Irish Mathematical Olympiad

To say that the graph of $f$ is symmetric about the line $x = a$ is to mean that $f(x_1) = f(x_2)$ whenever $x_1, x_2$ are real numbers whose sum is $2a$. Hence, by hypothesis,
$$f\left(\frac{1}{4} - x\right) = f\left(\frac{1}{4} + x\right) \quad \text{and} \quad f\left(\frac{3}{4} - x\right) = f\left(\frac{3}{4} + x\right) \quad \text{for all } x \in \mathbb{R}.$$
Letting $y = \frac{1}{4} - x$, we deduce from the first condition that $f(y) = f(\frac{1}{2} - y)$ for all $y \in \mathbb{R}$. Letting $y = \frac{3}{4} - x$ in the second relation, we see that $f(y) = f(\frac{3}{2} - y)$ for all $y \in \mathbb{R}$. Therefore,
$$f\left(\frac{1}{2} - y\right) = f\left(\frac{3}{2} - y\right) \quad \text{for all } y \in \mathbb{R}.$$
Hence, letting $x = \frac{1}{2} - y$ we see that $f(x) = f(1 + x)$ for all $x \in \mathbb{R}$, as desired.

Consider the function
$$f(x) = \sin^2(2\pi x).$$
Then, for all $x \in \mathbb{R}$,
$$\begin{aligned} f\left(\frac{1}{4} - x\right) &= \sin^2\left(2\pi\left(\frac{1}{4} - x\right)\right) \\ &= \left(\sin\left(\frac{\pi}{2}\right) \cos(2\pi x) - \cos\left(\frac{\pi}{2}\right) \sin(2\pi x)\right)^2 \\ &= \cos^2(2\pi x) = f\left(\frac{1}{4} + x\right). \end{aligned}$$

$$\begin{aligned} f\left(\frac{3}{4}-x\right) &= \sin^2\left(2\pi\left(\frac{3}{4}-x\right)\right) \\ &= \left(\sin\left(\frac{3\pi}{2}\right) \cos(2\pi x) - \cos\left(\frac{3\pi}{2}\right) \sin(2\pi x)\right)^2 \\ &= \cos^2(2\pi x) = f\left(\frac{3}{4}+x\right). \end{aligned}$$
Since it's nonnegative and not a constant function, this $f$ does the job.