"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).

The text editor on this forum offers super[SUP]script[/SUP] and sub[SUB]script[/SUB], just by the by. Makes things a bit easier to read.

[QUOTE=TheMawn;349317]The text editor on this forum offers super[SUP]script[/SUP] and sub[SUB]script[/SUB], just by the by. Makes things a bit easier to read.[/QUOTE]
[TEX]Well...\ A\ real\ mat^h^{em}_a_{ti}cian\ would\ write\ in\ \TeX,\ in\ fact...[/TEX] :razz:

[QUOTE=devarajkandadai;349220]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).[/QUOTE]

can this be translated in pari ?

A.K. Devaraj

one-step backward...
 

[QUOTE=TheMawn;349317]The text editor on this forum offers super[SUP]script[/SUP] and sub[SUB]script[/SUB], just by the by. Makes things a bit easier to read.[/QUOTE]
Do I have to download a software for this?
tks

A.K. Devaraj

[QUOTE=LaurV;349372][TEX]Well...\ A\ real\ mat^h^{em}_a_{ti}cian\ would\ write\ in\ \TeX,\ in\ fact...[/TEX] :razz:[/QUOTE]

Shd I dl TEX? - tks

Devaraj

[QUOTE=devarajkandadai;349513]Do I have to download a software for this? [/QUOTE]No, it is built in to the on-line interface.
[QUOTE=devarajkandadai;349514]Shd I dl TEX? - tks[/QUOTE]You can use it by inserting the formatting mark-ups within the [ TEX ][ /TEX ] (spaces inserted to keep the command from being interpreted) tags.

[QUOTE=devarajkandadai;349220]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.

[COLOR="DarkRed"]The algo[/COLOR] 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).[/QUOTE]
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 [B]not[/B] 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

[COLOR="DarkRed"]# not a power of two; the "algo" is invalid[/COLOR]
[/CODE]
What made you question the Ramanujan - Nagell result, anyway? There exist only five solutions which is a clear contradiction to your handwaving.

[QUOTE=Uncwilly;349521](spaces inserted to keep the command from being interpreted) tags.[/QUOTE]there is a noparse command that skips interpretation of tags:
[NOPARSE][NOPARSE][TEX]anything[/TEX][/NOPARSE][/NOPARSE]

Will display as [NOPARSE][TEX]anything[/TEX][/NOPARSE]

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.

One step backward....
 

[QUOTE=devarajkandadai;350196]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.[/QUOTE]

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.