IMO 1985 Problem 6
Testing small initial values clarifies the mechanism.
Proposed by: -
Verified: no
Verdicts: FAIL + FAIL
Solve time: 12m28s
Problem
For every real number $x_1$, construct the sequence $x_1,x_2,\ldots$ by setting:
$x_{n + 1} = x_n(x_n + {1\over n}).$
Prove that there exists exactly one value of $x_1$ which gives $0 < x_n < x_{n + 1} < 1$ for all $n$.
Exploration
Testing small initial values clarifies the mechanism. For $n=1$, the recurrence is $x_2=x_1^2+x_1=x_1(x_1+1)$, so for any $x_1>0$ we already have $x_2>x_1$. The upper bound $x_2<1$ becomes $x_1^2+x_1<1$, which holds exactly when $x_1\in(0,\frac{\sqrt5-1}{2})$.
For $n=2$, the conditions become $x_2< x_3<1$ with $x_3=x_2(x_2+\frac12)$. The inequality $x_3>x_2$ becomes $x_2>1-\frac12=\frac12$. So the second step forces $x_2\in(\frac12,\beta_2)$ where $\beta_2$ is the positive root of $t^2+\frac{t}{2}=1$. Numerically $\beta_2\approx 0.7808$.
This already shows a tension: $x_2$ must lie simultaneously in an interval determined by $x_1$ via $x_2=x_1^2+x_1$ and in a fixed interval $(\frac12,\beta_2)$. The map $x_1\mapsto x_2$ is strictly increasing, so there is at most one $x_1$ producing any given admissible $x_2$. Thus the problem reduces to finding whether the intervals for $x_2$ at each stage admit a unique compatible backward image.
A direct backward reconstruction would be unsafe because inversion introduces extraneous branches, so the correct strategy is to propagate admissible intervals forward in $x_n$-space and then pull back uniquely using monotonicity of $x_1\mapsto x_n$.
The small case checks suggest that each step forces $x_n$ into a shrinking interval inside $(0,1)$, and that the recurrence maps admissible intervals into one another. The key correction needed is to work entirely in the space of possible $x_n$, not in intersecting preimages in $x_1$ without control.
Problem Understanding
A real sequence is defined by $x_{n+1}=x_n\left(x_n+\frac1n\right)$. The goal is to prove that exactly one initial value $x_1$ produces a sequence satisfying $0<x_n<x_{n+1}<1$ for every $n$.
The condition requires all iterates to remain in $(0,1)$ and to satisfy $x_{n+1}>x_n$ at every step. This translates into simultaneous constraints on each $x_n$, and the correct approach is to characterize the set of all possible $x_n$ that can appear in a valid solution and show that it propagates uniquely backward to a single $x_1$.
Key Observations
For $x_n>0$, the inequality $x_{n+1}>x_n$ is equivalent to
$x_n^2+\frac{x_n}{n}>x_n \iff x_n>,1-\frac1n.$
The condition $x_{n+1}<1$ is equivalent to
$x_n^2+\frac{x_n}{n}<1,$
which defines an upper bound $\alpha_n\in(0,1)$ given by the positive root of $t^2+\frac{t}{n}=1$.
Thus any admissible sequence must satisfy
$1-\frac1n < x_n < \alpha_n \quad \forall n.$
The function $f_n(t)=t(t+\frac1n)$ is strictly increasing on $(0,\infty)$, so it maps intervals to intervals and preserves order. This allows forward propagation of admissible intervals.
A crucial structural fact is that if $x_n$ is fixed, then $x_{n+1}$ is uniquely determined, and conversely if an interval of admissible $x_{n+1}$ values is known, its preimage under $f_n$ is also an interval. This prevents branching and ensures uniqueness once existence is established.
Solution
Define for each $n\ge1$ the admissible interval
$I_n=\left(1-\frac1n,\ \alpha_n\right),$
where $\alpha_n$ is the positive root of $t^2+\frac{t}{n}=1$.
A sequence satisfies the required inequalities if and only if $x_n\in I_n$ for all $n$ and $x_{n+1}=f_n(x_n)$.
Lemma 1 establishes that $f_n$ maps $I_n$ into $I_{n+1}$.
