 4, [-3] = -3, [-3.5] = -4... And {x} = x - [x]

i am sorry,but i don't know which part do you want me to prove

newcomer wrote: we can easily prove that: if $[a_{m}]=s<-1$,then there must exsist a $k$,St $[a_{k}]<[a_{m}]$ I cannot understand why is this so easy... during the test I wasn't able to solve this step.

Note that if $a_{l} \in \mathbb{Z}$, then $a_{i}=0$ for all $i > l$ and the claim obviously holds. Suppose $a_0\not \in \mathbb{Z}$ and consider the following two cases. Case1: If $a_{0}<0$, then $a_0 \in (-k, -k + 1)$ for some $k \in \mathbb{N}$. We'll use induction on $k$. $(1)$ If $k=1$, then $a_{0}=-1+\alpha$ for some $0<\alpha<1$. Then $a_{1}=-\alpha$ and $a_{2}=\alpha-1=a_{0}$. This means $a_{i}$ is periodic with period $2$. $(2)$ Suppose the claim holds up to some $k> 1$. $(3)$ Let $a_{0}\in (-k,-k+1)$ and write $a_{0}=-k+\alpha_{0}$ for some $\alpha_{0}\in (0,1)$. Here we have two cases. $(a)$ If $\alpha_{0}\leq \frac{k-1}{k}$, then $a_1 = -k\alpha_0 \geq -k+1$ which means $a_{1}$ is either integer or it is in one of the intervals $(-1,0), (-2,-1), \ldots, (-k+1,-k+2)$. So by induction hypothesis and above discussion we are done. $(b)$ Now, suppose $\frac{k-1}{k}< \alpha_{0}$. It is obvious that $a_{j}\geq-k$ for all $j$. If one of these numbers $\geq -k+1$, then by induction hypothesis we are done. So suppose that $a_{j}\in (-k,-k+1)$ for all $j$ and let $a_j = -k +\alpha_j$. Then from recursion we get $$\alpha_{j}=k-k\alpha_{j-1}\, .$$This gives $$\alpha_{2l}=k-k^{2}+k^{3}-\cdots-k^{2l}+k^{2l}\alpha_{0}<1$$and $$\alpha_{2l+1}=k-k^{2}+k^{3}-\cdots+k^{2l+1}-k^{2l+1}\alpha_{0}<1\, .$$From the first inequality \begin{align*} \alpha_{0}<1- \frac{1}{k} + \cdots +\frac{1}{k^{2l}}\, . \end{align*}By letting $l \to \infty$, we obtain $\alpha_0 \leq \frac{k}{k+1}$. Similarly, from the second inequality $$\alpha_{0}>1-\frac{1}{k}+\frac{1}{k^{2}}-\cdots-\frac{1}{k^{2l+1}}\, .$$Again by letting $l\to \infty$, we get $\alpha_{0} \geq \frac{k}{k+1}$. Therefore $\alpha_0 =\frac{k}{k+1}$ which means $a_{0}=\frac{-k^{2}}{k+1}$. Plugging this into recursion we see that $a_{i}=\frac{-k^{2}}{k+1}$ for all $i \geq 0$. This completes induction. Case 2: If $a_{0}>0$, then $a_0 \in (k, k+1)$ for some $k\geq 0$. We'll again use induction on $k$. $(1)$ If $k=0$, then $0<a_{0}<1 \Longrightarrow a_{i}=0$ for all $i \geq 0$. $(2)$ Suppose the claim holds up to $k$. $(3)$ Let $a_{0}\in (k,k+1)$ and write $a_{0}=k+\alpha$ for some $\alpha \in (0,1)$. Then $0 < a_{1}=k\alpha <k$ which means $a_{1}$ is in one of the intervals $(0,1), (1,2), \ldots, (k-1,k)$. So by induction hypothesis the claim also holds for $k$.

cefer wrote: $a_{2l}=k-k^{2}+k^{3}-...-k^{2l}+k^{2l}\alpha_{0}<1$ $a_{2l+1}=k-k^{2}+k^{3}-...+k^{2l+1}-k^{2l+1}\alpha_{0}<1$ $(l \rightarrow \infty)$ I think $a_{2l}$ should be changed to $\alpha_{2l}$ and $a_{2l+1}$ to $\alpha_{2l+1}$ The rest is OK

barasawala wrote: cefer wrote: $a_{2l}=k-k^{2}+k^{3}-...-k^{2l}+k^{2l}\alpha_{0}<1$ $a_{2l+1}=k-k^{2}+k^{3}-...+k^{2l+1}-k^{2l+1}\alpha_{0}<1$ $(l \rightarrow \infty)$ I think $a_{2l}$ should be changed to $\alpha_{2l}$ and $a_{2l+1}$ to $\alpha_{2l+1}$ The rest is OK You are right,i made typo mistake .

