Minimum period of function such that $f\left(x+\frac{13}{42}\right)+f(x)=f\left(x+\frac{1}{6}\right)+f\left(x+\frac{1}{7}\right) $ Let $ f$ be a function from the set of real numbers $ \mathbb{R}$ into itself such for all $ x \in \mathbb{R},$ we have $ |f(x)| \leq 1,f(x)\neq constant $ and 

$$f\left(x+\dfrac{13}{42}\right)+f(x)=f\left(x+\dfrac{1}{6}\right)+f\left(x+\dfrac{1}{7}\right)$$
Prove that $ f$ is a periodic function (that is, there exists a non-zero real number $ c$ such $ f(x+c)=f(x)(c>0)$  for all $ x \in \mathbb{R}$),
  I have prove it by $c=1$,

also see this 1996 IMO Shortlist problem :http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1218567&&
and My question:
How find $c_{\min{}}$(and this question are very famous in china,But at last I can't any have prove  it.and this is nice problem 
in china this problem  are discussed somewhere:such
http://iask.sina.com.cn/b/10777191.html
http://tieba.baidu.com/p/6208699
http://bbs.pep.com.cn/thread-148404-1-1.html
http://tieba.baidu.com/p/1737938488
http://zhidao.baidu.com/link?url=fn0nKG7FaQwdPB1wLeMTajqXF8kpWbiMFIya4BnQB6YXHRb7-Fv8MOj_oHWs-k_Sf-c5N1p56SXDRfh8awQmca
http://emuch.net/html/201205/4454159.html
and so on
I guess $c=\dfrac{1}{7},c=\dfrac{1}{6}$ is true? ,and How find this $c$ minum value?
Thank you 
 A: Let $y=42x$ and $F(x)=f(\frac{x}{42})$.  The relation is 

$F(y+13) - F(y+7) - F(y+6) + F(y)=0$

or
$(T^6-1)(T^7-1)F=0$ 
where $T$ is the operator that takes a function $q(y)$ to $q(y+1)$.  That is,

$(T-1)^2 \displaystyle \prod_i (T - \omega_i) = 0$ for some $11$ distinct roots of unity $\omega_i$ of orders $7, 6, 3$ and $2$.  

The behavior of $F$ on sets $y + \mathbb{Z}$ is independent for $y \in [0,1)$.  In every such set, by the theory of linear recurrence relations, the solution is $F(y+n)=a(y)(1) + b(y)n + \sum g_i(y)\omega_i^n$ where the functions $a,b,g_i$ are coefficients that depend only on the value of $y$ and not on $n$.  Because $F$ is bounded, $b(y) = 0$ for all $y$.  Therefore, $F(y+42)=F(y)$ for all $y$, or in terms of $f$,

$f(x + 1) = f(x)$ 

and period $1$ can be realized by taking nonzero coefficients for two of the $\omega_i$ that are a primitive $6$th and $7$th root of $1$.
This proves that $c$, the maximum value of the minimum period, is $1$.
It follows from the same reasoning that  $F$ satisfies the stronger relation $(T^5 + T^4 + T^3 + T^2 + T +1)(T^7-1)F=0$, or

$$F(y+12) + F(y+11) + F(y+10) + F(y+9) + F(y+8) + F(y+7) = F(y+5) + F(y+4) + F(y+3) + F(y+2) + F(y+1) + F(y)$$

This degree 12 recurrence implies the degree 13 recurrence given in the problem, and solutions of the degree 12 are the same as bounded solutions of the degree 13 relation. Every polynomial in $T$ that evaluates to $0$ on $F$ is a multiple of the degree $12$ polynomial.  The smallest $e$ for which the polynomial $T^e - 1$ is divisible by $(T^5 + T^4 + T^3 + T^2 + T +1)(T^7-1)$ is $e=42$, which is another way of stating the question. 
