 mersenneforum.org "one step backward _two steps forward "
 Register FAQ Search Today's Posts Mark Forums Read 2013-08-12, 04:22 #1 devarajkandadai   May 2004 22·79 Posts "one step backward _two steps forward " This refers to the Ramanujan - Nagell theorem. i.e. there are only 5 solutions to the Diaphontine eqn: x^2 + 7 = 2 ^n. When x = 1, f(x) = 8 = 2^3 = 2^s, say. Then x = 1 + 2^(s-1), when substituted in f(x) yields 32 = 2 ^5. The algo x_i+1 = x_i + 2^(s-1) always yields a number such that f(x_i+1), whose power of 2 component is always greater than s. Here s is the power of 2 in f(x_i).   2013-08-12, 23:26 #2 TheMawn   May 2013 East. Always East. 32778 Posts The text editor on this forum offers superscript and subscript, just by the by. Makes things a bit easier to read.   2013-08-13, 06:04   #3
LaurV
Romulan Interpreter

"name field"
Jun 2011
Thailand

10,273 Posts Quote:
 Originally Posted by TheMawn The text editor on this forum offers superscript and subscript, just by the by. Makes things a bit easier to read.    2013-08-14, 05:40   #4

May 2004

22·79 Posts Quote:
 Originally Posted by devarajkandadai This refers to the Ramanujan - Nagell theorem. i.e. there are only 5 solutions to the Diaphontine eqn: x^2 + 7 = 2 ^n. When x = 1, f(x) = 8 = 2^3 = 2^s, say. Then x = 1 + 2^(s-1), when substituted in f(x) yields 32 = 2 ^5. The algo x_i+1 = x_i + 2^(s-1) always yields a number such that f(x_i+1), whose power of 2 component is always greater than s. Here s is the power of 2 in f(x_i).
can this be translated in pari ?

A.K. Devaraj   2013-08-14, 10:25   #5

May 2004

22×79 Posts one-step backward...

Quote:
 Originally Posted by TheMawn The text editor on this forum offers superscript and subscript, just by the by. Makes things a bit easier to read.
tks

A.K. Devaraj   2013-08-14, 10:30   #6

May 2004

22×79 Posts Quote:
 Originally Posted by LaurV  Shd I dl TEX? - tks

Devaraj   2013-08-14, 12:19   #7
Uncwilly
6809 > 6502

"""""""""""""""""""
Aug 2003
101×103 Posts

2A8516 Posts Quote:
 Originally Posted by devarajkandadai Do I have to download a software for this?
No, it is built in to the on-line interface.
Quote:
 Originally Posted by devarajkandadai Shd I dl TEX? - tks
You can use it by inserting the formatting mark-ups within the [ TEX ][ /TEX ] (spaces inserted to keep the command from being interpreted) tags.   2013-08-14, 17:08   #8
Batalov

"Serge"
Mar 2008
Phi(4,2^7658614+1)/2

274316 Posts Quote:
 Originally Posted by devarajkandadai This refers to the Ramanujan - Nagell theorem. i.e. there are only 5 solutions to the Diaphontine eqn: x^2 + 7 = 2 ^n. When x = 1, f(x) = 8 = 2^3 = 2^s, say. Then x = 1 + 2^(s-1), when substituted in f(x) yields 32 = 2 ^5. The algo x_i+1 = x_i + 2^(s-1) always yields a number such that f(x_i+1), whose power of 2 component is always greater than s. Here s is the power of 2 in f(x_i).
Ok, let's be generous and trust that your statement is true.
Code:
~/> gp
? f(x)=x^2+7
? ilog2(t)={s=0;while(t%2==0,t/=2;s++);return(if(t==1,s,-1))}
# returns -1 if input is not a power of two

? f(1)
%1 = 8
? s=ilog2(8)
%2 = 3
? f(1+2^(s-1))
%3 = 32
? s=ilog2(32)
%4 = 5
? f(1+2^(s-1))
%5 = 296
? ilog2(296)
%6 = -1

# not a power of two; the "algo" is invalid
What made you question the Ramanujan - Nagell result, anyway? There exist only five solutions which is a clear contradiction to your handwaving.   2013-08-14, 17:28   #9
only_human

"Gang aft agley"
Sep 2002

2×1,877 Posts Quote:
 Originally Posted by Uncwilly (spaces inserted to keep the command from being interpreted) tags.
there is a noparse command that skips interpretation of tags:
[NOPARSE][TEX]anything[/TEX][/NOPARSE]

Will display as [TEX]anything[/TEX]   2013-08-20, 05:09 #10 devarajkandadai   May 2004 1001111002 Posts one-step backward... I am afraid some members have not understood the algorithm. I will clarify with a few examples relevant to the R-N theorem. The mother function is f(x)=x^2+7 The algo: x_i+1 = x_i + 2^(s-1) where s is the power of two in (x_i)^2+7. result: f(x_i+1) is a number in which the power of 2 is always greater than that in f(x_i). f(1) = 2^3. x_i+1 = 1 + 2^(s-1) = 5. f(5)= 2 ^5. next x_i+1 = 5 + 2^4 = 21. f(21) = 7*2^6. This goes on & on. A.K. Devaraj p.s I had presented an alternative proof of R-N Theorem based on the above algo. at a con of Ramanujan mathematical society a few years ago.   2013-08-20, 12:57   #11

May 2004

22×79 Posts One step backward....

Quote:
 Originally Posted by devarajkandadai I am afraid some members have not understood the algorithm. I will clarify with a few examples relevant to the R-N theorem. The mother function is f(x)=x^2+7 The algo: x_i+1 = x_i + 2^(s-1) where s is the power of two in (x_i)^2+7. result: f(x_i+1) is a number in which the power of 2 is always greater than that in f(x_i). f(1) = 2^3. x_i+1 = 1 + 2^(s-1) = 5. f(5)= 2 ^5. next x_i+1 = 5 + 2^4 = 21. f(21) = 7*2^6. This goes on & on. A.K. Devaraj p.s I had presented an alternative proof of R-N Theorem based on the above algo. at a con of Ramanujan mathematical society a few years ago.
There is a simple school-algebra explanation for this; however the algo is interesting because it generates all the potential solutions and those excepting the known five are filtered by applying a criterion; that is the crux of my alternate proof.  Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post MooMoo2 Other Chess Games 5 2016-10-22 01:55 wildrabbitt Miscellaneous Math 11 2015-03-06 08:17 That Don Guy Math 10 2012-02-03 18:02 nitai1999 Software 7 2004-08-26 18:12

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