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# Algebra manipulation

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Hi.

I recently stumbled upon this question and still can't figure out how to prove it. This is a GCSE question and it has never been this hard before.

$$k=2^p - 1$$

$$N = k^2 -1$$

Show that 2^(p+1) is a factor of N

bqrs01

bqrs01  May 27, 2017
edited by bqrs01  May 28, 2017
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#1
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You say "Show that N is a factor of 2^(p+1)" ??. Are you sure it is not the reverse? That is:

Show that 2^(p+1) is a factor of N?

Guest May 27, 2017
#2
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ok :)

$$k=2^p-1\\ N=k^2-1\\ N=(2^p-1)^2-1\\ N=2^{2p}-2*2^p+1-1\\ N=2^p*2^p-2*2^p\\ N=2^p(2^p-2)\\ N=2^{p+1}(2^{p-1}-1)\\ so\\ 2^{p+1}\text{ is a factor of }N$$

So I have shown that  2^(p+1) is a factor of N

Which is the other way around from what you asked.

Melody  May 28, 2017

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