This is probably a really trivial question but I just cannot see the answer...
Suppose $p$ is prime. Then $x^2 \equiv 1 \pmod{p}$ has only solutions $x \equiv \pm 1 \pmod{p}$ for $x \in \mathbb{Z}$.
How do I prove this?
$x^2$ congruent to $1 \pmod {p}$
0
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elementary-number-theory
modular-arithmetic
finite-fields
3 Answers
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It is certainly clear that $x \equiv \pm 1$ are solutions.
On the other hand: The equation $x^2 -1 \equiv 0$ cannot have more than 2 solutions because of the degree of this polynomial.
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1But if p is not a prime and e.g. 8, the argument doesnt work as 1 , 3, 5, and 7 are solutions. – 2017-02-11
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0Right but you asked about $p$ prime i.e. those congruences are equalities in $\mathbb Z/ p \mathbb Z = \mathbb F _p$. – 2017-02-11
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Basically if $x$ is its own inverse modulo $p$, then $x^2 \equiv 1\ (\textrm{mod}\ p)$. So $p\, |\, (x+1)(x-1)$. Further $p$ is prime so either $x \equiv 1\ (\textrm{mod}\ p)$ or $x \equiv -1\ (\textrm{mod}\ p)$.
If on the other hand, $x \equiv 1\ (\textrm{mod}\ p)$ or $x \equiv -1\ (\textrm{mod}\ p)$, then $x \cdot x \equiv (\pm1)\cdot (\pm1)\ (\textrm{mod}\ p)$, i.e., $x^2 \equiv 1\ (\textrm{mod}\ p)$. $x$ is its own inverse modulo $p$.
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0[Euclid's Lemma](https://en.wikipedia.org/wiki/Euclid's_lemma). – 2017-02-12
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Since $p$ is prime then the ring $(\Bbb Z_p,+,\times)$ is a finite field and then it's an integral domain. So in $\Bbb Z_p$ we have: $$x^2=1\iff (x-1)(x+1)=0\implies \big((x-1)=0\big)\vee\big((x+1)=0\big)\iff x=\pm1$$