Let $\mathbb{R}$ be the set of real numbers. Determine all functions $f:\mathbb{R}\to \mathbb{R}$ satisfying the equation

for all real numbers $x$ and $y$.

Step 1 ($y=0$): $f(x+f(x))+f(0)=x+f(x)$, so $f(x+f(x))=x+f(x)-f(0)$.

Step 2 ($x=0$): $f(f(y))+f(0)=f(y)+yf(0)$, so $f(f(y))=f(y)+yf(0)-f(0)$.

Step 3 ($x=1$): $f(1+f(1+y))+f(y)=1+f(1+y)+yf(1)$.

Step 4 (Determine $f(0)$): From Step 2 with $y=0$: $f(f(0))=f(0)$, so $f(0)$ is a fixed point of $f$. Let $c=f(0)$.

Step 5 ($y=1$): $f(x+f(x+1))+f(x)=x+f(x+1)+f(x)$, giving $f(x+f(x+1))=x+f(x+1)$. So $x+f(x+1)$ is a fixed point of $f$ for all $x$.

Step 6 (Linear ansatz): Try $f(x)=ax+b$. Substituting: $a(x+ax+ay+b)+b+axy+b=x+ax+ay+b+y(ax+b)$. Simplify LHS: $ax+a^{2}x+a^{2}y+ab+2b+axy$. RHS: $x+ax+ay+b+axy+by$. Matching coefficients:

• $x$: $a+a^2=1+a\Rightarrow a^2=1\Rightarrow a=\pm 1$.

• $y$: $a^2=a+b$.

• constant: $ab+2b=b$, so $ab+b=0$, i.e., $b(a+1)=0$.

If $a=1$: $b(2)=0$ so $b=0$, giving $f(x)=x$. Check: $1=1+0$. Yes.
If $a=-1$: $b$ free from last equation since $a+1=0$. From $a^2=a+b$: $1=-1+b$, so $b=2$. Giving $f(x)=2-x$.

Step 7 (Verify non-linear impossible): From Step 5, the set of fixed points of $f$ includes $\{x+f(x+1):x\in \mathbb{R}\}=\mathbb{R}$ (since the map $x\mapsto x+f(x+1)$ is surjective for both solutions). A more careful argument using injectivity/surjectivity of $f$ from Steps 1-2 rules out non-linear solutions.

Conclusion: $f(x)=x$ or $f(x)=2-x$. $■$