Well,THis is the first problem of IMO shortlist 2006

Here is a shorter proof: If $ a_{n}>1$, we have $ 0\le\lfloor a_{n+1}\rfloor<\lfloor a_{n}\rfloor$, as $ a_{n+1}$ is $ \lfloor a_{n}\rfloor$ times a nonnegative less than 1. And if $ 0\le a_{n}<1$, then $ a_{n+1}$ and all terms thereafter are 0. So if $ a_{0}\ge 0$, the sequence of floors is strictly decreasing, so it eventually becomes constant at 0 and we are done. Suppose $ a_{n}<0$. Again, $ 0\ge\lfloor a_{n+1}\rfloor\ge\lfloor a_{n}\rfloor$, for the same reason. Hence if $ a_{0}<0$, the sequence of floors is nondecreasing, so either it eventually reaches 0 and we are done or it eventually becomes constant at $ m$. Suppose that for all $ n\ge j$, $ \lfloor a_{n}\rfloor=m$. Also suppose that $ a_{j}=m+x$, for an integer $ m$ and $ x\in [0,1)$. Then we prove by induction that for all $ i\ge 0$, $ a_{j+i}=a+bm^{i}$, where $ a=\frac{m^{2}}{m-1}$ and $ b=\frac{xm-x-m}{m-1}$. It is obvious for $ i=0$. Suppose it is true for a certain $ i$. Then $ a_{j+i+1}=\lfloor a_{j+i}\rfloor\cdot \{a_{j+i}\}=m(a_{j+i}-m)=m(a+bm^{i}-m)$ $ =(am-m^{2})+bm^{i+1}=a+bm^{i+1}$, as it may easily be verified that $ am-m^{2}=a$. Hence $ a_{j+i}=a+bm^{i}$ for all $ i\ge 0$. Also, $ b\neq 0$ since if it did, $ xm-x-m=0$ and $ (x-1)(m-1)=0$, but neither $ x$ nor $ m$ may equal 1. Now suppose that $ m<-1$. Then the absolute value of $ a_{j+i}$ becomes arbitrarily large, which is a contradiction to the fact that $ \lfloor a_{j+i}\rfloor=m$. Hence $ m=-1$. But then we are done, since $ a_{j+i+2}=a+b(-1)^{i+2}=a+b(-1)^{i}=a_{j+i}$.

you seem to reach a contradiction in the case of the seed value a_0 = -3.2 when there is no contradiction there

Oops, you're right. I made a small mistake. jb05 wrote: Also, $ b\neq 0$ since if it did, $ xm-x-m=0$ and $ (x-1)(m-1)=0$, but neither $ x$ nor $ m$ may equal 1. So this is wrong, but it patches up easily. If $ xm=x+m$, then $ a_{j+1}=xm=x+m=a_{j}$, so the sequence is constant which is good.

If $ \lfloor a_{j}\rfloor =-m \;\forall j \geq i_{0}$, then $ a_{i_{0}}= a_{i_{0}+1}= ... =-(m-1)-\frac{1}{m+1}= \frac{-m^{2}}{m+1}$. Infact, all sequences eventually take on a constant value except for those sequences which end up between $ -1$ and $ 0$, which oscillate between two values, $ -f$ and $ -(1-f)$ where $ 0 < f < 1$.

