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- #1
- Apr 13, 2013
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Hey!!! 
I am given the following exercise: Show that $$15|2^{4n}-1$$
How can I do this??
I am given the following exercise: Show that $$15|2^{4n}-1$$
How can I do this??
Hai!Hey!!!
I am given the following exercise: Show that $$15|2^{4n}-1$$
How can I do this??![]()
$2^{4n}-1 \pmod 3=0 $ and $2^{4n}-1 \pmod 5=0$.So,we have writen $15$ as a product of prime numbers $3 \cdot 5$,so the remainder of the division of $2^{4n}-1$ with both of these prime numbers should be $0$??Hai!
What is $2^{4n}-1 \pmod 3$?
And $2^{4n}-1 \pmod 5$?
Yep!$2^{4n}-1 \pmod 3=0 $ and $2^{4n}-1 \pmod 5=0$.So,we have writen $15$ as a product of prime numbers $3 \cdot 5$,so the remainder of the division of $2^{4n}-1$ with both of these prime numbers should be $0$??![]()
Great!!!Thank you very much!!!Yep!![]()
I'd probably use induction. To show $\displaystyle \begin{align*} 15 | \left( 2^{4n} - 1 \right) \end{align*}$, that means we have to show $\displaystyle \begin{align*} 2^{4n} - 1 = 15p \end{align*}$, where $\displaystyle \begin{align*} p \in \mathbf{Z} \end{align*}$ for all $\displaystyle \begin{align*} n \in \mathbf{N} \end{align*}$.Hey!!!
I am given the following exercise: Show that $$15|2^{4n}-1$$
How can I do this??![]()