Cauchy's equation f(x + y) = f(x) + f(y)

Step 1: basic values. Taking $x=y=0$ gives $f(0)=2f(0)$, so $f(0)=0$. Taking $y=-x$ gives $0=f(x)+f(-x)$, so $f$ is odd.

Step 2: integers. By induction $f(nx)=nf(x)$ for every integer $n\ge1$, and by oddness for every integer $n$. With $x=1$: $f(n)=na$.

Step 3: rationals. Let $\frac pq$ with $p\in\mathbb{Z}$, $q\ge1$. Then $q\,f\!\left(\frac pq\right)=f(p)=pa$, so $f\!\left(\frac pq\right)=a\,\frac pq$.

Step 4: all reals. The functions $x\mapsto f(x)$ and $x\mapsto ax$ are continuous and agree on $\mathbb{Q}$. By the problem *A continuous function that vanishes on the rationals* they agree everywhere.

Answer. $f(x)=ax$ with $a=f(1)$. Conversely every such function works.

Remark. Without continuity there are many weird (non-measurable) solutions, built with the axiom of choice. So here a single hypothesis, continuity, turns a "vast" family of solutions into just the lines through the origin.