mattilgale wrote: $ a_{0},\ a_{1},\ a_{2},\dots$ is a sequence of real numbers such that \[ a_{n + 1} = \left[a_{n}\right]\cdot \left\{a_{n}\right\} \] prove that exist $ j$ such that for every $ i\geq j$ we have $ a_{i + 2} = a_{i}$. Proposed by Harmel Nestra, Estionia A bit long and complicated solution. Assume that $a_0 \ge 0$. Then for all $i \ge 1$ we have $a_i \ge 0$, because $\lfloor a_{i-1} \rfloor \ge 0$ and $\{a_{i-1}\} \in [0;1)$. Moreover, \[\lfloor a_{i+1}\rfloor \le a_{i+1}=\lfloor a_i \rfloor \{a_i\}<\lfloor a_i \rfloor, \forall i \in \{0,1,2...\}.\] Therefore $\lfloor a_i \rfloor=0$ for some $i$. But then $a_j=0$ for all integer $j \ge i+1$. So let $a_0<0$. Then again for all $i \ge 1$ $a_i<0$, because $\lfloor a_{i-1} \rfloor<0$, $\{a_{i-1}\} \in [0;1)$. Let's immediately cosider $a_0 \notin \mathbb{Z}$, since otherwise $\{a_0\}=0$ and $a_i=0$ for all $i \ge 1$. We note that if for some $i$ $\lfloor a_i \rfloor=-1$, then $a_j=-\{a_i\}$ for $j=i+1, i+3, i+5,...$, and $a_{j'}=-1+\{a_i\}$ for $j'=i+2, i+4,...$. Now let $\lfloor a_0 \rfloor <-1$. It is clear that $a_{i+1}>\lfloor a_i \rfloor$ $\forall i \ge 1$, so the sequence $\{a_i\}^{+\infty}_{i=0}$ is bounded below by the constant $\lfloor a_0 \rfloor$ and above by zero. Let's make some notes previously: 1) If $a_{i+1} \ge \lfloor a_i \rfloor +1$, then $\lfloor a_i \rfloor\{a_i\} \ge \lfloor a_i \rfloor +1 \Longleftrightarrow \{a_i\} \le 1+\frac{1}{\lfloor a_i \rfloor}$. Therefore if $\{a_i\} \in (0;1+\frac{1}{\lfloor a_i \rfloor}]$, then $a_{i+1} \ge \lfloor a_i \rfloor +1$. We call this semi-interval transitional. So if $\{a_i\}$ falls into transitional semi-interval, then either $a_{i+1}$ becomes an integer (and then $a_j=0$ $\forall j \ge i+2$), or $a_{i+1}$ falls into the interval $(\lfloor a_i \rfloor +j; \lfloor a_i \rfloor +j+1)$ for some integer $j \ge 1$. 2) If $a_{i+1} \ge a_i$, then $\lfloor a_i \rfloor\{a_i\} \ge \lfloor a_i \rfloor +\{a_i\} \Longleftrightarrow \{a_i\} \le \frac{\lfloor a_i \rfloor}{\lfloor a_i \rfloor -1}$. Therefore if $\{a_i\} \in (1+\frac{1}{\lfloor a_i \rfloor}; \frac{\lfloor a_i \rfloor}{\lfloor a_i \rfloor -1}]$, then $\lfloor a_i \rfloor +1 > a_{i+1} \ge a_i$. We call this semi-interval large. So if $\{a_i\}$ falls into large semi-interval, then either $a_i= \frac{\lfloor a_i \rfloor}{\lfloor a_i \rfloor -1}$ (and then $a_{j+1}=a_j$ for all $j \ge i$), or $a_{i+1}$ becomes larger then $a_i$ and at the same $a_{i+1} \in (\lfloor a_i \rfloor; \lfloor a_i \rfloor+1)$. 3) If $\{a_i\} \in (\frac{\lfloor a_i \rfloor}{\lfloor a_i \rfloor -1}; 1)$, then $a_{i+1}<a_i$. We call this interval small. So if $\{a_i\}$ falls into small interval, then $a_{i+1}$ becomes smaller than $a_i$. Let's choose the smallest index $i$, for which $\{a_i\}$ lies in the small interval (if it is impossible then the sequence $\{a_i\}^{+\infty}_{i=0}$ does not decrease and so either $a_i=a_{i+1}=...$ for some $i$, or $\lfloor a_i \rfloor=-1$ for some $i$). Let $\{a_i\}=\frac{\lfloor a_i \rfloor -k}{\lfloor a_i \rfloor -1}$, where $k \in (0; 1)$. We have: \[\{a_{i+1}\}=a_{i+1}-\lfloor a_{i+1} \rfloor=\lfloor a_i \rfloor\{a_i\}-\lfloor a_i \rfloor=\lfloor a_i \rfloor(\{a_i\}-1)=\] \[=\lfloor a_i \rfloor \cdot \frac{\lfloor a_i \rfloor -k-\lfloor a_i \rfloor+1}{\lfloor a_i \rfloor-1}=\frac{\lfloor a_i \rfloor(1-k)}{\lfloor a_i \rfloor-1} \notin (\frac{\lfloor a_i \rfloor}{\lfloor a_i \rfloor -1}; 1).\] So $\{a_{i+1}\}$ falls either into the large, or into the transitional semi-interval. Let $\{a_{i+1}\}$ falled into the large semi-interval. Then: \[a_{i+2}=\lfloor a_{i+1} \rfloor\{a_{i+1}\}=\lfloor a_i \rfloor\{a_{i+1}\},\] \[\{a_{i+2}\}=\lfloor a_i \rfloor(\{a_{i+1}\}-1)=\lfloor a_i \rfloor\frac{\lfloor a_i \rfloor -k\lfloor a_i \rfloor-\lfloor a_i \rfloor+1}{\lfloor a_i \rfloor-1}=\] \[=\frac{\lfloor a_i \rfloor-\lfloor a_i \rfloor^2\cdot k}{\lfloor a_i \rfloor-1}>\{a_i\}.\] and besides $\{a_{i+2}\}$ falls into the small interval. If $\{a_{i+3}\}$ will not fall into the transitional semi-interval again, then $\{a_{i+4}\}$ will fall into the small interval and besides $\{a_{i+4}\}>\{a_{i+2}\}$ (by analogiousy). But since $\{a_{i+2k}\}$, $k \in \mathbb{N}$ can't increase infinetely (it is bounded above by $1$), then for some $k$ $\{a_{i+2k+1}\}$ will fall into the transitional semi-interval. Then $a_{i+2k+1} \in (\lfloor a_i \rfloor +j; \lfloor a_i \rfloor +j+1)$ for some integer $j \ge 1$, and the situation will repeat in this interval. Therefore for some sufficiently large $j$ it wiil be $\lfloor a_j \rfloor=-1$, as desired.

