IMO 1968: f(x + a) = ½ + √(f(x) − f(x)²)

Lower bound. The square root is defined, so $f(x)-f(x)^2\ge0$ and the equation shows $f(x+a)\ge\frac12$ for every $x$. Hence $f(t)\ge\frac12$ for all real $t$ (put $x=t-a$).

Squaring. From the equation, $\left(f(x+a)-\frac12\right)^2=f(x)-f(x)^2=\frac14-\left(f(x)-\frac12\right)^2$. Thus
$$\left(f(x+a)-\tfrac12\right)^2+\left(f(x)-\tfrac12\right)^2=\tfrac14\qquad\text{for all }x.$$

Shifting by $a$. Replace $x$ by $x+a$:
$$\left(f(x+2a)-\tfrac12\right)^2+\left(f(x+a)-\tfrac12\right)^2=\tfrac14.$$
Subtracting the two displayed relations,
$$\left(f(x+2a)-\tfrac12\right)^2=\left(f(x)-\tfrac12\right)^2.$$
Since $f(x+2a)-\frac12\ge0$ and $f(x)-\frac12\ge0$, we get $f(x+2a)=f(x)$.

Example. $f(x)=\dfrac12+\dfrac12\left|\sin\dfrac{\pi x}{2a}\right|$. Put $\theta=\frac{\pi x}{2a}$. Then $f(x+a)=\frac12+\frac12|\cos\theta|$, and $f(x)-f(x)^2=\frac14-\frac14\sin^2\theta=\frac14\cos^2\theta$, whose square root is $\frac12|\cos\theta|$ ✓.

Lesson. When a relation contains a square root, isolate it and square: the resulting symmetric identity in $f(x)$ and $f(x+a)$ often yields periodicity by shifting once.