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### Abstract Algebra: An Introduction

Book edition 3rd
Author(s) Thomas W Hungerford, David Leep
Pages 608 pages
ISBN 9781111569624

# Show that ${a}^{p-1}=1\left(modp\right)$ for the given p and a: $a=2,p=5$ .$a=4,p=7$ .$a=3,p=11$ .

1. It is proved that ${2}^{5-1}\equiv 1\left(mod5\right)$ .
2. It is proved that ${4}^{7-1}\equiv 1\left(mod7\right)$ .
3. It is proved that ${3}^{11-1}\equiv 1\left(mod11\right)$ .
See the step by step solution

## Step 1: Write the definition of Congruence

Consider that a, b, n as an integer with $n>0$ . If n divides $a-b$, then a is congruent tob modulo n (expressed as $a\equiv b\left(modn\right)$ ).

## Step 2: Show that ap-1=1modp for the given p and a

a)

Put $a=2$ and $p=5$ in the expression ${a}^{p-1}$ as shown below:

$\begin{array}{rcl}{2}^{5-1}& =& {2}^{4}\\ & =& 18\end{array}$

Divide 16 by 5 as follows:

$16=5\left(3\right)+1$

Then, ${2}^{5-1}$ can be expressed as follows:

${2}^{5-1}=1\left(mod5\right)$

Thus, it is proved that ${2}^{5-1}=1\left(mod5\right)$ .

## Step 3: Show that ap-1=1modp for the given p and a

b)

Put $a=4$ and $b=7$ in the expression ${a}^{p-1}$ as shown below:

$\begin{array}{rcl}{4}^{7-1}& =& {4}^{6}\\ & =& 4096\end{array}$

Divide 4096 by 7 as follows:

$4096=7\left(585\right)+1$

Then, ${4}^{7-1}$ can be expressed as follows:

${4}^{7-1}=1\left(mod7\right)$

Thus, it is proved that ${4}^{7-1}=1\left(mod7\right)$ .

## Step 4: Show that ap-1=1modp  for the given p and a

c)

Put $a=3$ and $p=11$ in the expression ${a}^{p-1}$ as shown below:

$\begin{array}{rcl}{3}^{11-1}& =& {3}^{10}\\ & =& 59049\end{array}$

Divide 59049 by 11 as follows:

$59049=5368\left(11\right)+1$

Then, ${3}^{11-1}$ can be expressed as follows:

${3}^{11-1}\equiv 1\left(mod11\right)$

Thus, it is proved that ${3}^{11-1}\equiv 1\left(mod11\right)$ .