All the questions of the test Partition reals onto disjoint arithmetic progressions with difference $a$. Then we may define $f$ in every progression $\{x_0+na\}$, by chosing $f(x_0)$ arbitrary. So, we have a lot of such functions. Maybe, I missed something? Or there are additional conditions, like continiuty?

Same idea as Fedor Petrov: Let $g(x) = f(x) - \frac{x^2}{2a}+\frac{x}{2}$ Then $g(a+x)=g(x)$. So it is enough to define arbitrarily $g(x)$ on $[0,a)$ and it will be periodic. From $g(x)$ we construct $f(x)$.

There are no additional conditions. This is the problem.

The function $f(x)=-\frac{x(x-a)}{2a}$ satisfys the condition Is it the only one ??

I found something : Putting $\alpha =f(0)$ we can show by induction that $f(n)=-\frac{n(n-a)}{2a} +\alpha$ for all $n \in \mathbb N$ So by using the method of Cauchy we can prove that $f(x)=-\frac{x(x-a)}{2a} +\alpha$ for all $x \in \mathbb R$

u can find it only for n=ak where k is a positive integer

prowler wrote: Let $g(x) = f(x) - \frac{x^2}{2a}+\frac{x}{2}$ Then $g(a+x)=g(x)$. $g(a+x)= f(a+x) - \frac{(a+x)^2}{2a}+\frac{a+x}{2}$ $= f(a+x) - \frac{a^2+2ax+x^2}{2a}+\frac{a+x}{2}$ $= f(a+x) - \frac{a^2}{2a}-\frac{2ax}{2a}-\frac{x^2}{2a}+\frac{a+x}{2}$ $= f(a+x) - \frac{a}{2}-x-\frac{x^2}{2a}+\frac{a}{2}+\frac{x}{2}$ $= f(a+x) -x-\frac{x^2}{2a}+\frac{x}{2}$ $= f(a+x) -\frac{x^2}{2a}-\frac{x}{2}$ $= f(x)-x -\frac{x^2}{2a}-\frac{x}{2}$ $= f(x)-\frac{x^2}{2a}-\frac{3x}{2}$ $= f(x)-\frac{x^2}{2a}+\frac{x}{2}-2x$ $= g(x)-2x$ Did I do anything wrong?

No I don't think it's wrong. So what prowler said was wrong

anyway, i found out what the error was. the function was actually $g(x) = f(x) + \frac{x^2}{2a}-\frac{x}{2}$ Then $g(a+x)=g(x)$.

I think it is easier to observe $f(x) = g(x) - x[\frac{x}{a}]$. g is the same as above.