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 3 of 5: Case

f(0)=0

: a second family of fixed points

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In plain words

If two consecutive integers are both fixed points, so is the next one up; combining this with the two known families of fixed points bridges a gap of $2$ .

f(0)=0 \ \Rightarrow\ f(-1)=-1,\ f(1)=1,\ f\big(x+f(x-1)\big)=x+f(x-1)

Detailed analysis

Now suppose $f(0)=0$ . Setting $y=0$ in the equation with $x$ replaced by $x+1$ gives $f\big(x+1+f(x+1)\big)+f(0)=x+1+f(x+1)$ , so (since $f(0)=0$ ) $x+1+f(x+1)$ is also a fixed point, for every $x$ . Setting $x=-1$ in the first step's fact ( $x+f(x+1)$ is fixed) gives $-1+f(0)=-1$ is fixed, i.e. $f(-1)=-1$ . Setting $x=1,y=-1$ in the original equation and using $f(-1)=-1$ gives $f(1)+f(-1)=1+f(0)-f(1)$ , i.e. $f(1)-1=1-f(1)$ , so $f(1)=1$ . Now for every $x$ , both $y_0=x+f(x+1)$ and $y_0+1=x+1+f(x+1)$ are fixed points; setting $x=1,y=y_0$ in the original equation and simplifying with $f(1)=1$ shows that whenever $y_0$ and $y_0+1$ are both fixed points, so is $y_0+2$ . Hence $x+f(x+1)+2$ is a fixed point for every $x$ ; replacing $x$ by $x-2$ turns this into: $x+f(x-1)$ is a fixed point for every $x$ , i.e. $f\big(x+f(x-1)\big)=x+f(x-1)$ .