How many elements of have square roots in a field of 13 elements?

Question

Initially, I thought since this field was isomorphic to $({0,...,12})$ , the elements $(0,4,9)$ would have square roots. However, when I checked the solutions, the answer was different. Thank you in advance.

Answer 1

Hint: Consider the group homomorphism $\mathbb F_{13}^\times \to \mathbb F_{13}^\times$ given by $x \mapsto x^2$. What is the size of its kernel?

Answer 2

Hint:

Those elements definitely have square roots. However, $4^2 = 16\equiv 3$, so clearly $3$ has a square root too.

Answer 3

Let $F$ be the field with $13$ elements.

Consider the map

$$f : F \to F, x \mapsto x^2.$$

Then $f(x) = f(-x)$ for all $x$, so every element except $0$ that is in the image of $f$ has at least two pre-images. Can you show that it can't have more than two?
Now we can count: $0$ is in the image, and of the $12$ remaining elements, each one in the image has exactly two different pre-images. Then how many elements are there in $f(F)$?