Find all functions that satisfy $\frac{f(x+y)+f(x)}{ky+f(x)}=\frac{kx+f(y)}{f(x+y)+f(y)}$

Question

Find all functions $f:\mathbb{N}\to\mathbb{N}$ for which there exists $k\in\mathbb{N}$, such that for any $x,y\in\mathbb{N}$,$$\frac{f(x+y)+f(x)}{ky+f(x)}=\frac{kx+f(y)}{f(x+y)+f(y)}.$$ I already get the function when it is even by substituting $P(x,x) = f(2x)=kx$ but how to get when it is odd thank you

Answer 1

Let $P(x, y)$ denote the assertion in question

$P(x,x) = (f(2x)+f(x))^2 = (kx+f(x))^2$

since both of them are natural

$f(2x)+f(x) = kx+f(x)$

$f(2x)=kx$

for odd

$P(2x-1,1) = k(x-1)(-f(2x-1)+f(1)+k(x-1))=0$

since k positive and for all x hence

$f(1)+k(x-1) = f(2x-1)$

$P(2x,1)= (f(2x+1)+f(1))(f(2x+1)+f(2x))=(2kx+f(1))(k+f(2x))$

$P(2x,1) = (2f(1)+kx)(2kx+f(1)) = (2kx+f(1))(k+kx)$ $P(2x,1) = 2k^2x+kf(1)=4kxf(1)+2f^2(1)$ giving us k = 2f(1)

therefore

$f(2x) = f(1)*2x $

$f(2x-1) = f(1)+2f(1)(x-1)=f(1)*f(2x-1)$

showing that f(x) is in the form of cx .QED $\square$