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Old 2013-08-12, 04:22   #1
devarajkandadai
 
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Default "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).
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Old 2013-08-12, 23:26   #2
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The text editor on this forum offers superscript and subscript, just by the by. Makes things a bit easier to read.
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Old 2013-08-13, 06:04   #3
LaurV
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Quote:
Originally Posted by TheMawn View Post
The text editor on this forum offers superscript and subscript, just by the by. Makes things a bit easier to read.
Well...\ A\ real\ mat^h^{em}_a_{ti}cian\ would\ write\ in\ \TeX,\ in\ fact...
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Old 2013-08-14, 05:40   #4
devarajkandadai
 
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Quote:
Originally Posted by devarajkandadai View Post
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
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Old 2013-08-14, 10:25   #5
devarajkandadai
 
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Default one-step backward...

Quote:
Originally Posted by TheMawn View Post
The text editor on this forum offers superscript and subscript, just by the by. Makes things a bit easier to read.
Do I have to download a software for this?
tks

A.K. Devaraj
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Old 2013-08-14, 10:30   #6
devarajkandadai
 
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Quote:
Originally Posted by LaurV View Post
Well...\ A\ real\ mat^h^{em}_a_{ti}cian\ would\ write\ in\ \TeX,\ in\ fact...
Shd I dl TEX? - tks

Devaraj
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Old 2013-08-14, 12:19   #7
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Quote:
Originally Posted by devarajkandadai View Post
Do I have to download a software for this?
No, it is built in to the on-line interface.
Quote:
Originally Posted by devarajkandadai View Post
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.
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Old 2013-08-14, 17:08   #8
Batalov
 
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Quote:
Originally Posted by devarajkandadai View Post
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.
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Old 2013-08-14, 17:28   #9
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Quote:
Originally Posted by Uncwilly View Post
(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]
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Old 2013-08-20, 05:09   #10
devarajkandadai
 
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Default 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.
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Old 2013-08-20, 12:57   #11
devarajkandadai
 
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Default One step backward....

Quote:
Originally Posted by devarajkandadai View Post
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.
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