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Topic: Conjecture Prime
Replies: 33   Last Post: Apr 26, 2013 7:21 AM

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ab

Posts: 49
Registered: 6/23/09
Re: Conjecture Prime
Posted: Jun 25, 2009 2:57 AM
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On Jun 25, 3:25 pm, Martin Michael Musatov <marty.musa...@gmail.com>
wrote:
> > 2n * 2 (n+1) - 1 is prime
>
> > it's pretty easy to show if n = 1 or 5 (mod 7) then 7
> > will divide the
> > whole thing

>
> I never wrote this:


you did, in words, in your very first post. this was the problem
before your revisions. n is any integer, 2n any even integer, 2 (n+1)
the next even integer.


but back to the main problem. it's true, and true that any number,
times that number plus 2, take away 1, will not be a square number. it
will always be 2 less than a square number. eg 8 * 10 - 1 = 79, 2 less
than 81.

this is shown by "completing the square"

n (n+2) - 1 = (n+1)(n+1) - 2, in other words the number will always be
two less than a square number.

by the way squares are distributed this can never be a square number.

for your expression [[N*N]*([N*N]+2)]-1 = (N*N+1)(N*N+1) - 2.

i liked the original problems you posted, but now everything has
become trivial






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