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 4 of 5: Case

f(0)=0

f

is odd

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In plain words

The freshly-found fixed point cancels almost the whole equation at $y=-1$ , leaving exactly the oddness of $f$ .

P(x,-1):\ f\big(x+f(x-1)\big)+f(-x)=x+f(x-1)-f(x)\ \Rightarrow\ f(-x)=-f(x)

Detailed analysis

Setting $y=-1$ in the original equation gives $f\big(x+f(x-1)\big)+f(-x)=x+f(x-1)+(-1)f(x)$ . By the previous step $f\big(x+f(x-1)\big)=x+f(x-1)$ , so this simplifies to $x+f(x-1)+f(-x)=x+f(x-1)-f(x)$ , i.e. $f(-x)=-f(x)$ for every real $x$ : $f$ is odd.