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 5 of 5: Case

f(0)=0

: finishing with

f(1)=1

and oddness

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In plain words

Adding the equation at $(1,y)$ to its "mirror image" at $(-1,-y)$ cancels the unknown term $f(1+f(y+1))$ and leaves $f(y)=y$ directly.

P(1,y):\ f\big(1+f(y+1)\big)+f(y)=1+f(y+1)+y \qquad P(-1,-y)+\text{oddness}\ \Rightarrow\ f(y)=y

Detailed analysis

Setting $x=1$ in the original equation and using $f(1)=1$ gives, for every $y$ : $f\big(1+f(y+1)\big)+f(y)=1+f(y+1)+y$ . Now setting $(x,y)\mapsto(-1,-y)$ and using $f(-1)=-1$ gives $f\big(-1+f(-1-y)\big)+f(y)=-1+f(-1-y)+y$ . Since $f$ is odd, $f(-1-y)=-f(y+1)$ , and applying oddness once more to the outer value, $f\big(-1-f(y+1)\big)=-f\big(1+f(y+1)\big)$ ; substituting these gives $-f\big(1+f(y+1)\big)+f(y)=-1-f(y+1)+y$ . Adding this to the first displayed equation, the terms $f\big(1+f(y+1)\big)$ cancel: $2f(y)=2y$ , so $f(y)=y$ for every real $y$ . One checks directly that $f(x)=x$ satisfies the original equation.