Additive functions with f(x²) = f(x)²

Positivity. For every $t\ge0$, $t=x^2$ for some $x$, so $f(t)=f(x)^2\ge0$.

Monotonicity. If $x\le y$ then $f(y)=f(x)+f(y-x)\ge f(x)$.

Linearity. As in Cauchy's equation over $\mathbb{Q}$, $f(q)=cq$ for every rational $q$, with $c=f(1)\ge0$. Take $x\in\mathbb{R}$ and rationals $p<x<q$. By monotonicity, $cp\le f(x)\le cq$. Letting $p,q\to x$ gives $f(x)=cx$.

The constant. The condition $f(1)=f(1)^2$ gives $c=c^2$, so $c\in\{0,1\}$.

Answer. $f\equiv0$ and $f(x)=x$, which clearly satisfy both conditions.

Lesson. Wild additive functions are killed by any condition that gives a sign or boundedness: look for a hidden square.