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 5 of 8: Show

f(1)=1

Previous step Next step

a:=f(1/f(1)):\quad f(a)=1,\ \ af(a)=f(af(a))=1 \implies a=1 \implies f(1)=1

Detailed analysis

Let $a=f(1/f(1))$ . Setting $x=1$ , $y=1/f(1)$ in $yf(x)=f(xf(y))$ gives $f(a)=1$ . Since $f(a)=1$ , trivially $a=af(a)$ and $f(af(a))=f(a)=1$ ; but the relation $yf(x)=f(xf(y))$ at $x=y=a$ gives $af(a)=f(af(a))$ , so $a=f(af(a))=1$ . As $f(a)=1$ and $a=1$ , we conclude $f(1)=1$ .