$$f(x^2 * f(y)^2) = f(x^2) * f(y)$$ $$f:{\mathbb Q}^+ \rightarrow {\mathbb Q}^+$$ $x,y \in {\mathbb Q}^+$
I have been given the following equation and information and am supposed to find all functions which satisfy this equation. I have found $f(x)=1$ and $f(x)=\sqrt{x}$ work. How can I prove more exist/ do not exist?
Observe, we do the following manipulation: let $x = 1$ : $$f(f(y)^2) = f(x^2 \cdot f(y)^2) = f(x^2) \cdot f(y) = f(1) \cdot f(y) = f(y \cdot f(1)^2) $$
Case: Assume the function is left-invertible (or simply invertible), which tends to be a desirable quality for solutions of functional equations, you get:
$$f(y)^2 = y \cdot f(1)^2$$
Which, since $f(y) \geq 0$, becomes:
$$f(y) = f(1) \sqrt{y}$$
However, in order for $f(y) \in \mathbb{Q}^{+}$, the only valid choice is $f(1) = 0$. That is, $$f(y) = 0$$
Note: The zero function isn't invertible, but I added it effectively falls into the next case, as well as your discovery that $f(x) = 1$ is a solution as well.
Case: $f$ possesses no left inverse. So $f$ is not injective.
This is where everything starts to break apart. You can start going down the Rabbits hole of tossing away assumptions.
Observe, let $x = 1$ and $y = 0$, then:
$$f(1) \cdot f(0) = f(f(0)^2) = f(0 \cdot f(1)^2) = f(0) $$
And letting $x = 0 = y$:
$$f(0) = f(0) \cdot f(0) \in \{ 0, 1 \}$$
Case: $f(0) = 1 = f(1)$. Then the equation becomes:
$$f(f(y)^2) = f(y)$$
This is simply a recurrence relation over $\mathbb{Q}^{+}$. Given a value $y$, choose the value $f(y) = m \in \mathbb{Q}^{+}$. Then:
$$f(m^2) = m$$
So this takes care of all the values $\{ x \in \mathbb{Q}^{+} : \exists y \in \mathbb{Q}^{+} \text{ such that } y^2 = x\}$. For the remainder of the values, we can make a variety of choices. Specifically, we can let those values go to one. That is, we can have the function:
$$f(x) = \begin{cases} \sqrt{x} & \sqrt{x} \in \mathbb{Q}^{+} \\ 1 & \text{otherwise} \end{cases}$$
But we can also try many other potential functions. You might not be able to construct a function that is continuous however. This because if you have a sequence of numbers from the right that approach $y$ that are rationally square rootable, and a sequence from the left that are rationally square rootable, you can try to choose $f(y)$ to such that
$$x_n \rightarrow y \leftarrow z_n$$ $$\sqrt{x_n} \rightarrow f(y) \leftarrow \sqrt{z_n}$$
But due to the incompleteness of $\mathbb{Q}$, I don't remember if there is a way of choosing a satisfactory alternative $f(y)$.
Case: $f(0) = 0$. I'll let you take this from here. It's very similar to the function we found in the previous case, except $f(0) = 0$ is stated explicitly.
Edit: Totally assumed $0 \in \mathbb{Q}^{+}$. If not, replace with an $\epsilon \in \mathbb{Q}^{+}$.