Proof that a certain function that satisfy a specific equation is constant

Question

I am currently doing Problem 2, Chapter 1, from the "Functional Equations and How to Solve Them" book by Christopher G. Small.
It is as follows:

Let $f(x)$ be a function that satisfy $f(x + y) = f(xy)$ for all positive $x$ and $y$.
Prove that $f(x)$ is a constant.

Here are my 2 possible solutions:

1. While the problem constraints us to positive values, if we assume it is continuous at $0$ and use a limit argument, then substituting $y = 0$: $$f(x + y) = f(xy) \\ \Rightarrow f(x + 0) = f(0x) \Rightarrow f(x) = f(0)$$ for a certain value of $f(0)$. This means that $f(x)$ is equal to a constant $f(0)$ and hence we proved the preposition.

2. Define $s = x + y, p = xy$. $x$ and $y$ are therefore solutions of the quadratic equation $t^2 - st + p = 0$ (with $t$ as the variable). Clearly for such a solution to exist $s$ and $p$ needed to satisfy $\Delta = s^2 - 4p \geq 0$. With that, we have $f(s) = f(p)$ for all values $s$ and $p$ that satisfy $s^2 - 4p \geq 0$. We'll call this condition $P(s, p)$.

With this, we take three values, $s_1$, $s_2$, and $p$, assuming $s_1 < s_2$ and ${s_1}^2 - 4p \geq 0$. Since $s_2 > s_1$, this also means that ${s_2}^2 - 4p \geq 0$. This means that $P(s_1, p)$ and $P(s_2, p)$ holds, which results in $f(s_1) = f(p) = f(s_2)$ for appropriate choices of those numbers. And since we can pick any arbitrary $p \leq \frac{{s_1}^2}{4}$ for a choice of two variables $s_1$ and $s_2$, this means that $f(s_1) = f(s_2)$ for all $s_1 < s_2$, meaning that $f$ is constant.

Are those 2 correct and/or need extra work? Thanks for your help!

Answer 1

Solution (1) requires $f(0)$ so it's not correct. Solution (2) is right. You don't need an arbitrary number though.
We have $$f\left(2a+\dfrac{2}{a}\right) = f\left(4\right)= f\left (b\left(4-b\right)\right)$$ $2a+\dfrac{2}{a}$ has the range $[4,+\infty)$ while $b(4-b)$ has the range $(0,4]$ for $a \gt 0$, $0 \lt b \leq 2$, hence they could cover the entire $\mathbb{R}^+$

Answer 2

Solution (1) has the hitch that we have to use $f(0)$ which might not be available or which might not exist. We will then have to conclude that $f(x)$ itself might not exist.

Solution (2) looks good & it is right. My Opinion is that it looks a little more complicated than necessary.

My Solution (3) might be something like this :
[[ I think , this is essentially what you are trying out too , though that was a little more complex ]]
Let $y=1$ , then $f(x + 1) = f(x)$ , then we see that $f(x)$ for large $x$ will eventually reduce to $x$ values in $(0,1]$.
Eg $f(5.5)=f(4.5)=f(3.5)=f(2.5)=f(1.5)=f(0.5)$

Now , let $(x+y)=n$ where $n$ is sufficiently large , like $n=10$.
We have $y=(n-x)$
Hence $f(x + y) = f(n) = f(xy) = f(x(n-x)) = f(nx-x^2)$

We can see that $(nx-x^2)$ takes all values from $0^+$ to $n^2/2$

In that range , $f(nx-x^2) = f(n)$ is constant.

Hence , $f(x)$ is a Constant though-out the Positive real number range.