Problem 5

Let

\mathbb{R}

be the set of real numbers. Determine all functions

f:\mathbb{R}\to\mathbb{R}

satisfying the equation

f(x+f(x+y))+f(xy)=x+f(x+y)+yf(x)

for all real numbers

x

and

y

Step 2 of 5: Case

f(0)\ne0

Previous step Next step

In plain words

If $f(0)\ne0$ , every fixed point of $f$ turns out to equal $1$ , and the family from the previous step must always equal it.

f(0)\ne0 \ \Rightarrow\ f(x)\equiv 2-x

Detailed analysis

Setting $x=0$ in the functional equation gives $f\big(f(y)\big)+f(0)=f(y)+yf(0)$ for every $y$ . If $y_0$ is any fixed point (so $f(y_0)=y_0$ ), substituting $y=y_0$ gives $f(y_0)+f(0)=y_0+y_0f(0)$ , i.e. $y_0+f(0)=y_0+y_0f(0)$ , i.e. $f(0)=y_0f(0)$ . Since $f(0)\ne0$ , this forces $y_0=1$ : the only fixed point of $f$ is $1$ . By the previous step, $x+f(x+1)=1$ for every $x$ , i.e. $f(x+1)=1-x$ , i.e. $f(u)=2-u$ for every real $u$ . One checks directly that $f(x)=2-x$ satisfies the original equation.