Let $ n \ge 2 $ be an integer. Show that there exists a function $ f : \mathbb R \to \mathbb R $ such that $$ f ( x ) + f ( 2 x ) + \dots + f ( n x ) = 0 $$ for all $ x \in \mathbb R $, and $ f ( x ) = 0 $ if and only if $ x = 0 $.
Attemp:
Let's consider Cauchy's functional equation: $$ f ( 0 + 0 ) = f ( 0 ) + f ( 0 ) \implies f (0) = 0 \text . $$ $$ f ( 2 x ) = f ( x + x ) = f ( x ) + f ( x ) = 2 f ( x ) \text . $$ By induction it follows that $ f ( n x ) = n f ( x ) $. Let's get to the problem: $$ f ( x + 2 x + \dots + n x ) = f ( x ) + f ( 2 x ) + \dots + f ( n x ) = \\ = f ( x ) + 2 f ( x ) + \dots + n f ( x ) = ( 1 + 2 + \dots + n ) f ( x ) = 0 \text . $$ So $ f ( x ) = 0 \implies x = 0 $. So there is a function that satisfies the given conditions.
Am I right?