Understanding a step in a functional equation

Question

I was trying to solve a functional equation and I while I was pretty close to the answer there is a remaining case that has stumped me for a while now. Here is the problem

Determine all functions $f: \mathbb{N}^* \to \mathbb{N}^*$ satisfying $$ \frac{f(x)+y}{x+f(y)} + \frac{f(x)y}{xf(y)} = \frac{2(x+y)}{f(x+y)}, \qquad \forall x,y \in \mathbb{N}^*. $$

It's easy to see that the function $f(x)=x$ satisfies the condition, but then remains the issue of proving that it is the only function that satisfies the condition. I was able to prove that $f{(2x)}=2x$ so naturally I then considered the odd numbers, however I ended up with two very nasty expressions, if we let $x=1$ and $y=2n$ we end up with this $$ \frac{2(1+2n)^2}{a(2+2n)}=f(2n+1)$$ Where $a=f(1)$

Or if we let $x=2n$ and $y=1$ Then we end up with: $$ \frac{a(2n+1)(2n+a)}{a(n+1)+n}=f(2n+1)$$ (In fact the intended solution uses the second substitution but I couldn't understand the rest of the explanation).

Since we have a fraction the denominator should divide the numerator, since we are dealing with integers, but I don't know how to prove that. Please help.

Answer 1

After proving that $f(2n)=2n$, your claim that substituting $x=1$ and $y=2n$ implies that $$ \frac{2(1+2n)^2}{a(2+2n)}=f(2n+1) $$ is false, and I see where you made a mistake with your algebraic manipulations. Using $f(2n)=2n$ and $a=f(1)$, I obtain instead that $$ \frac{a+2n}{1+2n}+a=\frac{2(2n+1)}{f(2n+1)}, $$ which expands to $$ \frac{a+2n+(2n+1)a}{2n+1}=\frac{2(2n+1)}{f(2n+1)}, $$ which rearranges to $$ f(2n+1)=\frac{2(2n+1)^2}{a+2n+(2n+1)a}\qquad (\star). $$ Note that this differs from your expression because $$ a+2n+(2n+1)a=2a+2n(a+1) $$ whereas your expression has the $2a+2n(a+1)$ replaced by $a(2+2n)$.

Once we obtain the corrected version of $(\star)$, to conclude the argument boils down to finding the value of $a$, and this can be done by substituting $x=1$ and $y=3$ into the original functional equation to obtain that $$ \frac{a+3}{1+f(3)}+\frac{3a}{f(3)}=2\qquad (\star\star). $$ On the other hand setting $n=1$ in $(\star)$ yields that $$ f(3)=\frac{18}{4a+2}=\frac{9}{2a+1}, $$ and substituting into $(\star\star)$ yields $$ \frac{a+3}{1+9/(2a+1)}+\frac{a(2a+1)}{3}=2, $$ which expands out to $$ \frac{4a^3+28a^2+31a+9}{a+5}=12, $$ thus $$ 4a^3+28a^2+19a-51=0. $$ We find one root is $a=1$ and dividing this cubic polynomial by $a-1$ leaves the quadratic equation $$ 4a^2+32a+51=0, $$ which has no integer solutions (easy to see either directly by the quadratic formula since the discriminant $32^2-4\cdot 4\cdot 51=208$ is not a perfect square, or by the rational root test). Thus $a=1$ is the only solution and substituting back into $(\star\star)$ yields that $f(2n+1)=2n+1$ and thus $f(x)=x$ when combined with your starting point that $f(2n)=2n$.

Answer 2

The observation, that the right hand side of the functional equation is invariant under exchanging $x$ with $y$, leads to the "flipped" equation

$$\frac{f(y)+x}{y+f(x)} + \frac{f(y)x}{yf(x)} = \frac{2(x+y)}{f(x+y).}$$

Now one can realize that each summand on the left side here is exactly the multiplicative inverse of the same summand in the original equation. That means, if we set

$$s_1=\frac{f(y)+x}{y+f(x)},\; s_2=\frac{f(y)x}{yf(x)},$$

we now know that $s_1+s_2 = \frac1{s_1}+\frac1{s_2}$. Multiplication with $s_1s_2>0$ leads to $s_1s_2(s_1+s_2)=s_2+s_1=s_1+s_2$ and since $s_1+s_2>0$ we get $s_1s_2=1$, which leads to

$$ \frac{f(y)+x}{y+f(x)} = \frac{f(x)y}{xf(y)}.$$

Using $x=2n$ and the already known $f(2n)=2n$, we get

$$ \frac{f(y)+2n}{y+2n} = \frac{2ny}{2nf(y)} = \frac{y}{f(y)}.$$

Now, for a given $y$, in the above equation the left hand side increases when $f(y)$ increases, but the right hand side decreases when $f(y)$ increases. That means there can be at most one possible value of $f(y)$ and of course $f(y)=y$ fits the bill and actually produces equality.