
6. Find all functions $f: \mathbf{R} \rightarrow \mathbf{R}$ such that

$$
f\left(x^{2}+y f(z)\right)=x f(x)+z f(y)
$$

for all $x, y, z$ in $\mathbf{R}$. (Here $\mathbf{R}$ denotes the set of all real numbers.)


Solution: Taking $x=y=0$ in (1), we get $z f(0)=f(0)$ for all $z \in \mathbf{R}$. Hence we obtain $f(0)=0$. Taking $y=0$ in (1), we get

$$
f\left(x^{2}\right)=x f(x)
$$

Similarly $x=0$ in (1) gives

$$
f(y f(z))=z f(y)
$$

Putting $y=1$ in (3), we get

$$
f(f(z))=z f(1) \quad \forall z \in \mathbf{R}
$$

Now using (2) and (4), we obtain

$$
f(x f(x))=f\left(f\left(x^{2}\right)\right)=x^{2} f(1)
$$

Put $y=z=x$ in (3) also given

$$
f(x f(x))=x f(x)
$$

Comparing (5) and (6), it follows that $x^{2} f(1)=x f(x)$. If $x \neq 0$, then $f(x)=c x$, for some constant c. Since $f(0)=0$, we have $f(x)=c x$ for $x=0$ as well. Substituting this in (1), we see that

$$
c\left(x^{2}+c y z\right)=c x^{2}+c y z
$$

or

$$
c^{2} y z=c y z \quad \forall y, z \in \mathbf{R}
$$

This implies that $c^{2}=c$. Hence $c=0$ or 1 . We obtain $f(x)=0$ for all $x$ or $f(x)=x$ for all $x$. It is easy to verify that these two are solutions of the given equation.