It can be easily noted that $a_n\ge 0$ for all $n$ iff $a_0\ge 0$. First let $a_n\ge 0$. Check that the sequence then is bounded below by $0$ and monotonically decreasing. Thus by Monotone Convergence Theorem the sequence is convergent. However, check that limit must be $0$. But then with some work we can prove that the sequence is eventually constant. Now, if $a_0<0$. Check that the sequence is monotonically increasing unless we arrive at some $-1\le a_i<0$. Also the sequence is bounded up by $0$. If such $i$ doesn't exist then the sequence is convergent by the above theorem. However the limit must be $-1$. Clearly then we can construct such $i$, which leads to a contradiction. Thus, after some finite number of terms we get hold of such $a_i$. Easy to see then $a_{j+2}=a_j$ for all $j\ge i$. $Q.E.D$

Just like many others, my solution to $a_0\ge 0$ is short but $a_0 < 0$ is quite lengthy.

We split this into two cases. Case 1: $a_0 \geq 0$. It is easy to prove with induction that all $a_i \geq 0$. Thus $a_{i+1} = \lfloor a_i \rfloor \langle a_i \rangle < \lfloor a_i \rfloor \implies a_{i+2} < \lfloor a_i\rfloor$. So $\{a_i\}$ is constantly decreasing. Thus for some sufficiently large $k-1$, $\lfloor a_{k-1} \rfloor= 0 \implies a_{k} = a_{k+2} = 0$. Case 2: $a_0 < 0$. Again, it is easy to prove with induction that all $a_i \leq 0$. Similar to above, $a_{i+1} > \lfloor a_i \rfloor$. So $\{a_i\}$ is constantly increasing. Thus for some sufficiently large $k$, $\lfloor a_k \rfloor = -1$. $$a_{k+1} = -\langle a_k \rangle = -(-1+a_k) = 1 - a_k$$and $$a_{k+2} = -\langle a_{k+1} \rangle = -(-1+a_{k+1}) = 1 - a_{k+1} = 1 - (1 - a_k) = a_k$$and we're done. $\square$

