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 2006-10-20, 23:12 #1 roger     Oct 2006 4048 Posts Powers and numbers in sequence I've been working on various forms to create primes, and came up with one that doen't create primes itself (though a similar one does), but has interesting properties: n^2-1 = (n-1)(n+1). I expanded this into n^2-x^2 = (n-x)(n+x), and also found n^2+x^2 = (n-x)(n+x)+x. The cubed form of n then is expressed as n^3-nx^2 = (n-x)(n)(n+x) and n^3+nx^2 = (n-x)(n)(n+x)+2nx^2 . The first n^4 equation is n^4-4*n^2*x^2-n^2+n-(n-4) = (n-2x)(n-x)(n+x)(n+2x), but I haven't found the equation beginning with n^4+4*n^2*x^2. Can anyone shed some light on the problem? Thanks, Roger PS - equations may not be completely accurate - will verify by next post [signs etc]
2006-10-23, 09:58   #2
mfgoode
Bronze Medalist

Jan 2004
Mumbai,India

22·33·19 Posts
Expansions

Quote:
 Originally Posted by roger I've been working on various forms to create primes, and came up with one that doen't create primes itself (though a similar one does), but has interesting properties: n^2-1 = (n-1)(n+1). I expanded this into n^2-x^2 = (n-x)(n+x), and also found n^2+x^2 = (n-x)(n+x)+x. The cubed form of n then is expressed as n^3-nx^2 = (n-x)(n)(n+x) and n^3+nx^2 = (n-x)(n)(n+x)+2nx^2 . The first n^4 equation is n^4-4*n^2*x^2-n^2+n-(n-4) = (n-2x)(n-x)(n+x)(n+2x), but I haven't found the equation beginning with n^4+4*n^2*x^2. Can anyone shed some light on the problem? Thanks, Roger PS - equations may not be completely accurate - will verify by next post [signs etc]
Kindly correct your expressions as some of them are wrong then we can get somewhere.

Quote:
 Originally Posted by roger "(though a similar one does), but has interesting properties:"
Which one??

Mally

 2006-10-23, 15:55 #3 roger     Oct 2006 22×5×13 Posts First of all, sorry for the erroneous data to you who have worked on this. Here are all the ones I have that are proven: ^2 : n^2-x^2 = (n-x)(n+x); n^2+x^2 = (n-x)(n+x)+x ^3 : n^3-n*x^2 = (n-x)(n)(n+x); n^3+nx^2 = (n-x)(n)(n+x)+2n*x^2 ^4 : n^4-5n^2*x^2+4x^4 = (n-2x)(n-x)(n+x)(n+2x); n^4+5n^2*x^2+4x^4 = (n-2x)(n-x)(n+x)(n+2x)+10n*x^2 ^5 : n^5-5n^3*x^2+4n*x^4 = (n-2x)(n-x)(n)(n+x)(n+2x); n^5+5n^3*x^2+4n*x^4 = (n-2x)(n-x)(n)(n+x)(n+2x)+10n^3*x^2 At the moment, ^6 is not on there because, even though I have double and checked again, whenever I enter a value for the equation I came up with (double checking the work behind it too), I get a negative or zero value, or at least ones that don't fit the approximated values. I will continue trying to fix this one, and will write back quicker than last time. Thanks to all those who attempted and wrote back, and I hope I haven't taken up too much of your time with my bad equations. Roger PS - mfgoode - n^2+1 often/always comes up with primes. Will look back into this also Last fiddled with by roger on 2006-10-23 at 15:56
 2006-10-24, 22:50 #4 roger     Oct 2006 22·5·13 Posts I checked back on n^6, and discovered I'd been thinking backwards, so confused after working pages of bad data that (obviously) gave no results. For n^6 though, I have come to another stumbling block. All I have discovered is that for at least x<4 (exclusive), the difference between the right hand side of the equation [(n-x)(n+x) for n^2] and the left [n^2*x^2 for n^2] is divisible by 7. My method for finding n^a+ (probably slow) is to obtain n as well as the right side of the equation, giving a value, and subtracting that value from the left side to obtain a new value (I'll call this b). a is then some form or other (the right-most bit of the right side), but with n^6+, all I know/can see is that b/y is a whole number (and then can be further broken down into a form of some kind, like the other a's) Can anyone advise? Thanks, Roger
2006-10-25, 01:37   #5
wblipp

"William"
May 2003
New Haven

22·32·5·13 Posts

Quote:
 Originally Posted by roger n^2+x^2 = (n-x)(n+x)+x
I quit reading at the first mistake I found.

 2006-10-26, 22:31 #6 roger     Oct 2006 22·5·13 Posts Sorry Sorry about all that, I realized how all my attempts were clouded by forgetting math class (factoring etc.). Next post I make will have relevence. Roger
2006-10-27, 11:52   #7
mfgoode
Bronze Medalist

Jan 2004
Mumbai,India

22×33×19 Posts
The Binomial.

Quote:
 Originally Posted by roger Sorry about all that, I realized how all my attempts were clouded by forgetting math class (factoring etc.). Next post I make will have relevence. Roger

Roger, study the masters first before you can become one yourself.
I would advise you to study and master the Binomial Theorem in its entirety.
I gather you dont have a clue about it.

In passing, it is not obligatory to post a thesis of your own. You can always learn by asking questions until you can stand on your own two feet.
Read every post as far as possible and whatever is in your range of interests.

The forum has a wide variety of topics so you can pick your posts

Best of luck,

Mally

 2006-10-27, 21:54 #8 roger     Oct 2006 22×5×13 Posts Thanks, mally I'll look into it, and take your advice Roger

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