[QUOTE=mfgoode;93697]The last term of the expansion ( ¼)^ 66 is less numerically than – 1.

Therefore the expansion is not fully divisible.

Therefore by FLT 2047 is not a prime[/QUOTE]

BOOM

That was the sound of my head asploding :rolleyes:

Mally, are you claiming that 2 ^ (2047 – 1) – 1 is not divisible by 2047?!

[QUOTE=mfgoode;93697]Now 2^1023 =[ 2^30 * 2] ^ 33 (Since 1023 = 31 * 33)[/QUOTE]

This works for a little number but not for a big one where you will no be able to factorise half of your number easily (1023 coming from 2047)

One thing is certain to me : for a number like 2047 this method is a bit ... complicated ?

[QUOTE=axn1;93698]Mally, are you claiming that 2 ^ (2047 – 1) – 1 is not divisible by 2047?![/QUOTE]

I think he is claiming that 2[sup]1023[/sup]-1 is not divisible by 2047.

But that''s wrong, too. I haven't taken the time figure out where, though.

Misinterprtation of calculation
 

[QUOTE=wblipp;93716]I think he is claiming that 2[sup]1023[/sup]-1 is not divisible by 2047.

But that''s wrong, too. I haven't taken the time figure out where, though.[/QUOTE]
:smile:

[Quote=axn1]Mally, are you claiming that 2 ^ (2047 – 1) – 1 is not divisible by 2047?![/QUOTE]

Yes. Try it out fellah on your p.c. if you like. Thats the beauty of the algorithm.

William you forgot the last line

[Quote= Mally]Squaring both sides and subtracting 1 we get2^2046 – 1 = (1/4)^66 (5N + 1)^66 – 1][/QUOTE]

I admit it is complicated no doubt but now I will try out 67 which is also a mersenne pseudoprime. I have the factors of 2^67 which I can present any time

Thanks for all the replies and Jacob it is, as you say, but involves at most one division of a known number which is 2^30.

In haste;

Mally :coffee:

[QUOTE=axn1;93698]
Mally, are you claiming that 2 ^ (2047 – 1) – 1 is not divisible by 2047?![/QUOTE]
[QUOTE=mfgoode;93728]
Yes. Try it out fellah on your p.c. if you like. Thats the beauty of the algorithm.
[/QUOTE]

My PC says that 2047 is a divisor of 2[sup]2046[/sup]-1. Anybody can check this on alpertron's Java factoring applet at

[url]http://www.alpertron.com.ar/ECM.HTM[/url]

copy and paste (2^2046-1)/2047

When the division is not exact, the applet reports a non-integer division.

Or better, because it looks up the divisors of full Mersenne numbers, just put in 2^2046-1 and observe that both 23 and 89 show up.

Mally - does this divisibility change your stance on the validity of your proof?

[QUOTE=mfgoode;93697]
Squaring both sides and subtracting 1 we get2^2046 – 1 = (1/4)^66 (5N + 1)^66 – 1

The last term of the expansion ( ¼)^ 66 is less numerically than – 1.

Therefore the expansion is not fully divisible.[/QUOTE]

Here is the mistake. Explicated more fully, the logic is that after binomial expansion we have 67 terms, and the first 66 are integers, but the last term is not an integer, so the entire expression is not an integer. The error is the unstated assumption that the first 66 terms are integers.

[QUOTE=wblipp;93716]I think he is claiming that 2[sup]1023[/sup]-1 is not divisible by 2047.

But that''s wrong, too. I haven't taken the time figure out where, though.[/QUOTE]
Of course it is wrong : 2[sup]1023[/sup]-1 is divisble by 2047 because 11 divides 1023 and that 2[sup]11[/sup]=2048=2047+1 and (a+1)[sup]n[/sup] leaves a reminder of 1 when divided by a if a and n are integers.

Mistake located.
 

[QUOTE=wblipp;93730]My PC says that 2047 is a divisor of 2[sup]2046[/sup]-1. Anybody can check this on alpertron's Java factoring applet at

[url]http://www.alpertron.com.ar/ECM.HTM[/url]

copy and paste (2^2046-1)/2047

When the division is not exact, the applet reports a non-integer division.

Or better, because it looks up the divisors of full Mersenne numbers, just put in 2^2046-1 and observe that both 23 and 89 show up.

Mally - does this divisibility change your stance on the validity of your proof?[/QUOTE]
:smile:
Thank you William and all the others for pointing out a flaw. I was wondering what was wrong in the calculation as I tried Dario's applet myself. What a remarkable device indeed Hats of to you alpertron. There is no doubt now that 2047 indeed divides the expression. Hence since it is composite and satisfies the FLT it is a pseudoprime.

But what went wrong with the calculation? This morning I got the answer.
The remainder 256 has been rounded of. It is actually 256 decimal and that makes a world of a difference.

No I dont change my stance on the proof. Mathematically it is correct.
I'm trying to devise a method on the same lines that shows that 2047 though composite passes the test and is indeed a pseudoprime. Unless 2047 is factorised it can not be certain that it is a prime or not but a pseudoprime.

I will try some actual small primes to test the algorithm and let you know my findings
Regards,

Mally :coffee:

Disregard.
 

[QUOTE=mfgoode;93959]:smile:

But what went wrong with the calculation? This morning I got the answer.
The remainder 256 has been rounded of. It is actually 256 decimal and that makes a world of a difference.

Regards,
Mally :coffee:[/QUOTE]

:sad:
Kindly disregard my post immediatey above.
The remiander 256 is correct and not ending in a fraction. How can it be when the two divisors are integers?
How ever your points lead me to another interpretation of the division which may prove useful and better than the original theory.

Mally :coffee: