Question

Show that if pis prime, the only solution of${\mathit{x}}^{\mathbf{2}}\mathbf{\equiv }\mathbf{1}\left(\mathrm{mod}p\right)$ are integer x such that role="math" localid="1668595929447" $\mathit{x}\mathbf{\equiv }\mathbf{1}\mathbf{(}\mathbf{mod}\mathbf{}\mathbf{p}\mathbf{)}\mathbf{}\mathbf{or}\mathbf{}\mathit{x}\mathbf{\equiv }\mathbf{-}\mathbf{1}\mathbf{(}\mathbf{mod}\mathbf{}\mathbf{p}\mathbf{)}$.

Short answer

It is proved that$x\equiv 1\left(\mathrm{mod}p\right)\mathrm{or}x\equiv -1\left(\mathrm{mod}p\right)$.

Step-by-step solution

Congruence relation

Let n be fixed positive integer. Two integer a and b are said to be congruent modulo n denoted by $\mathit{a}\mathbf{\equiv }\mathit{b}\mathbf{(}\mathbf{mod}\mathit{n}\mathbf{)}$, if n divides the difference$\mathit{a}\mathbf{-}\mathit{b}$ that is provided that$\mathit{a}\mathbf{-}\mathit{b}\mathbf{=}\mathit{k}\mathit{n}$ for some integer$\mathit{k}$

Showing that x≡1(modp)  or  x≡−1(modp)

We have$p$is any prime and$x^2\equiv 1(modp)$

So, by definition of congruence relation,

We can say that,

$p$Divides$x^2-1$

Which means$p/x^2-1$

So,$p/(x-1)\cdot (x+1)$

Then we have,

$p/(x-1)\mathrm{or}p/(x+1)$

So, again by definition of congruence relation,

$x\equiv 1(modp)\mathrm{or}x\equiv -1(modp)$

Hence proved.