Why does $10^{\frac{p-1}{2}}\equiv \pm1 \pmod p$ when $p \nmid 10$?

Question

By using Fermat's little theorem, the following: $$10^{p-1}\equiv 1 \text{(mod p)}$$ is true, given that p does not divide 10.

By substituting in prime numbers $7,11,13,15,21,23$ etc. and picking random larger prime numbers such as 127, 199 etc., I found that $10^{\frac{p-1}{2}}\equiv \text{always either }1 \text{ or }-1\text{(mod p)}$, and nothing else, so I wondered why and tried to prove by contradiction:

Assume $10^{\frac{p-1}{2}}\not \equiv ±1 \text{(mod p)}$, then $10^{\frac{p-1}{2}}\equiv(p±k)^2\equiv k^2 \text{(mod p)}$ where $k^2≠1,$ but $k^2$ still can be some large number such that $k^2=M\cdot p±1, M\in \mathbb{Z}$, so I cannot prove it still.

Is there any method to prove that $k^2=M\cdot p±1, M\in \mathbb{Z}$ cannot be true, or can this actually be possible? Or is there better approaches to proving?

Answer 1

We have $10^{(p-1)/2}\equiv1{\pmod{p}}$ so that $10^{(p-1)/2}$ is a solution to $$x^2\equiv1\pmod{p}\iff(x-1)(x+1)\equiv0\pmod{p}$$

Since $p$ is prime, $p|(x+1)$ or $p|(x-1).$

Answer 2

The reason why this is true is very simple: set $x=10^{\tfrac{p-1}2}$. You have $x^2=1$, so as $\mathbf Z/p\mathbf Z$ is a field since $p$ is prime, this quadratic equation can't have more than two solutions.