- #1

mae

- 6

- 0

**Prove that if n is a perfect square, then (2^n) -1 is not prime.**

All I can get is that 2^n is some even number. I can't work in the perfect square part.

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In summary, the conversation discusses trying to prove that if n is a perfect square, then (2^n) - 1 is not prime. The conversation explores various methods and approaches, including factoring and using induction. The final attempt is to use the contrapositive statement, but it does not lead to a conclusive answer.

- #1

mae

- 6

- 0

All I can get is that 2^n is some even number. I can't work in the perfect square part.

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- #2

NateTG

Science Advisor

Homework Helper

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2^4-1=15=3*5

2^9-1=511=7*73

2^16-1=323767=7*31*151

You could at least write the expression as:

[tex]2^{k \times k}-1[/tex]

There's an easy answer when k is even.

- #3

mae

- 6

- 0

I feel like I've tried everything. The last thing I tried was contrapositive : If (2^n)-1 is prime, then n is not a perfect square. No dice... at least for me.

- #4

morphism

Science Advisor

Homework Helper

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By the way, primes of the form 2^n - 1 are called Mersenne primes. They're relatively well-known, and have many unsolved problems associated to them. For example, are there infinitely many Mersenne primes?

- #5

mae

- 6

- 0

I've been up all night and I can't really think straight anymore. I'm not sure that would help anyway though.

- #6

qspeechc

- 844

- 15

n = k^2

It is true for k=1, k=2. Assume it is true for some k. So 2^(k^2)-1 is not prime

For the next number k+1, n = k^2 + 2k +1, we have:

2^n - 1= 2^(k^2 + 2k +1) - 1 = 2^(k^2) + 2^(2k) + 1 = 2^n -1 + 2^(2k) +2

Er, then, I don't know what.

- #7

mae

- 6

- 0

- #8

mae

- 6

- 0

- #9

mae

- 6

- 0

Holy crap, I think I just got it.

Edit: I did not.

This is for contrapositive, which seems more promising.

I have 2^n -1 = p (some prime)

so... 2^n = p+1 (some even)

now, if only log[2](p+1) was something easy and neat.

Edit: I did not.

This is for contrapositive, which seems more promising.

I have 2^n -1 = p (some prime)

so... 2^n = p+1 (some even)

now, if only log[2](p+1) was something easy and neat.

Last edited:

- #10

NateTG

Science Advisor

Homework Helper

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[tex]x^2-1=(x-1)(x+1)[/tex]

[tex]x^3-1=(x-1)(x^2+x+1)[/tex]

[tex]x^4-1=(x-1)(x^3+x^2+x+1)[/tex]

...

The significance of this proof lies in the fact that it helps us better understand the patterns and properties of prime numbers. It also has applications in cryptography and number theory.

There is ongoing research on this topic, with new discoveries and proofs being made. However, the proof for proving that (2^n) - 1 is not prime for perfect squares has been established and accepted by the mathematical community.

The proof involves using mathematical techniques such as modular arithmetic, algebraic manipulations, and number theory concepts. It requires a deep understanding of prime numbers and their properties.

One of the main challenges in this proof is finding a general formula or method that can be applied to all values of n to prove the statement. Additionally, the proof may involve complex mathematical concepts that require a high level of expertise to understand and apply.

The proof has implications for other areas of mathematics, such as number theory, cryptography, and algebra. It also opens up new avenues for further research and exploration in the field of prime numbers and their properties.

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