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Old 2010-06-29, 05:30   #1
Carl Fischbach
 
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Default The composite conjecture

For the equation

2^(2*n)+1=K=x*y


For every whole value of n and composite value of K there exists
a pair of whole factors x and y where x+y= 2^z +2.
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Old 2010-06-29, 06:38   #2
xilman
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Quote:
Originally Posted by Carl Fischbach View Post
For the equation

2^(2*n)+1=K=x*y


For every whole value of n and composite value of K there exists
a pair of whole factors x and y where x+y= 2^z +2.
And what is z?

Paul
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Old 2010-06-29, 10:26   #3
Carl Fischbach
 
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Good point, z is a whole.
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Old 2010-06-29, 10:47   #4
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Isn't y=1 always a solution?
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Old 2010-06-29, 13:08   #5
ATH
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If you mean fermat numbers 22[sup]n[/sup] + 1, then factors are of the form k*2n+2 + 1, so:

x+y = (k1*2n+2 + 1) + (k2*2n+2 + 1) = (k1+k2)*2n+2 + 2

So your z = n+2 , but (k1+k2) != 1 so you miss a factor infront of 2z.

Last fiddled with by ATH on 2010-06-29 at 13:09
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Old 2010-06-30, 06:28   #6
Carl Fischbach
 
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y=1 and x=K is only one solution for a composite, most composites
will produce several unique factor pairs, my conjecture is saying
that K will have an unique factor pair x and y where x and y will
add up to 2^z +2.

No I do not mean fermat numbers which are, 2^2+1, 2^4+1, 2^8+1 ...
I mean the following numbers 2^2+1, 2^4+1, 2^6+1, 2^10+1 ...

The following are x and y examples:

2^2+1=prime

2^4+1= prime

2^6+1= 65= 5*13
5+13=18=2^4+2

2^8+1=prime

2^10+1=1025= 25*41
25+41=66+2^6+2

2^12+1=4097=17*241
17+241=258=2^8+2

2^14+1=16385=145*113
145+113=258=2^8+2

2^16+1=prime

2^18+1=262145=65*4033
65+4033=4098=2^12+2


Using sucessive approximation for all possible z values you can easily
solve x and y for any large K value, if the conjecture can be proven
then K can also be proven to be prime.
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Old 2010-06-30, 08:11   #7
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Quote:
Originally Posted by Carl Fischbach View Post
y=1 and x=K is only one solution for a composite, most composites
will produce several unique factor pairs, my conjecture is saying
that K will have an unique factor pair x and y where x and y will
add up to 2^z +2.

No I do not mean fermat numbers which are, 2^2+1, 2^4+1, 2^8+1 ...
I mean the following numbers 2^2+1, 2^4+1, 2^6+1, 2^10+1 ...

The following are x and y examples:

2^2+1=prime

2^4+1= prime

2^6+1= 65= 5*13
5+13=18=2^4+2

2^8+1=prime

2^10+1=1025= 25*41
25+41=66+2^6+2

2^12+1=4097=17*241
17+241=258=2^8+2

2^14+1=16385=145*113
145+113=258=2^8+2

2^16+1=prime

2^18+1=262145=65*4033
65+4033=4098=2^12+2


Using sucessive approximation for all possible z values you can easily
solve x and y for any large K value, if the conjecture can be proven
then K can also be proven to be prime.
2^32+1 = 641*6700417
641+6700417=6701058=2^14*409+2

2^40+1=257*4278255361
257+4278255361
=2^16*97*673+2

2^92+1 = 17*291280009243618888211558641
17+291280009243618888211558641
=28*241*3361*15790321*88959882481+2


Maybe here is what you see
2^2h+1=L.M, L=2^h-2^k+1, M=2^h+2^k+1, h=2k-1.
L+M=2^(h+1)+2
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Old 2010-07-02, 06:39   #8
Carl Fischbach
 
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Good work on those numbers.
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Old 2010-07-02, 08:03   #9
Batalov
 
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Quote:
Originally Posted by wreck View Post
Maybe here is what you see
2^2h+1=L.M, L=2^h-2^k+1, M=2^h+2^k+1, h=2k-1.
L+M=2^(h+1)+2
wreck (and Aurifeuille) made OP trivially true for odd h (n in original post).

You could have the same fun with the powers of three:
for all odd h, there exist L<N<M such that
33h+1=L.N.M, and L+N+M=3h+1+3.
(and L,N,M make an arithmetic progression, too)
E.g. 315+1=217.244.271, and 217+244+271=36+3.

3999+1=
760988023132059809720425867265032780727896356372077865117010037035791631439306151832670880090020513365489579531305695215863254325114203194140903811357733748937.
760988023132059809720425867265032780727896356372077865117010037035791631439306199613044145649378522557935351570949952010001833769302566531786879537190794573524.
760988023132059809720425867265032780727896356372077865117010037035791631439306247393417411208736531750381123610594208804140413213490929869432855263023855398111,
and their sum is 3334+3.
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