# Show that k is an odd integer, except when k=2

• Math100
In summary, To show that k is an odd integer means to prove that k can only be divided by 2 with a remainder of 1. To prove this, one can use the definition of an odd number which states that an odd number can be expressed as 2n+1, where n is any integer. It is important to show that k is an odd integer in certain mathematical proofs and calculations, and to identify patterns and relationships between numbers. When k=2, it is a special case and does not fit the definition of an odd integer. There are no exceptions to the rule that k is an odd integer, except when k=2.
Math100
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 ##.

usermaths
You've got to tidy that up!

PeroK said:
You've got to tidy that up!
But how?

Math100 said:
But how?
Review and edit. It's something you need to learn to do.

Math100 said:
But how?
Rule of thumb: write it in a way that convinces yourself, then delete all but every third row

Concentrate on the crucial lines:

E.g.: Assume ##p=2^k-1## is prime and assume ##k=2m >3.##

This gets you directly to your fifth line and everything is said, including the contradictional assumption.

Now you get ##p=4^m-1=(4-1)\cdot (\ldots) ## and so on.

You get the dots by using long division on ##(4^m-1):(4-1)## or in general ##(x^m-1):(x-1)## which is a valuable exercise anyway, since the formula is often used, and long division a method you should know.

Math100
Math100 said:
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. ##

$2^{2a} - 1$ is a difference of squares. Do you recall that $x^2 - y^2 = (x+y)(x - y)$?

usermaths and Math100
pasmith said:
$2^{2a} - 1$ is a difference of squares. Do you recall that $x^2 - y^2 = (x+y)(x - y)$?
Yes.

fresh_42 said:
Rule of thumb: write it in a way that convinces yourself, then delete all but every third row

Concentrate on the crucial lines:

E.g.: Assume ##p=2^k-1## is prime and assume ##k=2m >3.##

This gets you directly to your fifth line and everything is said, including the contradictional assumption.

Now you get ##p=4^m-1=(4-1)\cdot (\ldots) ## and so on.

You get the dots by using long division on ##(4^m-1):(4-1)## or in general ##(x^m-1):(x-1)## which is a valuable exercise anyway, since the formula is often used, and long division a method you should know.
So you're saying that after fifth line in my proof, I should revise it into another method?

I only said what I thought. It is a reflex if I see ##x^m -1.## It is the formula that is used for interest rates:
$$\sum_{k=0}^{m-1} r^k = \dfrac{r^m-1}{r-1}$$
As I said, this formula and the method of long division are valuable tools.

A faster method has been given to you in post #6.

Your method is the same, only that you did not provide evidence for ##3\,|\,4^m-1.## My method and the one in post #6 provide this evidence. Other methods are a proof by induction or modular arithmetics:
$$4^m-1 \mod 3 \equiv (4\mod 3)^m-1 =1^m-1=0 \Longrightarrow 3\,|\,4^m-1$$

Once you have ##3\,|\,p## you automatically have ##3=p## since ##p## is prime and so ##k=2##, contradicting the assumption.

All you have written can be compressed into that line.

Math100
I wouldn't write this as a proof by contradiction. You derive a contradiction. It's a problem here because you want to derive a contradiction from:

Math100 said:
Suppose for the sake of contradiction that p=2k−1 is prime
but k is not an odd integer.
You never derive a contradiction to the complete statement.

It would be much simpler to start with "k is even and p = prime)" and derive k=2 from that.

PeroK

## 1. How do you prove that k is an odd integer?

To prove that k is an odd integer, we need to show that it can be expressed as 2n+1, where n is any integer. This means that k is one more than an even number, making it an odd integer.

## 2. Why is k=2 an exception to being an odd integer?

When k=2, it can be expressed as 2(1), which is an even number. Therefore, it does not fit the definition of an odd integer and is considered an exception.

## 3. Can you provide an example of an odd integer other than k=2?

Yes, for example, k=5 can be expressed as 2(2)+1, making it an odd integer. Any integer that can be expressed as 2n+1, where n is an integer, is considered an odd integer.

## 4. How does knowing that k is an odd integer help in solving equations?

Knowing that k is an odd integer can help in solving equations by narrowing down the possible solutions. It eliminates the need to consider even numbers, making the solution process more efficient.

## 5. Can k be a negative odd integer?

Yes, k can be a negative odd integer. For example, k=-3 can be expressed as 2(-2)+1, making it an odd integer. The definition of an odd integer does not specify whether it is positive or negative, as long as it can be expressed as 2n+1.

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