# if a^2 \equiv 1, then a \equiv \pm 1 (mod p)

• Jun 27th 2010, 01:16 PM
dwsmith
if a^2 \equiv 1, then a \equiv \pm 1 (mod p)
if $\displaystyle a^2 \equiv 1$, then $\displaystyle a \equiv \pm 1 \ \mbox{(mod p)}$

Not sure how to prove this.
• Jun 27th 2010, 02:15 PM
jamix
Well $\displaystyle a^2 \equiv 1 mod(p)$ if and only if $\displaystyle p | (a-1) \cdot (a+1)$ so.....
• Jun 27th 2010, 02:17 PM
dwsmith
I wasn't sure if I could assume the mod p in if part because it wasn't written in the book.
• Jun 27th 2010, 02:22 PM
Bruno J.
Also : in general, a polynomial of degree $\displaystyle n$ has at most $\displaystyle n$ roots $\displaystyle \mod p$. In this case, we know $\displaystyle \pm 1$ are two roots, so there can't be any others.
• Jun 27th 2010, 03:41 PM
mathman88
Quote:

Originally Posted by Bruno J.
Also : in general, a polynomial of degree $\displaystyle n$ has at most $\displaystyle n$ roots $\displaystyle \mod p$. In this case, we know $\displaystyle \pm 1$ are two roots, so there can't be any others.

Why is this true?
• Jun 27th 2010, 03:43 PM
dwsmith
Quote:

Originally Posted by mathman88
Why is this true?

$\displaystyle x^3=8$ has at most 3 real solutions; however, we know it has $\displaystyle x=2$ and a conjugate pair of complex numbers.