Question

Find all the functions f (from rational numbers to rational numbers) such that
f(x+y) + f(x-y) = 2f(x) + 2f(y), for all rationals x,y.

Answer 1

Find all the functions $f$ (from rational numbers to rational numbers) such that

$$f(x + y) + f(x - y) = 2f(x) + 2f(y), \text{ for all rationales } x, y.$$

Solution.

Find all continuous $f: \mathbb{R} \to \mathbb{R}$ which satisfy

$$f(x + y) + f(x - y) = 2[f(x) + f(y)] .$$

Fix $\delta > 0$ and let $C = \int_0^\delta 2f(y)dy$.

Then

$$\begin{aligned} 2\delta f(x) + C &= \int_0^\delta 2[f(x) + f(y)] dy = \int_0^\delta (f(x + y) + f(x - y) dy) = \int_x^{x + \delta} f(y) dy + \\ \int_{x - \delta}^x f(y) dy &= \int_{x - \delta}^{x + \delta} f(y) dy \end{aligned}$$

Now since $f$ is continuous, the last expression is a differentiable function of $x$ and thus the first expression must also be differentiable; hence $f$ is differentiable. By induction, $f$ is infinitely differentiable.

Differentiating (1) first with respect to $y$, we arrive at:

$$f'(x + y) - f'(x - y) = 2f'(y) .$$

Differentiating once more with respect to $x$, we have:

$$f''(x + y) = f''(x - y),$$

so $f''$ is constant.

It follows that $f(x) = ax^2 + bx + c$ are the only potential solutions.

Substituting $x = y = 0$ in (1) and (2) implying $f(0) = f'(0) = 0$.

Hence

$$f(x) = ax^2(a \in \mathbb{R})$$

It is easy to check that all such $f$ are indeed solutions.

Answer:

$$f(x) = ax^2(a \in \mathbb{R})$$