Functional equation: $(x+y)f(x,y)=yf(x,x+y)$

Question

Find functions $f$ on pairs of positive natural numbers satisfying:

1. $f(x,x)=x$
2. $f(x,y)=f(y,x)$
3. $(x+y)f(x,y)=yf(x,x+y)$

It is quite easy to find that $f(1,k)=k$ for all $k$ by induction: if $f(k,1)=k$,
then $(1+k)f(k,1)=(1+k)k=k(1,k+1)$ by condition (3) leading to the result.
Then also $f(k,k+1)=k(1+k)$.
After this I'm a bit stuck. $x^2=f(x,x+1)-f(x,x)$ by the above, but I don't know where to go to find a general solution. Hints?

Answer 1

Ok so here is an idea. I have the answer but I think it is a fun problem. Come up with a formula for f(x,kx) using induction. Then think about what f(x,y) might be.

Answer 2

Letting $g(x,y)=\frac{1}{y}f(x,y)$ we get conditions:

1. $g(x,x)=1$
2. $yg(x,y)=xg(y,x)$
3. $g(x,y)=g(x,x+y).$

Given any $m=nq+r_1$ you get that:

$$g(n,m)=g(n,r_1)=\frac{n}{r}g(r_1,n)$$

If we do the repeat the Euclidean algorithm until $r_k=r_{k-1}=\gcd(m,n)$ you get $g(m,n)=\frac{n}{r_1}\frac{r_1}{r_2}\cdots\frac{r_{k-1}}{r_k}g(d,d)=\frac{n}{d}.$

Then $f(n,m)=\frac{mn}{\gcd(m,n)}.=\operatorname{lcm}(m,n)$

This doesn't mean that we've got the correct answer, only that if it exists, it must be this $f.$

We can easily verify that this $f$ satisfies these conditions:

$$f(x,x)=x\\f(x,y)=f(y,x)\\(x+y)f(x,y)=\frac{xy(x+y)}{\gcd(x,y)}=yf(x,x+y)$$ The last since $\gcd(x,y)=\gcd(x,x+y).$