Question

Let a and n be two positive integers with a,n => 2. Prove that if a^n - 1
is prime then a = 2 and n is prime.

Answer 1

\[a,n \ge 2\]

Answer 2

what does _> INDICATE

Answer 3

wrote it above just now

Answer 4

is equal to or greater than

Answer 5

yes mate

Answer 6

it is prime

Answer 7

i need to prove it

Answer 8

oh ok

Answer 9

any ideas anyone?

Answer 10

2 power anything - 1 is always prime try it.

Answer 11

sorry 2 power any prime number.

Answer 12

but i need to prove it for all numbers aboove and equal to 2

Answer 13

2^11-1 is not prime

Answer 14

this is actually really easy...where are you stuck?

Answer 15

please show how to prove.

Answer 16

I need to prove for all numbers equal to and above the number 2, that when n is prime and a is 2, the equation \[2^{n}-1\] is always prime. how do i prove this?

Answer 17

\[2^n-1\] is not always prime...see my counter example above.

Answer 18

then im lost, dont get the question, my first post shows the whole question as its written

Answer 19

that one is stated correctly...your recent statement is not

Answer 20

this question confuses me, can you explain what im supposed to do please?

Answer 21

is says that if \[2^n-1\] is prime then a must be 2 and n must be prime

Answer 22

yes i need to prove that

Answer 23

i get my mistake previously i was working the wrong way around, sorry

Answer 24

start by showing that is \[2^n-1\] is prime then n is prime

Answer 25

how do i show that?

Answer 26

what happens if n is not prime

Answer 27

when n is not prime the equation is not prime

Answer 28

but that doesn't prove it for all results?

Answer 29

prove it is not prime..one step at a time...geesh

Answer 30

can you explain it fully?

Answer 31

if n is not prime then n=x*y

\[2^n-1=2^{xy}-1=(2^x)^y-1\]

this is factorable...you can show that.

thus n must be prime

Answer 32

after you show it is factorable...
assume a is odd then a^n-1 is even and not prime so it is not prime

so a must be even

assume a>2 then a=2*m

show \[(2m)^n-1\] is divisible by (2m-1)

thus a=2 is the only number that works

Answer 33

thanks very much i get the proving by contradiction. i dont get how to show the last part, that (2m)^n -1 is divisible by (2m-1). appreciate the help

Answer 34

prove it by induction

Answer 35

proving by induction, subbed in n = 1 for base which works, then n+1 so (2m)^(n+1) -1 but dont know how to prove its divisible from there

Answer 36

or use

\[x^n-1=(x-1)(x^{n-1}+x^{n-2}+\cdots+x+1)\]

Answer 37

id rather use induction, need to learn how to do that for my test

Answer 38

\[(2m)^{n+1}-1=(2m)^n(2m)-1=(2m)^n(1+(2m-1))-1\]

\[=(2m)^n-1+(2m-1)(2m)^n\]

Answer 39

is that the whole proof?

Answer 40

you then state that \[(2m-1)|((2m)^n-1)\]

by our assumption and

\[(2m-1)|(2m-1)(2m)^n\]

since 2m-1 is a factor. Thus it dives the sum.

Answer 41

*divides

Answer 42

thanks i get it now. how does this show that a = 2 is the only number that works when the equation is prime?