Question

Use Fermat’s little theorem to show that if $\mathit{p}$is prime and $\mathit{p}\mathit{\chi }\mathit{a}$, then role="math" localid="1668657209564" ${\mathit{a}}^{\mathbf{p}\mathbf{-}\mathbf{2}}$is an inverse of a modulo $\mathit{p}$

Short answer

Using the Fermat’s little theorem,

${\mathit{a}}^{\mathbf{p}\mathbf{-}\mathbf{2}}$is an inverse of $\mathit{a}$modulo $\mathit{p}$

Step-by-step solution

Step 1

Given

is prime and not divisible by

To proof:

${\mathit{a}}^{\mathbf{p}\mathbf{-}\mathbf{2}}$is an inverse of $\mathit{a}$modulo$\mathit{p}$

Proof

Fermat’s little theorem states$a^{p-1}\equiv 1(modp)$if $p$is prime and $a$not divisible by$p$

$a^{p-1}\equiv 1(modp)$

Step 2

Since$a^{p-1}=a\cdot a^{p-2}$

$a\cdot a^{p-2}\equiv 1\left(\mathrm{modp}\right)$

The inverse of role="math" localid="1668657544210" $a$modulo $m$is an integer $b$for which $ab\equiv 1\left(\mathrm{modm}\right)$

Thus, the inverse of $a$modulo $p$is then $a^{p-1}$