Question

Show with the help of Fermat’s little theorem that if n is a positive integer then 42 divides${\mathbf{n}}^{\mathbf{7}}\mathbf{-}\mathbf{n}$

Short answer

$\mathbf{42}$divides ${\mathbf{n}}^{\mathbf{7}}\mathbf{-}\mathbf{n}$is proved by Fermat’s little theorem

Step-by-step solution

Fermat’s little theorem    

FERMAT’S LITTLE THEOREM tells us that If p, is prime and ais an integer not divisible by p

Then ${\mathbf{a}}^{\mathbf{p}\mathbf{-}\mathbf{1}}\mathbf{\equiv }\mathbf{1}\mathbf{(}\mathbf{mod}\mathbf{p}\mathbf{)}$

To prove 42 divides n7 - n

Let p = 7anda = n

${\mathit{n}}^{\mathbf{7}\mathbf{-}\mathbf{1}}\mathbf{=}{\mathit{n}}^{\mathbf{6}}\mathbf{\equiv }\mathbf{1}\mathbf{mod}\mathbf{7}$

$\mathit{a}\mathbf{\equiv }\mathbf{b}\mathbf{mod}\mathbf{m}$, If role="math" localid="1668680041705" $\mathbf{m}$divides $\mathbf{a}\mathbf{-}\mathbf{b}$then,

$\mathbf{7}\mathbf{\mid }\left(n^6-1\right)$

Now since ${\mathbf{n}}^{\mathbf{7}}\mathbf{-}\mathbf{n}\mathbf{=}\mathbf{n}\left(n^6-1\right)$,

$\mathbf{7}\left|\left(n^6-1\right)=7\right|\left(n^7-n\right)$

Factorize n7−n

By factorising

$$\begin{gathered} {\mathbf{n}}^{\mathbf{7}}\mathbf{-}\mathbf{n}\mathbf{=}\mathbf{n}\left(n^6-1\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\begin{array}{r}\mathbf{=}\mathbf{n}\left(n^3-1\right)\left(n^3+1\right)\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\\ \mathbf{=}\mathbf{n}\mathbf{(}\mathbf{n}\mathbf{-}\mathbf{1}\mathbf{)}\left(n^2+n+1\right)\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}\left(n^2-n+1\right)\end{array} \end{gathered}$$

Here$\left(n-1\right)$and $\mathbf{n}$are two consecutive integers and one of them is divisible by

$\mathbf{2}\mathbf{\mid }\mathbf{(}\mathit{n}\mathbf{-}\mathbf{1}\mathbf{)}$or $\mathbf{2}\mathbf{\mid }\mathit{n}$

Thus, $\mathbf{2}\mathbf{\mid }\mathbf{(}{\mathit{n}}^{\mathbf{7}}\mathbf{-}\mathit{n}\mathbf{)}$

Factorize n7−n

By factorising

$$\begin{gathered} {\mathbf{n}}^{\mathbf{7}}\mathbf{-}\mathbf{n}\mathbf{=}\mathbf{n}\mathbf{(}{\mathbf{n}}^{\mathbf{6}}\mathbf{-}\mathbf{1}\mathbf{)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\begin{array}{r}\mathbf{=}\mathbf{n}\mathbf{(}{\mathbf{n}}^{\mathbf{3}}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{(}{\mathbf{n}}^{\mathbf{3}}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\\ \mathbf{=}\mathbf{n}\mathbf{(}\mathbf{n}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{(}{\mathbf{n}}^{\mathbf{2}}\mathbf{+}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{(}{\mathbf{n}}^{\mathbf{2}}\mathbf{-}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}\end{array} \end{gathered}$$

Here$\mathbf{(}\mathit{n}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{}\mathbf{,}\mathbf{}\mathbf{(}\mathit{n}\mathbf{+}\mathbf{1}\mathbf{)}$and $\mathit{n}$ are three consecutive integers and one of them is divisible by 3

$\mathbf{3}\mathbf{\mid }\mathbf{(}\mathit{n}\mathbf{-}\mathbf{1}\mathbf{)}\text{or}\mathbf{3}\mathbf{\mid }\mathbf{(}\mathit{n}\mathbf{+}\mathbf{1}\mathbf{)}\text{or}\mathbf{3}\mathbf{\mid }\mathit{n}$

Thus,$\mathbf{3}\mathbf{\mid }\left(n^7-n\right)$

Proof

The factors are $\mathbf{7}\left|\left(n^7-n\right),2\right|\left(n^7-n\right)$and $\mathbf{3}\mathbf{\mid }\left(n^7-n\right)$

Now,

$$\begin{gathered} \mathbf{(}\mathbf{7}\mathbf{.2.3}\mathbf{)}\mathbf{\mid }\left(n^7-n\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{42}\mathbf{\mid }\left(n^7-n\right) \end{gathered}$$

Therefore $\mathbf{}\mathbf{42}$divides${\mathbf{n}}^{\mathbf{7}}\mathbf{-}\mathbf{n}$

Thus proved.