IMO 1968 Problem 5
The functional equation applies a transformation
Proposed by: -
Verified: yes
Verdicts: PASS + PASS
Solve time: 4m05s
Problem
Let $f$ be a real-valued function defined for all real numbers $x$ such that, for some positive constant $a$, the equation $$ f(x + a) = \frac{1}{2} + \sqrt{f(x) - (f(x))^2} $$ holds for all $x$.
(a) Prove that the function $f$ is periodic (i.e., there exists a positive number $b$ such that $f(x + b) = f(x)$ for all $x$).
(b) For $a = 1$, give an example of a non-constant function with the required properties.
Exploration
The functional equation applies a transformation
$$T(y)=\frac12+\sqrt{y-y^2}.$$
The expression under the square root becomes
$$y-y^2=\frac14-(y-\tfrac12)^2,$$
so the map is confined to the interval where this is nonnegative, hence $y\in[0,1]$ for all relevant values of $f$.
A natural attempt is to reparameterize $f(x)$ using a trigonometric substitution because $\frac14-(y-\tfrac12)^2$ suggests a circle. Writing
$$2f(x)-1 = u(x)$$
transforms the equation into
$$u(x+a)=\sqrt{1-u(x)^2}.$$
This strongly resembles the relationship between sine and cosine.
If $u(x)=\cos\theta(x)$ with $\theta(x)\in[0,\frac{\pi}{2}]$, then
$$\sqrt{1-u(x)^2}=\sin\theta(x),$$
so the iteration becomes a rotation of the angle by $\frac{\pi}{2}$ up to identification of sine and cosine. This suggests that two steps return to the original value, producing periodicity with period $2a$.
A delicate point is the sign choice in the square root, which forces nonnegativity and constrains the range of $u$ and hence $f$.
For construction when $a=1$, defining $\theta(x+1)=\frac{\pi}{2}-\theta(x)$ suggests a piecewise constant function on an interval of length $2$ that alternates between two complementary values.
Problem Understanding
This is a Type B problem for part (a), asserting that any function satisfying a nonlinear functional equation must be periodic. The second part requests an explicit nonconstant example when $a=1$.
The key difficulty is that the equation does not give a linear recurrence but a nonlinear transformation that involves a square root, so standard periodicity arguments do not apply directly. The structure becomes visible only after recognizing a hidden trigonometric form.
The expected conclusion is that every such function must satisfy
$$f(x+2a)=f(x),$$
so periodicity follows with period $2a$.
Proof Architecture
Lemma 1 asserts that $0\le f(x)\le 1$ for all real $x$, derived from the nonnegativity of the square root and invariance under the functional equation.
Lemma 2 asserts that $2f(x)-1\ge 0$ for all $x$, hence $f(x)\in[\frac12,1]$, obtained by propagating nonnegativity through the functional equation.
Lemma 3 introduces the substitution $u(x)=2f(x)-1$ and proves that it satisfies
$$u(x+a)=\sqrt{1-u(x)^2}.$$
Lemma 4 constructs an angle function $\theta(x)\in[0,\frac{\pi}{2}]$ such that $u(x)=\cos\theta(x)$ and proves that it is well-defined for all $x$.
Lemma 5 proves the recurrence $\theta(x+a)=\frac{\pi}{2}-\theta(x)$.
Lemma 6 deduces $\theta(x+2a)=\theta(x)$ and hence periodicity of $f$ with period $2a$.
The most delicate point is Lemma 4, where the correct choice of trigonometric parametrization must remain consistent for all $x$.
Solution
Lemma 1
For all real $x$, one has $0\le f(x)\le 1$.
The expression under the square root in the defining equation is $f(x)-f(x)^2=f(x)(1-f(x))$, which must be nonnegative for every $x$ since it equals the argument of a real square root. This forces $f(x)\in[0,1]$ for all $x$.
This establishes that the functional equation never leaves the unit interval, preventing divergence outside the domain of the square root.
Lemma 2
For all real $x$, one has $f(x)\ge \frac12$.
From the functional equation,
$$f(x+a)=\frac12+\sqrt{f(x)-f(x)^2},$$
the right-hand side is at least $\frac12$, hence $f(x+a)\ge \frac12$ for all $x$. Replacing $x$ by $x-a$ yields $f(x)\ge \frac12$ for all $x$.
This shows that the lower half of the interval is excluded for all inputs.
Lemma 3
Defining $u(x)=2f(x)-1$, one has $u(x+a)=\sqrt{1-u(x)^2}$.
Substituting $f(x)=\frac{1+u(x)}{2}$ gives
$$f(x)-f(x)^2=\frac{1+u(x)}{2}-\frac{(1+u(x))^2}{4} =\frac{1-u(x)^2}{4}.$$
Taking square roots yields
$$\sqrt{f(x)-f(x)^2}=\frac12\sqrt{1-u(x)^2}.$$
Multiplying the functional equation by $2$ gives
$$1+u(x+a)=1+\sqrt{1-u(x)^2},$$
hence
$$u(x+a)=\sqrt{1-u(x)^2}.$$
This reformulation isolates the essential transformation acting on a single bounded variable.
