Funky Functional Equation Part II

By setting $x=0$ , we see that $f(f(y))=y+f(0)^2$ . Because $f(0)^2$ is a constant, $f(f(y))$ must be linear.

The degree of $f(f(y))$ is the square of the degree of $f(y)$ , so the degree of $f(y)$ is 1. In other words, $f(y)$ is a linear polynomial.

Every linear polynomials obviously has one root. In other words, $f(y)=0$ for one $y$ . Now, setting $y=-f(x)^2$ gives $f(xf(x)+f(-f(x)^2)=0$ . Let $x$ be the value such that $f(x)=0$ . Then we have $f(f(0))=0$ . However, we have from earlier that $f(f(y))=y+f(0)^2$ . If we set $y=0$ in this equation, we get that $f(f(0))=f(0)^2$ . Hence, $f(0)^2=0$ , so $f(0)=0$ .

Write $f(y)$ as $ay$ . Then $f(f(y))=a^2y=y$ . Because the slope of the right hand side is 1, $a^2=1$ . Therefore, $a=\pm1$ .

We have now proven that $f(y)=y$ or $-y$ . This tells us that $f(10)=10$ or $-10$ . The absolute value of the product of these two numbers is $\boxed{100}$ .

On the LHS, the degree of the $x$ expression is $d_f^2 + d_f$ , where $d_f$ is the degree of $f\left(a\right)$ , and the degree of the $y$ expression is $d_f^2$

On the RHS, the degree of the $x$ expression is $2d_f$ and the degree of the $y$ expression is $1$ . From this we can therefore conclude that $d_f = \pm 1, \left(d_f+2\right)\left(d_f-1\right) = 0, \Rightarrow d_f = 1 \Rightarrow f\left(a\right) = ma + c$ for some constants $m, c$ .

Now we have $f\left(xf\left(x\right) + f\left(y\right)\right) = y + f\left(x\right)^2 \Rightarrow m^2x^2 + mcx + m^2y +mc + c = y + m^2x^2 + 2mcx +c^2$

$m^2y + c + mc = y + 2mcx + c^2$

As the coefficients of powers of $x,y$ and the constants must be equal on both sides, we therefore have:

$m^2 = 1$ , $2mc = 0$ , $mc + c = c^2$

Therefore, $m = \pm1$ , $c = 0$ , hence we have $f\left(a\right) = \pm a \Rightarrow f\left(10\right) = \pm10$

$|10 * \left(-10\right)| =$ $100$

let's plug in y=x we get: f(f(x) x+f(x))= f(x)^2 +x we see that there is a kind of symatry here. now we think what if the function was nothing really doing nothing to the number conneting the number to itself in that case: f(f(x) x+x)= f(x)*x+f(x)=f(x)^2+x. in that case f(x)=x. x^2+x=x^2+x. and f(10)=10. now we assume that the action is the opposite namely f(x)=-x and we see that it satisfies. therefore f(10)=+/-10 and the product is 100