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Algebra Difficulty 4.8 AIME Prove it India

Find all functions $f$ from the set of real numbers to itself satisfying
$f(x(1+y)) = f(x)(1+f(y))$
for all real numbers $x, y$ .

Solution

If $f$ is not identically zero, then by standard substitutions we get that $f(x) = x$ for $x = 0, \pm 1$ . Using these it is easy to see that $f$ is additive and multiplicative on the set of real numbers. It then follows by induction and continuity that $f(x) = x$ for all real $x$ . $\square$

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.