WakeUp wrote: Find all functions $f$ of two variables, whose arguments $x,y$ and values $f(x,y)$ are positive integers, satisfying the following conditions (for all positive integers $x$ and $y$): \begin{align*} f(x,x)& =x,\\ f(x,y)& =f(y,x),\\ (x+y)f(x,y)& =yf(y,x+y).\end{align*} Hello WakeUp! could you confirm us that there is a closed form for the solution and that you'll be able to post it when asked ? I have some doubts about existence of such closed form. Some results I've got up to now : a) there exist a unique solution b) $f(ax,ay)=af(x,y)$ c) $f(1,n)=n!$ And also, for example : $f(x,nx+1)=x!(1+\frac 1x)(2+\frac 1x)(3+\frac 1x)...(n+\frac 1x)$ And I wonder what closed form could give the above result :

pco wrote: Hello WakeUp! could you confirm us that there is a closed form for the solution and that you'll be able to post it when asked ? I have some doubts about existence of such closed form. Some results I've got up to now : a) there exist a unique solution b) $f(ax,ay)=af(x,y)$ c) $f(1,n)=n!$ And also, for example : $f(x,nx+1)=x!(1+\frac 1x)(2+\frac 1x)(3+\frac 1x)...(n+\frac 1x)$ And I wonder what closed form could give the above result : Hi pco, much to my embarrassment there was a typo that is now fixed I'm sure you can complete your solution now.

WakeUp wrote: Find all functions $f$ of two variables, whose arguments $x,y$ and values $f(x,y)$ are positive integers, satisfying the following conditions (for all positive integers $x$ and $y$): \begin{align*} f(x,x)& =x,\\ f(x,y)& =f(y,x),\\ (x+y)f(x,y)& =yf(x,x+y).\end{align*} Firstly, we can prove, that $\all a > 0 : f(x, ax) = af(x, x)$. Proof : $(x + (a - 1)x)f(x, (a - 1)x) = (a - 1)xf(x,ax); \Rightarrow f(x, ax) = \frac{a}{a - 1}f(x, (a - 1)x) =$ $=af(x, x) = ax$. After that: $f(x, px + q) = \frac{f(px, px + q)}{p} = \frac{(px + q)f(px, q)}{pq}$. (We take - $x' = px$ and $y' = q$;) $\frac{(px + q)f(px, q)}{pq} = \frac{px + q}{q}f(q, x)$. Let's watch on new function $g$ : $g(x, y) = \frac{xy}{f(x, y)}$. $\frac{xy}{f(x, y)} = \frac{x(px + q)}{f(x, px + q)} = \frac{x(px + q)}{\frac{px + q}{q}f(q, x)} = \frac{qx}{f(q, x)} = ... = \frac{d(kd)}{f(d, kd)} = d$. (Where d = (x, y).) From that - we can see, that $\frac{xy}{f(x, y)} = (x, y); \Rightarrow f(x, y) = [x, y]$! That's all! !