- #1

Math100

- 783

- 220

- Homework Statement
- From Fermat's theorem deduce that, for any integer ## n\geq 0, 13\mid (11^{12n+6}+1) ##.

- Relevant Equations
- None.

Proof:

Let ## n\geq 0 ## be any integer.

Applying the Fermat's theorem produces:

## a=11, p=13 ## and ## p\nmid a ##.

Then ## 11^{13-1}\equiv 1\pmod {13}\implies 11^{12}\equiv 1\pmod {13} ##.

Observe that

\begin{align*}

&11^{12n+6}+1\equiv [(11^{12})^{n}\cdot 11^{6}+1]\pmod {13}\\

&\equiv [1^{n}(-2)^{6}+1]\pmod {13}\\

&\equiv (64+1)\pmod {13}\\

&\equiv 65\pmod {13}\\

&\equiv 0\pmod {13}.\\

\end{align*}

Thus ## 13\mid (11^{12n+6}+1) ##.

Therefore, ## 13\mid (11^{12n+6}+1) ## for any integer ## n\geq 0 ##.

Let ## n\geq 0 ## be any integer.

Applying the Fermat's theorem produces:

## a=11, p=13 ## and ## p\nmid a ##.

Then ## 11^{13-1}\equiv 1\pmod {13}\implies 11^{12}\equiv 1\pmod {13} ##.

Observe that

\begin{align*}

&11^{12n+6}+1\equiv [(11^{12})^{n}\cdot 11^{6}+1]\pmod {13}\\

&\equiv [1^{n}(-2)^{6}+1]\pmod {13}\\

&\equiv (64+1)\pmod {13}\\

&\equiv 65\pmod {13}\\

&\equiv 0\pmod {13}.\\

\end{align*}

Thus ## 13\mid (11^{12n+6}+1) ##.

Therefore, ## 13\mid (11^{12n+6}+1) ## for any integer ## n\geq 0 ##.