Functions are awesome! - Part 2

Let $P(x,y,z)$ be the statement $f(x^2+yf(z))=xf(x)+zf(y)$ .

$P(0,0,0)$ gives $\begin{aligned} f(0^2+0f(0))&=0f(0)+0f(0)\\ f(0)&=0 \end{aligned}$ Now assume $f(a)=0$ for some $a \in \mathbb{R}, a \neq 0$ .

For any $x \in \mathbb{R}$ , $P(0,x,a)$ gives $\begin{aligned} f(xf(a))&=af(x)\\ f(0)&=af(x)\\ 0&=af(x)\\ f(x)&=0 \end{aligned}$ Note that $f(x)=0$ is a solution, as $P(x,y,z)$ gives $\begin{aligned} f(x^2+yf(z))&=xf(x)+zf(y)\\ 0&=0+0 \end{aligned}$ Now to find any other solutions, we will assume that for $x\neq 0$ , $f(x)\neq 0$ .

Let $f(1)=b$ . Then $P(0,1,1)$ gives $\begin{aligned} f(f(1))&=f(1)\\ f(b)&=b \end{aligned}$ Further, $P(0,1,b)$ gives $\begin{aligned} f(f(b))&=bf(1)\\ b&=b^2\\ b&=1 \end{aligned}$ as $b\neq 0$ . Therefore, $f(1)=1$ .

Note that $P(0,1,x)$ gives $f(f(x))=x$ . Now consider $P(x,0,0)$ and $P(0,x,x)$ for $x\neq 0$ . $\begin{aligned} f(x^2)&=xf(x) & f ( x f(x))=xf(x) \end{aligned}$ Therefore, $\begin{aligned} f(x^2)& = f ( x f(x))\\ f(f(x^2)) & = f(f(x(f(x)))\\ x^2 & = xf(x)\\ f(x)&=x \end{aligned}$ Note that $f(x)=x$ is a solution, as $P(x,y,z)$ gives $\begin{aligned} f(x^2+yf(z))&=xf(x)+zf(y)\\ x^2+yz&=x^2+yz \end{aligned}$ We have fully examined the case where there exists $a\neq 0$ such that $f(a)=0$ and the case where $x\neq 0$ implies $f(x)\neq 0$ . Therefore, the only two solutions are $f(x)=0\text{ and }f(x)=x$

When we put the values $y=z=0$ ,

We get $f(x^2)=xf(x)$

This is true only when $f(x)= \alpha x$ , where $\alpha$ is some constant

So from LHS, we get $f(x^2 + yf(x))=\alpha x^2 +\alpha^{2}yz$

And from RHS we get $xf(x) + zf(y)=\alpha x^2 +\alpha yz$

Equating LHS and RHS we get $\alpha^2=\alpha$

So, $\alpha\equiv 0,1$

Thus answer is $\boxed{2}$

Take $x=0$ , we have $f(yf(z))=zf(y)$ , and $y=1$ gives $f(f(z))=zf(1)$ .

If $f(1) = 0$ , then $f(f(z))=0$ for all $z$ . This imply $f(0)=f(yf(f(z)))=f(z)f(y)$ for all $y$ and $z$ . Take $y=z=0$ gives $f(0)=0$ or $f(0)=1$ . If $f(0)=1$ , then $f(1)=f(f(0))=0$ , but then $1=f(0)=f(1)f(1)=0$ , contradiction. So $f(0)=0$ and $f(y)f(z)=0$ for all $y$ and $z$ , which is $f(x)=0$ for all $x$ .

If $f(1)\neq 0$ then $f$ bijective .Take $a$ such that $f(a)=1$ , then $f(y)=af(y)$ . If $a \neq 1$ , then $f(y)=0$ for all $y$ , contradicts surjectivity of $f$ . If $a=1$ , $f(f(z))=z$ , then $f(x^2)=xf(x)$ , gives $f(f(x)^2)=f(x)f(f(x))=xf(x)=f(x^2)$ , by injectivity, $f(x)^2=x^2$ for all $x$ . If there exist $b$ such that $f(b)=-b$ , then $b^2=f(f(b^2))=f(bf(b))=bf(b)=-b^2 \Rightarrow b=0$ . So $f(x)=x$ is a solution.

$f(x)=x$ and $f(0)=0$ are only solutions.

That was easier than I thought :)

$x \rightarrow 0, z \rightarrow y; f(yf(y)) = yf(y)$

By induction, $\boxed{f(x) = x}$ .

By guessing; $\boxed{f(x) = 0}$ .

Therefore, number of solutions = $\boxed{2}$ .

Note: not sure if it's okay to do this.

Plugging in $y=z=0$ , we find that $f(x^{2})=xf(x)$ , or $f(0)=0$ . Looking at the degree of $f(x)$ in $f(x^{2})=xf(x)$ , its degree must be either 0 or 1, as in $f(x)=0$ or $f(x)=ax+b$ . Plugging in $f(0)=0$ , we find that $f(x)=ax$ . Plugging this into the original, $a=1$ , $f(x) = 0$ or $x$ . Thus the number of solutions we have is $\boxed{2}$ .

Indian MO 2005

Letting $x = 0$ gives $f(yf(z)) = zf(y)$ and $f(zf(y)) = yf(z)$ after flipping $y$ and $z$ . From the second equation, you get $f(f(zf(y))) = f(yf(z))$ , so $f(f(zf(y)) = zf(y)$ . This means for all $x$ , $f(f(x)) = x$ as long as there exists some $n$ such that $f(n) \ne 0$ .

If no $n$ exists satisfying that condition, then $f(n) = 0$ must be true for all $n$ . This function satisfies the equation.

If $y = z = 0$ , you have $f(x^2) = xf(x)$ . If $x = 0$ and $y = z$ , you have $f(yf(y)) = yf(y)$ . Replacing $y$ with $x$ , you get $f(xf(x)) = f(x^2)$ from those two equations, which can be rewritten as $f(f(xf(x))) = f(f(x^2)) = xf(x) = x^2$ . Factoring this gives $x(f(x)-x) = 0$ , or $f(x) = x$ for all $x \ne 0$ . You can prove if $x = 0$ , $f(x) = 0$ by letting $x=y=z=0$ in the original equation.

There are $\boxed{2}$ function that satisfy the equation.