if $\displaystyle a^2 \equiv 1$, then $\displaystyle a \equiv \pm 1 \ \mbox{(mod p)}$

Not sure how to prove this.

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- Jun 27th 2010, 01:16 PMdwsmithif 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 PMjamix
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 PMdwsmith
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 PMBruno 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 PMmathman88
- Jun 27th 2010, 03:43 PMdwsmith