Problem 5

Find all functions

f:\mathbb{R}^+\to\mathbb{R}^+

such that

(z+1)f(x+y)=f(xf(z)+y)+f(yf(z)+x)

for all positive real numbers

x,y,z

Step 8 of 8: Conclude

f(x)=x

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f(x+y)=f(x)+f(y) \implies f \text{ strictly increasing},\quad f(x)>x \text{ or } f(x)<x \text{ both impossible} \implies f(x)=x

Detailed analysis

With $b=0$ , $f(x+y)=f(x)+f(y)$ for all positive $x,y$ , so $f$ is strictly increasing. If $f(x)>x$ for some $x$ , applying the strictly increasing $f$ and using the involution gives $f(f(x))>f(x)$ , i.e. $x>f(x)$ , contradicting $f(x)>x$ ; symmetrically $f(x)<x$ is impossible. Hence $f(x)=x$ for all positive real numbers $x$ .