A question on modular arithmetics

[life is maths]

Homework Statement

Hi, my question is:

Show that if p is a prime, x²≡1(mod p) has exactly two solutions for p>2.

Homework Equations

The Attempt at a Solution

I did:
p|x²-1
Then for a number a;
pa=x²-1
pa+1=x²
x=$$\mp$$$$\sqrt{pa+1}$$

But I did not use the fact that p is prime and greater than 2. Do you have any ideas? Thanks in advance.

This is not a homework but a suggested exercise before exam.

 

[Stephen Tashi]

It's more likely the problem is based on
$$x^2 - 1 = 0 (mod\ p)$$
$$(x+1)(x-1) = 0 (mod\ p)$$
and the facts about the "zero divisors" of mod p arithmetic.

 

[life is maths]

Thank you, but we haven't learned zero divisors yet. Is it a subject related to rings?
Since I have no idea on that subject, do you have any ideas about how to solve it in another way? Thanks again :)

 

[Stephen Tashi]

Yes, it is a topic in rings. It's also one of main reasons that people are interested in have a prime modulus.

If you can have non-zero numbers such a,b such that ab = 0 (mod K) then the mod K arithmetic is said to have "zero divisors". For example if you are solving

(x)(x+1) = 0 (mod 12) , you get one solution due to x = 3 because ( 3)(4) = 0 (mod 12).
You also can have the solutions x = 0 and x = 11. That's more than 2 solutions to this quadratic equation.

When the modulus is a prime number, you can solve an equation like
(x)(x+1) = 0 (mod p) just as you would solve it in ordinary algebra by assuming that one of the factors must be zero. That gives only two solutions.

 

[life is maths]

Thanks, I think I understood the zero divisors, but I'm still a bit confused and couldn't actually constructed a good proof. I have (x+1)(x-1) = 0 (mod p) and what should I do next? Should I assume the factors are 0? How is it possible without knowing p? Thanks for your help.

 

[Stephen Tashi]

(x+1)(x-1) = 0 (mod p) and what should I do next? Should I assume the factors are 0?

Assume one or the other factor is zero. It's just like solving a quadratic equation by factoring.

How is it possible without knowing p?

Because p is a prime, so in its modular arithmetic there can't be two non-zero factors whose product is zero. Arithmetic modulo a prime has no zero divisors. And p > 2 so x+1 and x-1 can't be the same number.