document.write( "Question 409839: ) Show that if n is a natural number such that 2n-1 is prime, thenn n is a prime. Is the concerse true? Explain?\r
\n" ); document.write( "\n" ); document.write( "B) Show that if n is a natural number such that 2n+1 is prime, thenn n is a power of 2. Is the concerse true? Explain?\r
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Algebra.Com's Answer #288355 by richard1234(7193) $\"\"$ $\"About$

You can put this solution on YOUR website!
I sure hope you mean $\"2%5En+-+1\"$ instead of $\"2n-1\"$ . In fact I am working on a mathematics project regarding odd perfect numbers, and such a prime in the form $\"2%5En+-+1\"$ is important since there is a one-to-one correlation between these primes (called Mersenne primes) and the *even* perfect numbers.\r
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\n" ); document.write( "\n" ); document.write( "Suppose that n is not prime, i.e. n = ab for positive integers a and b, where a and b are not equal to 1. Then, $\"2%5En+-+1+=+2%5Eab+-+1\"$ , and this can be factored as $\"%282%5Ea+-+1%29%2Asum%282%5Eai%2C+i+=+1%2C+b-1%29\"$ (this comes from the geometric series). Since neither factor is 1, then $\"2%5En+-+1\"$ can be factored as a product of two integers, hence it is not prime. The contrapositive is true, i.e. if $\"2%5En+-+1\"$ is prime, then $\"n\"$ is prime. The converse is not necessarily true, and we can find counterexamples, such as n = 11, as $\"2%5E11+-+1+=+23%2A89\"$ .\r
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\n" ); document.write( "\n" ); document.write( "Part b is a little tricky, I spent a bit of time on the problem and couldn't come with a full solution. Perhaps you could use an inductive argument, or show that $\"2%5E%282%5Ek%29+%2B+1\"$ cannot be factored for any k (which might also use induction). \n" ); document.write( "

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