Question

Show that $a^{p-1}=1\left(modp\right)$ for the given p and a:

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

Short answer

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

Step-by-step solution

Write the definition of Congruence

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

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

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

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