" wrote: let $ a=\frac 16,b=\frac 17$ so $ a+b=\frac{13}{42}$ and: $ f(x+a+b)+f(x)=f(x+a)+f(x+b)$ now in this equation,if we plug in $ x+a,x+2a,\ldots ,x+5a$ instead of $ x$,and then sum the equations up,we get that: $ f(x+1+b)+f(x)=f(x+1)+f(x+b)$ now in this equation,plug in $ x+b,x+2b,\ldots ,x+6b$ instead of $ x$,and then sum them up to get that: $ f(x+2)-f(x+1)=f(x+1)-f(x)=c$ hence for every natural number $ n$,we have: $ f(x+n)-f(x)=nc$ now note that if $ c\neq 0$,then $ f(x+n)$ would tend to $ \infty$ as $ n$ gets bigger,so according to $ |f(x)|\leq 1$ we must have $ c=0$,so $ f(x+1)-f(x)=0$.i.e. $ f(x+1)=f(x)$ for all $ x\in\mathbb{R}$. QED

I claim that for all real $a$, $f(a+1) = f(a)$. If we define $f_a(x) = f(x-a)$, we see that $f_a$ trivially satisfies the functional equation of $f$, and that $f_a(0) = f_a(1)$ would imply that $f(a) = f(a+1)$. Let $c_i = f_a\left(\frac{i}{42}\right)$ for all $i \geq 0$. We will show that $c_{n+42} = c_n$ for all integers $n$. $\{c_i\}$ satisfies $|c_n| \leq 1$ and $c_{n+13} = c_{n+7} + c_{n+6} - c_n$ for all integers $n$. Its characteristic polynomial is $x^{13} - x^7 - x^6 + 1 = (x^7 - 1)(x^6 - 1)$, which has 1 as a double root and the 6th and 7th roots of unity not equal to 1 as single roots. Let $r_1, r_2, \ldots, r_{11}$ be the 6th and 7th roots of unity not equal to 1, in some order. $c_n$ must satisfy $c_n = an + a_01^n + a_1 r_1^n + a_2 r_2^n + \cdots + a_{11} r_{11}^n$, where $a, a_0, a_1, \cdots, a_{11}$ are some complex numbers. If $a \neq 0$, then we can pick some integer $n$ with $|an| > 1 + |a_1| + |a_2| + \cdots + |a_{11}|$, whence \begin{align*} |c_n| &= |an + a_1 r_1^n + a_2 r_2^n + \cdots + a_{11} r_{11}^n| \\ &\geq |an| - |a_1 r_1^n + a_2 r_2^n + \cdots + a_{11} r_{11}^n| \\ &\geq |an| - (|a_1| + |a_2| + \cdots + |a_{11}|) \\ &> 1, \end{align*} a contradiction. Since $r_i^{42} = r_i$, $c_{n+42} = c_n$. In particular, $c_{42} = c_0$, so $f_a(1) = f_a(0)$, i.e., $f(a+1) = f(a)$, as desired.

IMO shortlist 1996