MathLabs

Problem 5

Let R\mathbb{R} be the set of real numbers. Determine all functions f:R→Rf:\mathbb{R}\to\mathbb{R} satisfying the equation f(x+f(x+y))+f(xy)=x+f(x+y)+yf(x)f(x+f(x+y))+f(xy)=x+f(x+y)+yf(x) for all real numbers xx and yy.
Step 1 of 5: Every y=1y=1 substitution produces a fixed point
In plain words

Setting y=1y=1 collapses the product term xyxy into just xx, revealing a whole family of fixed points.

P(x,1): f(x+f(x+1))=x+f(x+1)P(x,1):\ f(x+f(x+1))=x+f(x+1)
Detailed analysis

Let P(x,y)P(x,y) denote the assertion f(x+f(x+y))+f(xy)=x+f(x+y)+yf(x)f(x+f(x+y))+f(xy)=x+f(x+y)+yf(x). Taking y=1y=1: P(x,1)P(x,1) gives f(x+f(x+1))+f(x)=x+f(x+1)+f(x)f(x+f(x+1))+f(x)=x+f(x+1)+f(x), so f(x+f(x+1))=x+f(x+1)f\big(x+f(x+1)\big)=x+f(x+1). That is, the number x+f(x+1)x+f(x+1) is a fixed point of ff, for every real xx.