Periodicity of this function Let $f(x)\colon \Bbb R\to\Bbb R$ be a bounded function satisfying the following condition:
$$f(x+\frac{13}{42})+f(x)=f(x+\frac16)+f(x+\frac17),   \forall x\in\Bbb R$$
Is the function $f(x)$ periodic?
 A: The answer is YES.
Put $g(x)=f(\frac{x}{42})$. Since
$f(\frac{x}{42}+\frac{13}{42})+f(\frac{x}{42})=
f(\frac{x}{42}+\frac16)+f(\frac{x}{42}+\frac17) $, we see that
for any $x$,
$$
g(x+13)=g(x+6)+g(x+7)-g(x) \tag{1}
$$
Let $h(x)=g(x+7)-g(x)$. Then (1) says that $h(x+6)=h(x)$. Now for 
$x\in{\mathbb R}$ and $n\in{\mathbb Z}$ put $g_{x}(n)=g(x+n)$ and
$h_x(n)=g_x(n+7)-g_x(n)$. We still have $h_x(n+6)=h_x(n)$.
So for $n\in{\mathbb Z}$, $h_x(n)$ only
depends on the value of $n$ modulo $6$ (and also depends on $x$, of course). 
For any $n\in{\mathbb Z}$, the set of values 
$n,n+7,n+2\times 7, \ldots ,n+5\times7$ modulo
$6$ is simply the set of all values modulo $6$.
We deduce
$$
\sum_{k=0}^5 h_x(n+7k)=\sum_{j=0}^5 h_x(j) \tag{2}
$$
If we denote by $C(x)$ the common value above, we have
$$
C(x)=\sum_{k=0}^5 h_x(n+7k)=
\sum_{k=0}^5 g_x(n+7(k+1))-g_x(n+7k)=g_x(n+42)-g_x(n) \tag{3}
$$
As $f$ is bounded, $g_x$ is bounded also, so $C(x)=0$ for any $x$.
Hence $g$ is $42$-periodic, and $f$ is therefore $1$-periodic.
Remark : this is false if $f$ isn’t bounded. 
Let $g_1$ be the map $\lbrace 1,2,3 \ldots, 42 \rbrace$ defined by the rules
$$
g_1(x)=\left\lbrace
\begin{array}{lcl}
2k-2, & \text{if } & 7\times (k-1) \leq x \leq 7\times k, \\
2k-1, & \text{if } & x=7\times k \
\end{array}\right.
$$
for $1\leq k\leq 6$. Now extend $g_1$ to a map $g_2$ defined on $\mathbb Z$,
by putting $g_2(42q+r)=2q+g_1(r)$. Then $g_2$ satisfies 
$$
g_2(x+13)=g_2(x+6)+g_2(x+7)-g_2(x) (x\in {\mathbb Z})
$$
Extend $g_2$ to a map $g_3$ defined on :${\mathbb R}$ by putting
$g_3(x)=0$ when $x\not\in{\mathbb Z}$. Then $f(x)=g(42x)$ satsifies the identity
but is not periodic (in fact $f(x+42)=f(x)+2$).
A: you may write the fucntion as
$$ f(x+13/42)+f(x)=f(x+7/42)+f(x+6/42)=g(x) $$
and $ 42=13.4$ $42=7.6 $ 
if this stuff is periodic there will be a number T so $ g(x)=g(x+T) $
this gives $ 42T+13 $ $42T+7$ and $42T+6 $
now if the normal period of the function $ f(x)$ is q so $ f(x+q)=f(x)$ then
