My Math Forum 2 elementary number theory problems....

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 October 2nd, 2010, 01:32 AM #1 Newbie   Joined: Jun 2009 Posts: 14 Thanks: 0 2 elementary number theory problems.... problem 1: suppose $P$ is a prime number such that (P-1)/4 and (P+1)/2 are also primes. Show that p = 13. problem 2: let $a_1,a_2,...,a_n$ be n numbers such that each $a_i$ is either $1$ or $-1$. If $a_1a_2a_3a_4 + a_2a_3a_4a_5 + ......+ a_na_1a_2a_3= 0$, then prove that $4$ divides $n$. HELP!
 October 2nd, 2010, 08:36 PM #2 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: 2 elementary number theory problems.... For #1, work mod 4.
 October 4th, 2010, 05:31 PM #3 Newbie   Joined: Sep 2010 Posts: 17 Thanks: 0 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. ;]
 October 8th, 2010, 06:42 AM #4 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,825 Thanks: 1051 Math Focus: Elementary mathematics and beyond 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. Thanks from agentredlum
October 9th, 2010, 01:19 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.

 October 9th, 2010, 05:54 AM #6 Math Team   Joined: Apr 2010 Posts: 2,778 Thanks: 361 Re: 2 elementary number theory problems.... Thanks, CRGreathouse, I'll use $\bmod 2$. I unfortunately can't edit anymore, so i'll replace a part: So we have: $(-1)^{(a+b+c+d)\;\bmod 2}+(-1)^{(b+c+d+e)\;\bmod 2}+(-1)^{(c+d+e+f)\;\bmod 2}+(-1)^{(d+e+f+g)\;\bmod 2}+...+(-1)^{(n+a+b+c)\; \bmod 2}=\\(-1)^{(a+b+c+d)\;\bmod 2}+(-1)^{(a+b+c+d+(e-a))\;\bmod 2}+\\ (-1)^{(a+b+c+d+(e-a)+(f-b))\;\bmod 2}+...+(-1)^{(a+b+c+d+(n-d))\;\bmod 2}$ And than factor out $(-1)^{(a+b+c+d)\;\bmod 2}$ mod 2 is in the exponent. Would that help? Or is there a "better" alternative? Hoempa

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