20100629, 05:30  #1 
Oct 2007
2·17 Posts 
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. 
20100629, 06:38  #2 
Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
2991_{16} Posts 

20100629, 10:26  #3 
Oct 2007
2×17 Posts 
Good point, z is a whole.

20100629, 10:47  #4 
"William"
May 2003
New Haven
3·787 Posts 
Isn't y=1 always a solution?

20100629, 13:08  #5 
Einyen
Dec 2003
Denmark
3071_{10} Posts 
If you mean fermat numbers 2^{2[sup]n}[/sup] + 1, then factors are of the form k*2^{n+2} + 1, so:
x+y = (k_{1}*2^{n+2} + 1) + (k_{2}*2^{n+2} + 1) = (k_{1}+k_{2})*2^{n+2} + 2 So your z = n+2 , but (k_{1}+k_{2}) != 1 so you miss a factor infront of 2^{z}. Last fiddled with by ATH on 20100629 at 13:09 
20100630, 06:28  #6 
Oct 2007
2·17 Posts 
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. 
20100630, 08:11  #7  
"Bo Chen"
Oct 2005
Wuhan,China
A7_{16} Posts 
Quote:
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^h2^k+1, M=2^h+2^k+1, h=2k1. L+M=2^(h+1)+2 

20100702, 06:39  #8 
Oct 2007
22_{16} Posts 
Good work on those numbers.

20100702, 08:03  #9  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
10010010101100_{2} Posts 
Quote:
You could have the same fun with the powers of three: for all odd h, there exist L<N<M such that 3^{3h}+1=L.N.M, and L+N+M=3^{h+1}+3. (and L,N,M make an arithmetic progression, too) E.g. 3^{15}+1=217.244.271, and 217+244+271=3^{6}+3. 3^{999}+1= 760988023132059809720425867265032780727896356372077865117010037035791631439306151832670880090020513365489579531305695215863254325114203194140903811357733748937. 760988023132059809720425867265032780727896356372077865117010037035791631439306199613044145649378522557935351570949952010001833769302566531786879537190794573524. 760988023132059809720425867265032780727896356372077865117010037035791631439306247393417411208736531750381123610594208804140413213490929869432855263023855398111, and their sum is 3^{334}+3. 

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