Lemma 4
There exists a function $\theta(x)\in[0,\frac{\pi}{2}]$ such that $u(x)=\cos\theta(x)$ for all $x$.
Lemma 2 gives $u(x)\in[0,1]$. The interval $[0,1]$ is exactly the range of the cosine function on $[0,\frac{\pi}{2}]$, so for each $x$ there exists a unique $\theta(x)\in[0,\frac{\pi}{2}]$ satisfying $u(x)=\cos\theta(x)$.
This defines $\theta(x)$ unambiguously for all $x$.
This step establishes a global angular parametrization of the system without ambiguity in sign or branch.
Lemma 5
The identity $\theta(x+a)=\frac{\pi}{2}-\theta(x)$ holds for all $x$.
From Lemma 3,
$$u(x+a)=\sqrt{1-u(x)^2}.$$
Writing $u(x)=\cos\theta(x)$ gives
$$u(x+a)=\sqrt{1-\cos^2\theta(x)}=\sin\theta(x).$$
Since $\theta(x)\in[0,\frac{\pi}{2}]$, both $\sin\theta(x)$ and $\cos(\frac{\pi}{2}-\theta(x))$ lie in $[0,1]$, and they are equal. The uniqueness of the representation in Lemma 4 implies
$$\theta(x+a)=\frac{\pi}{2}-\theta(x).$$
This converts the nonlinear functional equation into a linear affine transformation on the angle variable.
Lemma 6
For all $x$, one has $\theta(x+2a)=\theta(x)$.
Applying Lemma 5 twice gives
$$\theta(x+2a)=\frac{\pi}{2}-\theta(x+a)=\frac{\pi}{2}-\left(\frac{\pi}{2}-\theta(x)\right)=\theta(x).$$
Therefore $\theta$ is periodic with period $2a$, and since $u(x)=\cos\theta(x)$ and $f(x)=\frac{1+u(x)}{2}$, both $u$ and $f$ inherit the same period.
This completes the proof that every solution is periodic with period $2a$.
Construction for $a=1$
Define a function $\theta:\mathbb{R}\to[0,\frac{\pi}{2}]$ by specifying it on $[0,2)$ and extending by the rule $\theta(x+2)=\theta(x)$. On $[0,1)$ set $\theta(x)=0$, and on $[1,2)$ set $\theta(x)=\frac{\pi}{2}$.
For $x\in[0,1)$ one has $x+1\in[1,2)$, hence
$$\theta(x+1)=\frac{\pi}{2}=\frac{\pi}{2}-0=\frac{\pi}{2}-\theta(x).$$
For $x\in[1,2)$ one has $x+1\in[2,3)$, hence using periodicity,
$$\theta(x+1)=\theta(x-1)=0=\frac{\pi}{2}-\frac{\pi}{2}=\frac{\pi}{2}-\theta(x).$$
Thus $\theta(x+1)=\frac{\pi}{2}-\theta(x)$ holds for all $x$.
Define
$$f(x)=\frac{1+\cos\theta(x)}{2}.$$
Then $f$ is nonconstant since it takes the values $1$ and $\frac12$ on the two subintervals. The identity $\theta(x+1)=\frac{\pi}{2}-\theta(x)$ implies
$$\cos\theta(x+1)=\sin\theta(x),$$
and substituting into the original equation verifies that $f$ satisfies the required functional relation for $a=1$.
This completes the proof. ∎
Verification of Key Steps
The transition from $f(x)-f(x)^2$ to $\frac{1-u(x)^2}{4}$ relies on a direct algebraic expansion of $(1+u)^2$; any omission of cross terms would incorrectly alter the transformation law.
The deduction that $\theta(x)$ lies in $[0,\frac{\pi}{2}]$ is essential, since extending beyond this interval would introduce sign ambiguity in the square root and break the identification between $\sqrt{1-\cos^2\theta}$ and $\sin\theta$.
The periodicity argument depends critically on applying the involution $\theta\mapsto \frac{\pi}{2}-\theta$ twice, where any incorrect handling of branch choices would destroy the cancellation leading to $2a$-periodicity.
Alternative Approaches
A different approach replaces trigonometric parametrization with the invariant relation
$$(2f(x)-1)^2 + (2f(x+a)-1)^2 = 1,$$
which encodes a rotation on the unit circle. Iterating this relation produces a geometric rotation system whose period is $2a$.
Another approach interprets the transformation as a deterministic dynamical system on $[0,1]$ with a conjugacy to a piecewise rotation, avoiding explicit angle functions but requiring a more involved topological argument to control branch selection.