We consider two cases. Case 1: There is a $i$ such that $a_i\in\mathbb{Z}.$ Then $a_{k+1}=0$ for all $k\geq i$ and we have $a_{j+2}=a_j$ for all $j>i.$ Case 2: $a_i\not\in\mathbb{Z},\quad\forall i.$ Then $0<\{a_i\}<1$ and $[a_i]\not=0$ for all $i.$ +) If $a_0>0$ then $a_n>0$ for all $n$, therefore $[a_n]\geq 1$ for all $n.$ From hypothesis we have $$1\leq [a_{n+1}] <[a_n],\quad\forall n\geq 1,$$contradiction. +) If $a_0<0$ then $a_n<0$ for all $n$ and $[a_n]\leq -1$ for all $n.$ Therefore $[a_{n+1}]\geq [a_n]$ for all $n,$ so there is an integer $m\leq -1$ such that $[a_n]=m$ from some where. We can assume that $[a_n]=m\leq -2$ for all $n\geq 0,$ then $$\{a_{i+1}\}=m\{a_i\}-m,\quad\forall i\geq 0.$$If $\{a_0\}=\frac{m}{m-1}$ then $\{a_i\}=\frac{m}{m-1}$ for all $i$ and $a_i=\frac{m^2}{m-1}$ for all i, we are done. Else, $$\{a_{n+1}\}=m^{n+1}\{a_0\}-\frac{m(m^{n+1}-1)}{m-1},\quad\forall n,$$and we have $\{a_{n}\}>1$ for some $n,$ contradiction.

For $a_0 > 0$ the sequence $\{\lfloor a_i \rfloor\}$ strictly decreases, so there is nothing to prove. When $a_0 < 0$, it suffices to show that $\{\lfloor a_i \rfloor\}$ will increase after a finite amount of time or $a_i$ remains fixed. Let $n = \lfloor a_0 \rfloor$. Consider the disjoint set of intervals $I_k = [x_k, y_k]$ determined recursively by $I_1 = \left[n-\frac{n-1}n, n\right]$, $I_2 = \left[n-1, n-1 + \frac{(n-1)^2}{n^2}\right]$, and $$I_{2n+1} = \left[x_{2n-1} - \frac{(n-1)^2}{n^2} (y_{2n-1}-x_{2n-1}), x_{2n-1}\right] \text{ and } I_{2n} = \left[x_{2n-2}, x_{2n-2} + \frac{(n-1)^2}{n^2}(y_{2n-2}-x_{2n-2})\right].$$Every real number in $[n-1, n]$ belongs in one of these intervals except for $n - \frac n{n+1}$, which is a fixed point. On the other hand, for $a_0 \in I_n$, we will have $\left \lfloor a_n \right \rfloor > n$, as needed. Thus, eventually the sequence will reach a point where $\lfloor a_i \rfloor = 0$ for all $i > N$. It is clear that the sequence alternates now.

We claim that this sequence eventually goes to $0$ for positive real numbers $a_0$. Notice that if $a_0$ is positive and less than $1$, then we will have that $a_1=0$, proving our claim. Now we will prove that for all $a_k>1$, then we will have that $\lfloor{}a_{k+1}\rfloor{}<\lfloor{}a_k\rfloor{}$, which is clearly true, since $a_{k+1}=\lfloor{}a_k\rfloor{}\{a_k\}$. Therefore the sequence must go to zero, meaning that eventually, $a_i=a_{i+2}=0$. What about negative $a_0$? Notice that for $a_0<0$, we have that the fractional part of $a_k$ is constantly increasing, meaning that eventually, we will have that $a_{i+1}=1-a_i$. Therefore we have that $a_{i+2}=a_i$, and we are done. FS found by bobthegod78 What about negative $a_0$? Let $a_0=-k+\frac{k}{k+1}+x$ for some $k$. If $x\neq{}0$, we have that $a_1=-k+\frac{k}{k+1}-xk$, $a_2=-k+\frac{k}{k+1}+xk^2$, $a_3=-k+\frac{k}{k+1}-xk^3$, and so on and so forth until we are left with $a_i=-k+\frac{k}{k+1}-x*(-k)^i$. It is clear that this will eventually go to an integer, leaving us with a sequence of $n$, $-1-n$, and so on and so forth, proving that $a_i=a_i+2$. If $x=0$, it is easy to see that the sequence is constant anyways, and we are done.

For $a_0\geq 0,$ it is obvious that the sequence becomes all $0$s eventually. For negative $a_i,$ let $a_i = -n + \frac{n}{n+1} + \epsilon$. If $\epsilon =0$, the sequence is constant and we are done. Notice that $a_{i+1} = -n+\frac{n}{n+1} - \epsilon \cdot n$, and so on, so eventually, the value of $\epsilon \cdot n^k$ exceeds $\frac 1{n+1}$, and the value the floor is incremented by 1, unless $n=1$. If $n=1$, we see that it oscillates between $-x, x-1$, so $a_{i+2}=a_i$.

