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Expert-verifiedFound in: Page 30

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: **

** **

**$a=2,p=5$****.****$a=4,p=7$****.****$a=3,p=11$****.**

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

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

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

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

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

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