"one step backward _two steps forward "

[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).

[TheMawn]

The text editor on this forum offers super^script and sub_script, just by the by. Makes things a bit easier to read.

[LaurV]

Originally Posted by TheMawn

The text editor on this forum offers super^script and sub_script, just by the by. Makes things a bit easier to read.

[devarajkandadai]

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

[devarajkandadai]

Originally Posted by TheMawn

The text editor on this forum offers super^script and sub_script, just by the by. Makes things a bit easier to read.

Do I have to download a software for this?
tks

A.K. Devaraj

[devarajkandadai]

Originally Posted by LaurV

Shd I dl TEX? - tks

Devaraj

[Uncwilly]

Originally Posted by devarajkandadai

Do I have to download a software for this?

No, it is built in to the on-line interface.

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.

[Batalov]

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.

~/> 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.

[only_human]

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]

[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.

[devarajkandadai]

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.