TITLE: Is my solution correct in terms of proof? I can't prove directly that $f(x)=4$.
QUESTION [5 upvotes]: Given  $f:\mathbb{R}\rightarrow \mathbb{R}^+$ and the following conditions:

$f(x+2)\cdot f(x+3)=16$,
$f(x)+f(-x)=8$,

we have to find integral $\int_{-8}^8 f(x-2019)dx$. So, if we plug in $0$, then $f(0)+f(0)=8$. Therefore, $f(0)=4$. Then plug in $-2$ in first equation,  we get $f(0)\cdot f(1)=16$, therefore $f(1)=4$, analogically plug $x=-1$ and I think that $f(x)=4$, so answer will be  $64$ (value after integrating). Is this correct? How can i prove it in a better manner?

REPLY [3 votes]: Here's a more direct solution:
From 1 we have $\displaystyle f(1+x)=\frac{16}{f(x)}$ and from 2 we have $f(x) = 8-f(-x)$.
Then $$f(x) = f(1+x-1) = \frac{16}{f(x-1)} = \frac{16}{8-f(1-x)} = \cfrac{16}{8-\cfrac{16}{f(-x)}} = \cfrac{16}{8-\cfrac{16}{8-f(x)}}$$
Solving for $f(x)$ we get $f(x)=4$.