## FLT

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• Hello, Thanks to Chris Nash (!), David Broadhurst, Marcel Martin, Tom Hadley, Alan Wong and Michael Hartley for your comments on my FLT paper. Yes, I have
Message 1 of 6 , Sep 4, 2001
Hello,
Thanks to Chris Nash (!), David Broadhurst, Marcel Martin, Tom
paper. Yes, I have found a bunker with my tee shot. X can be
divisible by p in 3.37. But the news is that I have fixed it over
the weekend and I am now writing up version 2. So the second shot is
soaring towards the green.
I have now put up some comments on version 1, at the link in my
last post. I think you will enjoy the second shot. Stay tuned.

Regards
Paul Mills.
• Paul, You claim in your corrected proof that (a+b)^p is congruent to a^p + b^p (mod p) even when a and b are not rationals, given a definition of modularity
Message 2 of 6 , Sep 4, 2001
Paul,
You claim in your corrected proof that (a+b)^p is congruent to a^p + b^p
(mod p) even when a and b are not rationals, given a definition of
modularity that says for example t = 1/2 mod p means that t = 1/2 + k.p for
some integer k, which seems to me a reasonable definition.

This is not true. A counter-example is easy:

let a = 2, b = 1/5, p = 3
(a+b)^p mod p would be (2+1/5)^3
= (11/5)^3
= 1331/125
= 9 + 206/125
= 3*3 + 206/125
= 206/125 mod 3

However,

a^p + b^p would be 2^3 + (1/5)^3
= 8 + 1/125
= 3*2 + 2 + 1/125
= 3*2 + 251/125
= 251/125 mod 3

These are not equal. Your "Binomial prime theorem" is false. Sorry, but
it's back to the tee for you.

I agree with Phil that you should probably be working with Usenet's sci.math
group, not this list. I'm not going to spend any more time reviewing your
stuff. I'm an amateur. Even if I _don't_ find a flaw, that would mean
nothing.

• Typo: Second line should read even when a and b are not integers . Sorry, Tom ... From: Hadley, Thomas H (Tom), NLCIO Sent: Tuesday, September 04, 2001 10:29
Message 3 of 6 , Sep 4, 2001
Typo: Second line should read "even when a and b are not integers".
Sorry, Tom

-----Original Message-----
From: Hadley, Thomas H (Tom), NLCIO
Sent: Tuesday, September 04, 2001 10:29 AM

Paul,
You claim in your corrected proof that (a+b)^p is congruent to a^p + b^p
(mod p) even when a and b are not rationals, given a definition of
modularity that says for example t = 1/2 mod p means that t = 1/2 + k.p for
some integer k, which seems to me a reasonable definition.

This is not true. A counter-example is easy:

let a = 2, b = 1/5, p = 3
(a+b)^p mod p would be (2+1/5)^3
= (11/5)^3
= 1331/125
= 9 + 206/125
= 3*3 + 206/125
= 206/125 mod 3

However,

a^p + b^p would be 2^3 + (1/5)^3
= 8 + 1/125
= 3*2 + 2 + 1/125
= 3*2 + 251/125
= 251/125 mod 3

These are not equal. Your "Binomial prime theorem" is false. Sorry, but
it's back to the tee for you.

I agree with Phil that you should probably be working with Usenet's sci.math
group, not this list. I'm not going to spend any more time reviewing your
stuff. I'm an amateur. Even if I _don't_ find a flaw, that would mean
nothing.

Unsubscribe by an email to: primenumbers-unsubscribe@egroups.com
The Prime Pages : http://www.primepages.org

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• Hello Tom, A nice observation and I will amend the paper so that the Binomial prime `theorem is a definition. But just a small comment. Your example seems to
Message 4 of 6 , Sep 4, 2001
Hello Tom,
A nice observation and I will amend the paper so that the
Binomial prime `theorem' is a definition. But just a small comment.
Your example seems to be a contradiction but in fact you are
comparing a Binomial (2 + 1/5)^3 and a monomial (11/5)^3. :-). How
does one expand a monomial? How long is a piece of string? What
matters is that a consistent framework is developed and that is why
maths is fun. If you think that Andrew Wiles et al have approached
the problem of modulo rational arithmetic any other way you are in
for a pleasant surprise. If it ain't broke, don't fix it. If it is
broke, define! Stay tuned as you see the end of my proof in version
2 and can you guess where it ends? No, it doesn't use Logical
Resonance! Something much more visual.

Regards,
Paul Mills
• http://primes.utm.edu/top20/page.php?id=2 Around 1825 Sophie Germain proved that the first case of Fermat s Last Theorem is true for such primes. Soon after
Message 5 of 6 , Feb 7, 2005
http://primes.utm.edu/top20/page.php?id=2
"
Around 1825 Sophie Germain proved that the first case of Fermat's
Last Theorem is true for such primes. Soon after Legendre began to
generalize this by showing the first case of FLT also holds for odd
primes p such that kp+1 is prime, k=4, 8, 10, 14 and 16. In 1991 Fee
and Granville [FG91] extended this to k<100, k not a multiple of
three."

My question is: What about the converse? How many things does the
FLT now prove about primes? Not just about SG primes and the others
which were use to reduece the problem, I am thinking of what other
ways it may be used. Like: Is there any type of primes of a form
which includes x^n + y^n?
• So I was looking at the Mersenne numbers yesterday and found a curious property: 3^(2^(p-1)) = -3 (mod 2^p - 1) if (2^p - 1) divides S(p-1). For the sake of
Message 6 of 6 , Mar 1, 2005
So I was looking at the Mersenne numbers yesterday and
found a curious property: 3^(2^(p-1)) = -3 (mod 2^p -
1) if (2^p - 1) divides S(p-1).

For the sake of reference, I'm using S(1) = 4, S(n+1)
= S(n)^2 - 2. Also, by 'Mersenne number' I mean "let
p be prime, then 2^p - 1 is a Mersenne number, namely
M_p".

I got this result by taking the proof of the
sufficiency of the Lucas-Lehmer test (see
http://www.utm.edu/research/primes/notes/proofs/LucasLehmer.html)
and doing some squaring and modding. I haven't been
able to show (yet?) whether the property I found is a
sufficient condition for primality, but so far it
hasn't found any false positives or missed any actual
primes (tested up through 521 at the time of this
writing). It does seem to be (at least) a necessary
condition.

Of course... now that I look more closely at my code,
this test appears to be one cycle longer than the
Lucas-Lehmer test, so maybe it isn't really worth
pursuing. Hard numbers: it was 6 seconds less time to
check M_11213 (94 seconds for Lucas-Lehmer, 88 for my
method, both using algorithms written in Lisp, run
under SBCL 0.8.19 on a 566MHz Celeron), and a little
less memory too...

Joseph.

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