- #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 ##.