- #1

Math100

- 771

- 219

- Homework Statement
- Another unproven conjecture is that there are an infinitude of primes that are ## 1 ## less than a power of ## 2 ##, such as ## 3=2^{2}-1 ##.

If ## p=2^{k}-1 ## is prime, show that ## k ## is an odd integer, except when ## k=2 ##.

- Relevant Equations
- None.

Proof:

Suppose for the sake of contradiction that ## p=2^{k}-1 ## is prime

but ## k ## is not an odd integer.

That is, ## k ## is an even integer.

Then we have ## k=2a ## for some ## a\in\mathbb{Z} ##.

Thus ## p=2^{k}-1

=2^{2a}-1

=4^{a}-1. ##

Note that ## 3\mid 4^{n}-1 ## for all ## n\geq1 ##.

This means ## 3 ## divides ## p ##,

and so ## p ## must be ## 3 ##,

because ## p ## is a prime number.

Now we have ## p=2^{k}-1

=3 ##,

which implies that ## 2^{k}=3+1=4 ##,

so ## k=2 ##.

Since ## k\neq2 ##,

it follows that ## p\neq3 ##.

This is a contradiction because p cannot be a composite,

given the fact that ## p=2^{k}-1 ## is prime.

Therefore, if ## p=2^{k}-1 ## is prime,

then ## k ## is an odd integer, except when ## k=2 ##.

Suppose for the sake of contradiction that ## p=2^{k}-1 ## is prime

but ## k ## is not an odd integer.

That is, ## k ## is an even integer.

Then we have ## k=2a ## for some ## a\in\mathbb{Z} ##.

Thus ## p=2^{k}-1

=2^{2a}-1

=4^{a}-1. ##

Note that ## 3\mid 4^{n}-1 ## for all ## n\geq1 ##.

This means ## 3 ## divides ## p ##,

and so ## p ## must be ## 3 ##,

because ## p ## is a prime number.

Now we have ## p=2^{k}-1

=3 ##,

which implies that ## 2^{k}=3+1=4 ##,

so ## k=2 ##.

Since ## k\neq2 ##,

it follows that ## p\neq3 ##.

This is a contradiction because p cannot be a composite,

given the fact that ## p=2^{k}-1 ## is prime.

Therefore, if ## p=2^{k}-1 ## is prime,

then ## k ## is an odd integer, except when ## k=2 ##.