TITLE: Hard functional equation: $ f \big ( x y + f ( x ) \big) = f \big( f ( x ) f ( y ) \big) + x $
QUESTION [13 upvotes]: Let $ \mathbb R _ { > 0 } $ be the set of positive real numbers. Find all functions $ f : \mathbb R _ { > 0 } \to \mathbb R _ { > 0 } $ such that
$$ f \big ( x y + f ( x ) \big) = f \big( f ( x ) f ( y ) \big) + x $$
for all positive real numbers $ x $ and $ y $.

What I thought: We could change $ x $ by $ y $, and then subtract.
Source: Brazil National Olympiad 2019 #3

REPLY [2 votes]: $$f(xy + f(x)) = f(f(x)\cdot f(y)) + x.$$
Let us substitute $y = 1$:
$$f(x + f(x)) = f(f(x)\cdot f(1)) + x.$$
Now, let us substitute into the initial equation $x = 1$:
$$f(y + f(1)) = f(f(1)\cdot f(y)) + 1.$$
In the latter equation, let us replace $y$ by $x$:
$$f(x + f(1)) = f(f(x)\cdot f(1)) + 1.$$
Now, we have 
$$
f(x + f(x)) = f(f(x)\cdot f(1)) + x \\
f(x + f(1)) = f(f(x)\cdot f(1)) + 1
$$
Let $g(x) = f(f(x)\cdot f(1))$. Then, we have
$$
f(x + f(x)) = g(x) + x \\
f(x + f(1)) = g(x) + 1
$$
We see that a linear shift in the argument of the function $f(x)$ results in a linear shift in the values of function g(x).

Shifting by $f(x)$, i.e. $f(x + f(x))$ implies shifting by $x$.
Shifting by $f(1)$, i.e. $f(x + f(1))$ implies shifting by $1$.  

This is true if both $f(x)$ and $g(x)$ are linear functions, particularly if 
$$f(x) = x.$$
Let us check that $f(x) = x$ is a solution:
$$f(xy + f(x)) = f(f(x)\cdot f(y)) + x \Leftrightarrow 
f(xy + f(x)) = xy + x \text{ and }f(f(x)f(y)) + x = xy + x \text{ (TRUE). }$$