When $a_0$ is nonnegative, then $\lfloor a_{i+1}\rfloor <\lfloor a_{i}\rfloor$, so it will eventually go down to zero. If $a_0$ is an integer, then clearly it will just go to 0. For the rest of this solution, assume $a_0$ is negative and non-integer. Define a equlibrium point as a real number that is equal to $$-\frac{(n-1)^2}{n}$$for some positive integer $n$. The equilibrium points have the property that if $a_i$ is an equilibrium point, then $a_{i+1}=a_i$. Furthermore, there is exactly one equilibrium point between any two nonpositive integers, and that equilibrium point is the only fixed point between those two numbers. If $a_0$ starts between -1 and 0, then it will alternate between $a_0$ and $-1-a_0$ so the statement holds. From now on, assume $a_0$ is less than -1. Then, consider the equilibrium point in $a_0$'s integer interval. If $a_0$ is at the equilibrium point, it will just stay there forever so the problem statement holds. Otherwise, its distance away from the equilibrium point will multiply by $-\lfloor a_i \rfloor$ but go in the opposite direction. Since we are assuming less than -1, the distance to the equilibrium point will keep increasing, and it will eventually escape the current integer range to the right (it's not possible for it to escape to the left), into either a nonnegative range or a less negative range. Of course, if it goes into a nonnegative range, it will then eventually go to zero as shown earlier. Thus, for any range below -1, as long as it doesn't land into the range at an equilibrium point, it will escape further to the right. Thus, if we keep doing this, it will either get stuck at an equilibrium point, get between -1 and 0 and alternate, or eventually go to zero, so we are done.

It's easy to see that for $a_0\ge 0$, the problem is trivial since the sequence is strictly decreasing. For $a_0<-1$ ($-1\le a_0<0$ implies it iterates from a_i=c<0 to -1*(c-(-1))=-1-c and oscillates with period 2), if we let -b_i=floor[a_i]<-1, we want to show $$a_{i+1}=-b_i(a_i+b_i)\ge a_i\iff a_i\ge \frac{b_i^2}{-b_i-1}=-b_i+\frac{b_i}{b_i+1}.$$Because of this experimenting, we're motivated to set $$a_i =-b_i+\frac{b_i}{b_i+1}+c_i.$$Indeed, $$a_{i+1}=-b_i+\frac{b_i}{b_i+1}-b_ic_i$$and so on. Since b_i>1, for some even amount of iterations eventually $b_i^kc_i>\frac{1}{b_i+1}$, whence -b_i increases by 1 and so b_i decreases by 1 monotonically until b_i becomes 0, so all cases have been considered, and we are done! $\blacksquare$

For a_0>0 Just observe that [a_i+1]<[a_i].Therefore the sequence is decreasing.We will find an [a_i]=0

The sequence is strictly decreasing for $a_1 > 0$ and will eventually remain constant at 0. Hence we proceed by considering negative $a_1$. Consider the fixed points $a_i$ such that $a_i = a_{i+1}$, which can be determined to be 0 and $-k\tfrac{1}{k+2}$ for $k \in \mathbb{Z}_{\ge 0}$. Once we hit/start with this fixed point, it remains constant, which finishes. If we have $a_i$ greater than the corresponding fixed point, $a_{i+1}$ is less than the fixed point. If we have $a_{i+1}$ less than the fixed point, either $\lfloor a_{i+1} \rfloor$ strictly increases, or $\lfloor a_i \rfloor < a_{i+2} < a_i$ where $\lfloor a_i \rfloor \leq -2$, from which we eventually must increase $\lfloor a_{i+k} \rfloor$. In the end state, we either land on a fixed point (including 0), or $\lfloor a_i \rfloor = -1$, from which we always have $a_{i+2} = a_i$. $\blacksquare$

