If $p$ is prime and $x^2\equiv1\text{ }(\text{mod }p)$, then $x\equiv\pm1\text{ }(\text{mod }p)$.

$\text{Proof.}$ Let $p$ be prime and $x^2\equiv1\text{ }(\text{mod }p)$. Then$$\sqrt{x^2}\equiv \sqrt{1}\text{ }(\text{mod }p)$$$$\implies x\equiv\pm1\text{ }(\text{mod }p).\blacksquare$$

Can I do this?

Question

If $p$ is prime and $x^2\equiv1\text{ }(\text{mod }p)$, then $x\equiv\pm1\text{ }(\text{mod }p)$.

$\text{Proof.}$ Let $p$ be prime and $x^2\equiv1\text{ }(\text{mod }p)$. Then$$\sqrt{x^2}\equiv \sqrt{1}\text{ }(\text{mod }p)$$$$\implies x\equiv\pm1\text{ }(\text{mod }p).\blacksquare$$

Can I do this?

anonymous · Answer

Yes, You Can lol How would I know!?!

experimentx · Answer

yes ...

if p divides xy, then p either divides x or y

experimentx · Answer

but i wouldn't exactly do that ... instead i would add ... either ... or

anonymous · Answer

Oh, please ignore my reply (the previous one), I thought you were asking a question not posting your solution (It's not rendering, the latex).

anonymous · Answer

That makes sense. Thank you, experimentX.

experimentx · Answer

never trust me completely ... i usually make hunch

anonymous · Answer

No, it's true because$$x^2\equiv1\text{ }(\text{mod }p)\implies p|(x^2-1)\implies p|[(x-1)(x+1)],$$which, by Euclid's lemma, implies that $x\equiv1\text{ }(\text{mod }p)$ or $x\equiv-1\text{ }(\text{mod }p)$.

experimentx · Answer

i know that,

but 

what if x = 3, 5, ... , p=2

my 'OR' statement is a waste ... should have been thoughtful before posting

experimentx · Answer

for the rest cases .. it is valid, because no other prime would divide two consecutive no's of diff 2,  ..... excluding -2