Zeros of polynomial $ f(z)= z^2+ az+ p^{11}$

Question

Consider the polynomial $ f(z)= z^2+ az+ p^{11}$ where $ a \in \mathrm{Z}- \{0\}, \, p \geq 13$ is a prime. Suppose, $ a^2 \leq 4 p^{11},$ then which of the following is true?

1. $f$ has a zero on imaginary axis.
2. $f$ has a zero for which real and imaginary parts are equal.
3. $f$ has distinct roots.
4. $f$ has exactly one root.

For $a= 1$ and $p= 13$, I got two distinct roots for $f(z)$, so option $3$ is possibly true. Also $ f'(z)$ has a zero at $z= \frac{-a}{2}$, at which $f$ will have a root if $4p^{11} = a^2.$ How to approach for this question and discard other options?

Answer 1

The only significance of $p$ and $a$ being integers, with $p$ not a square, is that we know $$ a^2 \neq 4 p^{11} $$

Answer 2

Answer: (3) is true i.e. $f$ has distinct roots.

Use Quadratic equation formula for the root of the polynomial. $$\therefore z = \frac{-a \pm \sqrt{a^2-4p^{11}}}{2}$$ Now Option (1),(2) and (4) can be eliminated by using $a=4p^{11}$