<s>[INST] If $ f(x) = f(x+1) $ and $ f''(x) + f(x) = \frac {1} { f( x+ \frac{3}{4}) } $ then $ f(x) = f( x+ \frac{1}{4}) $ I was trying to solve following problem. 
Let $ f: \Bbb R \rightarrow \Bbb R $ be a twice differentiable function such that $ f(x) = f(x+1) $ for all $x$. If $$ f''(x) + f(x) = \frac {1} { f( x+ \frac{3}{4}) } $$ holds for all x . Then prove that $ f(x) = f( x+ \frac{1}{4})  $ for all $x$      To sove this problem i replaced $x$ by $x + \frac {1}{4} $ in $ f''(x) + f(x) = \frac {1} { f( x+ \frac{3}{4}) } $ so i got $ f''(x + \frac{1}{4}) + f(x + \frac{1}{4}) = \frac {1} { f( x+ 1) }$ =$ \frac{1}{ f( x) } $. I got stuck here how to proceed further. Thanx for your help.
 [/INST] Let me try it : We put $x=\ln(y)$ and $y\neq 0$
The equality :$$ f''(x) + f(x) = \frac {1} { f( x+ \frac{3}{4}) } $$
Becomes :
$$\frac{f''(\ln(y))-f'(\ln(y))}{y^2}+f(\ln(y))=\frac{1}{f(\ln(y)+0.75)}$$
Now we assume $$f''(\ln(y))-f'(\ln(y))\neq 0\quad \forall y\neq 0$$
But :
$$\frac{f''(\ln(y))-f'(\ln(y))}{y^2}+f(\ln(y))=\frac{1}{f(\ln(y)+0.75)}=\frac{1}{f(\ln(y)+0.75+1)}=\frac{f''(\ln(y)+1)-f'(\ln(y)+1)}{y^2e^2}+f(\ln(y)+1)$$
Or :
$$\frac{f''(\ln(y))-f'(\ln(y))}{y^2}+f(\ln(y))=\frac{f''(\ln(y)+1)-f'(\ln(y)+1)}{y^2e^2}+f(\ln(y)+1)$$
Wich implies $$\frac{1}{y^2e^2}=\frac{1}{y^2}$$
Wich is a contradiction .
So we deduce that :
$$\frac{f''(\ln(y))-f'(\ln(y))}{y^2}=0 \quad \forall y\neq 0$$
Wich implies $f(\ln(y))=Constant$
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