$P(0,0): f(0)=0$ $P(x,0): f(ax)=f(x)$ Thus, $f(x+y)=f(x)+f(y)$, which is Cauchy. Since $f(cx)=cf(x)$ for $c \in \mathbb{Q}$, $x \in\mathbb{R}$. $a \notin \mathbb{Q}$. Since $\mathbb{Q}$ is dense over $\mathbb{R}$, we can find $q_1<a<q_2 \implies c(q_1)=f(q_1)<cf(a)<f(q_2)=c(q_2)$ so eventually, by finding closer and closer pairs of rationals around $a$, we can find that $f(a)=c(a)$, i.e. only $a=1$ will lead to a nonconstant function.