Notice that if $a_0$ is nonnnegative, we can easily show $a_i$ is eventually $0$, if $a_i$ is a noninteger the floor clearly decreases, so eventually the floor must be $0$ and we win, or if $a_i$ is an integer we floor and get $0$ anyways. Continue to ignore all integer values since those instantly finish. So we take $a_i$ as negative, if $a_i = c$ is between $-1$ and $0$, we see that the next term is just $-1 - c$, and applying again give $c$, so the sequence alternates. Thus we show $a_i$ eventually goes to something in $(-1, 0]$. Take $a_i = x + (-k)$, where $x,k$ are positive, and $k > 2$. Then assume for contradictory purposes that the floor of $a_i$ never changes. If $x = \frac{k}{k + 1}$, the sequence is constant and we can ignore this case. Otherwise, if $x > \frac{k}{k + 1}$, $a_{i + 1}$ has floor $k$ and fractional part less than $\frac{k}{k + 1}$, so take $a_j = -k + x$ where $x < \frac{k}{k + 1}$, we then calculate the decrease in $x$ from $a_j$ to $a_{j + 2}$. Clearly $a_{j + 1} = -kx$, then since we assumed the floor never changes we have $a_{j + 2} = -k(-kx + k) = k^2(x - 1)$, the decrease is then $-k + x + k^2 - k^2x = -k^2x + x + k^2 - k = (1 - k^2)x + k^2 - k$, which is positive for $x < \frac{k}{k + 1}$, moreover this decrease is decreasing in $x$, so as we compute $a_{j + 2n}$ with the fixed floor being assumed, $x$ will continue to decrease at an ever faster rate, and eventually we will have negative $x$, impossible. Thus the floor must decrease eventually unless the sequence is constant, done.

If $a_0$ is positive, then the sequence is nonincreasing and will eventually become just $0$, so clearly $a_i = a_{i+2}$. If $a_0$ is negative, then we have $\lfloor a_i \rfloor \leq a_{i+1} \leq 0$, so eventually $\lfloor a_i \rfloor$ is constant. Let this constant value be $C$. Then, we have \[\left \langle a_{i+1} \right \rangle = C \left \langle a_i \right \rangle - C \implies \left( \left \langle a_{i+1} \right \rangle - \tfrac{C}{C-1} \right) = C \left( \left \langle a_i \right \rangle - \tfrac{C}{C-1} \right)\]for sufficiently large $i$. If $C \leq -2$, we have a contradiction since $\langle a_{i} \rangle - \frac{C}{C-1}$ is an unbounded geometric sequence, so $C = -1$ or $C = 0$. The latter case is trivial, and the former case implies that $\langle a_i \rangle + \langle a_{i+1} \rangle = 1$ for sufficiently large $i$, which also implies the problem statement.

If $a_0 \geq 0,$ it is clear that the sequence is strictly decreasing and thus will eventually hit $0,$ so this case is finished. Meanwhile, for $a_0 < 0,$ observe that $$a_{i+1} =\lfloor a_i \rfloor \{ a_i \} \geq \lfloor a_i \rfloor$$so $\lfloor a_i \rfloor$ is decreasing (not strictly). If the sequence eventually reaches zero, it stays there so assume otherwise. Then the sequence is always negative meaning that its floor will eventually converge to a negative integer, say $-n$ for some positive integer $n.$ Then for sufficiently large $i,$ let $a_i = x_i-n$ for $0 \leq x_i < 1.$ If at some point $x_i=0,$ the sequence stays at zero so assume otherwise. Then the recurrence becomes $$x_{i+1}-n = -nx_i \implies x_{i+1}=(1-x_i)n.$$Letting $r_i = \frac{x_i}{n^i}$ yields $r_{i+1} = \frac{1}{n^i} - r_i \implies r_{i+2}=r_i.$ Therefore $x_{i+2} = n^2 x_i$ for sufficiently large $i,$ but if $n \geq 2$ this is a contradiction as eventually $x_i > 1.$ Therefore, $n=1$ so $x_{i+2}=x_i,$ and thus $a_{i+2}=a_i.$ QED

When $a_0\geq 0$, we know that $\lfloor a_{i+1} \rfloor<\lfloor a_i \rfloor,$ therefore we eventually reach $0$ with sufficiently large $i$. If $a_0<0$, then $\lfloor a_i \rfloor \leq a_{i+1} \leq 0,$ therefore eventually $\lfloor a_i \rfloor$ will reach a constant value, say $c$. In this case, we see that $${a_{i+1}}=c{a_i}-c,$$which by algebraically manipulating gives $${a_{i+1}}-\frac{c}{c-1}=c ({a_i}-\frac{c}{c-1} ).$$We see that if $c\leq -2$, then the sequence of decimal parts is unbounded, a contradiction. Thus, $c=-1$ or $c=0$. If $c=0$, then we will eventually reach a sequence of 0s. On the other hand, $c=-1$ gives an eventual sequence of period 2, which proves $a_{i+2}=a_i.$