Ytukyg

$$f(x^2 * f(y)^2) = f(x^2) * f(y)$$
$$f:{mathbb Q}^+ rightarrow {mathbb Q}^+$$
$x,y in {mathbb Q}^+$

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 $epsilon in mathbb{Q}^{+}$.