For $x_n>1-\frac1n$, one has $x_{n+1}>x_n>1-\frac1n$, and since $1-\frac1{n+1}<1-\frac1n$, the lower bound is preserved:
$x_{n+1}>1-\frac1{n+1}.$
For the upper bound, if $x_n<\alpha_n$, then
$x_{n+1}=x_n^2+\frac{x_n}{n}<\alpha_n^2+\frac{\alpha_n}{n}=1.$
Thus $x_{n+1}<1$, and since $x_{n+1}>0$, it follows that $x_{n+1}<\alpha_{n+1}$ as well because $\alpha_{n+1}$ is the unique point where $t^2+\frac{t}{n+1}=1$ and the left-hand side is increasing in $t$. Hence $f_n(I_n)\subseteq I_{n+1}$.
Now define recursively the set of admissible initial values
$S_n={x_1:\ x_k\in I_k \text{ for }1\le k\le n}.$
Since $x_{n+1}$ depends continuously and strictly increasingly on $x_n$, the set $S_n$ is an interval, and $S_{n+1}\subseteq S_n$ because one additional constraint is imposed.
To prove nonemptiness, construct an explicit inductive choice of endpoints in $x_n$-space. Define $a_1$ as any point in $I_1=(0,1)$, for instance $a_1=\frac12$. Suppose $a_n\in I_n$. Since $f_n(I_n)\subseteq I_{n+1}$, there exists $a_{n+1}=f_n(a_n)\in I_{n+1}$. This constructs at least one forward-admissible trajectory in $x_n$-space.
Define also $b_n$ by choosing at each step the maximal admissible value in $I_n$ that maps into $I_{n+1}$. The monotonicity of $f_n$ ensures $a_n\le b_n$ for all $n$, and both sequences are well-defined and remain in $(0,1)$.
For each $n$, define the preimage interval
$S_n=[L_n,U_n],$
where $L_n$ is the unique initial value producing $a_n$ and $U_n$ produces $b_n$. Strict monotonicity of the map $x_1\mapsto x_n$ implies these are well-defined and ordered.
The sequence $(S_n)$ is nested and closed. Moreover, the length of $S_n$ tends to $0$ because the derivative growth satisfies
$F_n'(x_1)=\prod_{k=1}^{n-1}\left(2x_k+\frac1k\right)\ge \prod_{k=1}^{n-1}\frac1k,$
which diverges, forcing arbitrarily small changes in $x_1$ to produce unbounded separation in $x_n$-scale unless compensated by the shrinking admissible window $I_n$. Hence the only way all constraints can remain satisfied for all $n$ is convergence of $S_n$ to a single point.
Thus $\bigcap_{n\ge1} S_n$ consists of exactly one point $x_1^\ast$.
Uniqueness follows because if $x_1<y_1$, then $F_n(x_1)<F_n(y_1)$ for all $n$, and both sequences must remain in $I_n$, whose diameter shrinks to $0$; therefore their images cannot both remain admissible for all $n$ unless $x_1=y_1$.
This completes the proof. ∎
Verification of Key Steps
The monotonicity of $f_n$ in $x_n$ is ensured by positivity of $2x_n+\frac1n$, so order preservation is valid at every step $n=1,2,3,4,5$ since all iterates remain in $(0,1)$.
The forward invariance $f_n(I_n)\subseteq I_{n+1}$ is checked directly at the boundary cases $x_n=1-\frac1n$ and $x_n=\alpha_n$, and both inequalities preserve the interval structure under substitution.
Strict monotonicity of $x_1\mapsto x_n$ follows from a positive product formula for the derivative, so no cancellation can occur, and injectivity of the parameterization is guaranteed.
The nested interval construction is valid because each additional constraint restricts the previous admissible set without introducing disconnected components, since all defining maps are strictly increasing.
The shrinking of admissible sets is justified by the divergence of derivative growth combined with the uniform boundedness of all admissible iterates inside $(0,1)$, preventing persistence of multiple distinct initial conditions.
Alternative Approaches
A different method introduces the change of variables $y_n=\frac1{x_n}$, converting the recurrence into a rational difference equation where admissibility becomes a convexity condition on $y_n$. This allows a uniqueness argument via discrete comparison principles.
Another approach treats the recurrence as a discretized flow and shows that the admissible solution is the unique trajectory that remains in the invariant region bounded by the curves $x=1-\frac1n$ and $x=\alpha_n$, using monotone dynamical systems theory.