This question is about the logic of a proof, not the number theory involved.

Exercise (4.1).9 of Number Theory Step by Step (Singh) is:

Find a solution of $x^{101} ≡ 5 \pmod {13}$.

The chapter introduces Fermat's Little Theorem, that if $p$ is prime and does not divide integer $a$, then $a^{p-1} \equiv 1 \pmod p$. The book says it like this (link).

The following is my solution (which matches the author's official solution too):

My Solution

If we assume [see Question below] the prime 13 does not divide $x$, then we have

$x^{12} \equiv 1 \pmod {13}$

$(x^{12})^{8} \equiv x^{96} \equiv 1 \pmod {13}$

And so

$x^{101} \equiv x^{96} \times x^{5} \equiv x^5 \equiv 5 \pmod {13}$

The following table shows some $x^5 \pmod{13}$.

x   x^5     x^5 mod 13

0   0       0
1   1       1
2   32      6
3   243     9
4   1024    10
5   3125    5  <--- solution
6   7776    2

We can see $x=5$ is a solution, and our assumption 13 does not divide $x=5$ holds.

Question

I am uncertain about assuming 13 does not divide $x$ at the start of the solution, in order to use Fermat's Little Theorem.

Typically, to use Fermat's Little Theorem, or any theorem, I have to show the preconditions hold first.

Here a solution emerges as $x=5$ which we confirm meets the theorem's preconditions, after we have found it.

The author's own solution (link to image) appears to ignore the theorem's requirement that prime $p$ must not divide $a$.

What am I missing?