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October 2nd, 2010, 02:32 AM   #1
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2 elementary number theory problems....

problem 1:
suppose is a prime number such that (P-1)/4 and (P+1)/2 are also primes. Show that p = 13.

problem 2:
let be n numbers such that each is either or . If , then prove that divides .

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October 2nd, 2010, 09:36 PM   #2
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Re: 2 elementary number theory problems....

For #1, work mod 4.
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October 4th, 2010, 06:31 PM   #3
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Re: 2 elementary number theory problems....

If P, (P+1)/2 and (P-1)/4 are prime numbers then P=4x+1, so our numbers are 4x+1, 2x+1 and x.

Now, if x=3K then K must be 1, x=3.
if x=3K+1 then 2x+1=2(3K+1)+1=3(2K+1). Then 2K+1 must be 1, so x=1 which is not prime.
if x=3K+2 then 4x+1=4(3K+2)+1=3(4K+3). Then 4K+3 must be 1, and x isn't even an integer.

So the only solution is x=3, P=13.

;]
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October 8th, 2010, 07:42 AM   #4
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Re: 2 elementary number theory problems....

Clearly, p = 2, 3 are not solutions.

Assume p is of the form 6n - 1. p = 6n - 1, (p - 1)/4 = (3n - 1)/2, (p + 1)/2 = 3n; 3n can never be prime so there are no primes of the form 6n - 1 that are solutions.

Assume p is of the form 6n + 1. p = 6n + 1, (p - 1)/4 = 3n/2, (p + 1)/2 = 3n + 1; 3n/2 can be prime only when n = 2, therefore p = 13.
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October 9th, 2010, 02:19 AM   #5
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Quote:
Originally Posted by greg1313
3n can never be prime . . .
It's prime if n = 1, but then (p - 1)/4 = (3n - 1)/2 = 1, which isn't prime.
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October 9th, 2010, 06:54 AM   #6
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Re: 2 elementary number theory problems....

Thanks, CRGreathouse, I'll use . I unfortunately can't edit anymore, so i'll replace a part: So we have:

And than factor out
mod 2 is in the exponent.
Would that help? Or is there a "better" alternative?

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