Re: Computability

On Jan 14, 7:42 pm, The Dougster 22044 <DGo...@xxxxxxxxxxxx> wrote:
On Jan 14, 4:46 am, quasi <qu...@xxxxxxxx> wrote:



On Mon, 14 Jan 2008 04:14:39 -0500, quasi <qu...@xxxxxxxx> wrote:
On Sun, 13 Jan 2008 20:02:36 -0800 (PST), The Dougster 22044
<DGo...@xxxxxxxxxxxx> wrote:

If (x/y)^p == -1 then (x/y)^2p == 1 and o(x/y,z) = 2p But we
lose information when we write o(x/y,z) = 2p. That doesn't
specify that (x/y)^p == -1. (x/y)^p could be congruent to x for
all we know! It just doesn't say.

Right.

Let me try to restate your conjecture ...

For greater simplicity, I won't bother with the order function.
Instead, I'll just use the divisibility relation.

Dougster's conjecture:

There do not exist positive integers x,y,z such that

(1) x < y < z < x+y

(2) x,y,z, are pairwise coprime

(3) z - y is not a multiple of x

(4) For some prime p > 2,

x | z^p - y^p
y | z^p - x^p
z | x^p + y^p

Remarks:

(1) For p = 3, your conjecture holds for z <= 1000.

(2) As you've previously noted, if a proof of your conjecture could be
had, that would yield an instant proof of FLT, however the known truth
of FLT does not appear to yield a proof of your conjecture.

(3) As far as trying to prove your conjecture, I would start with a
fixed prime, for example p = 3.

quasi

==================================================
In the future, for new posts, please start the subject line with a
prefix of "-- ". By using this convention, such posts can be quickly
recognized as "non-spam". Thanks.
==================================================

Well, it appears that your conjecture fails.

Here's a counterexample:

(x,y,z) = (43, 638, 659)

with p = 7.

Well, I wrote a check(x,y,z) function just for this counterexample,
and no, it doesn't check out. I am flagging five conditions:

PASS: x < y < z < x+y
PASS: (x,y) = (y,z) = (z,x) = 1
PASS: ( o(x/y,z), o(z/x,y), o(z/y,x)) = (2p, p, p)
PASS: (x/y)^p mod z == -1
FAIL: x + y - z = 0 (because x == y == z mod p here)

Hm. Some days when I am riding my bicycle I seem to see an answer to
this problem, a contradiction buried here that would prove FLT. Maybe
one day it will come to me, or to one of you.

Note, 43 == 638 == 659 == 1 mod 7.

Do any readers here tinker with Bayesian statistics? You see, if

p(a union b) = p(a)*p(b)/p(a intersect b) (is this even right?)

in tests, as we climb higher and higher, beyond 659, toward 65535,
then we may start looking for a link between conditions a and b. That
is, say, a is "x < y < z < x+y", and b is "(x/y)^p == -1 mod z), or
some such. Like this pseudocode:

t:= 0
for x = 1 to lim
for y = 1 to lim
for z = 1 to lim
t := t + 1
a := ( x < y < z < z+y)
b := ( gcf(x,y) = gcf(y,z) = gcf(z,x) = 1)
c := ( ( p := o(x,y,z)/2) = o(z,x,y) = o(z,y,x) = p)
d := ( power(x,y,z, o(x,y,z)/2) = z - 1 )
(e := mod(x,p) + mod(y,p) - mod(z,p) = 0) if c: rem if signature
valid
f := any other derivable condition(s)
store ( p, x, y, z, a, b, c, d, e ): rem analyze later
display ( p, x, y, z, a/t, b/t, c/t, d/t, e/t ): rem look for
trends
next z
next y
next x
browse stored file...and sort a-e in decreasing order of probability
trends...

Doug

I'd like to understand the new condition.
The Dougster wrote:

) PASS: x < y < z < x+y
) PASS: (x,y) = (y,z) = (z,x) = 1
) PASS: ( o(x/y,z), o(z/x,y), o(z/y,x)) = (2p, p, p)
) PASS: (x/y)^p mod z == -1
) FAIL: x + y - z = 0 (because x == y == z mod p here)

Now the last condition is presumably not meant to be
z = x+y, since this is contrary to the stipulation of
z < x+y. Most likely z = x+y mod p is meant, because
elsewhere the Dougster notes x^p = x mod p (which is
a Corollary of Fermat's Little Theorem), etc. and
the underlying motive is to exclude x^p + y^p = z^p.
The parenthetical comment "because x == y == z mod p
here" probably refers to the fact that with x = 43,
y = 638, z = 659 we have x,y,z = 1 mod 7.

Although the additional condition obviously increases
the range of values we must search to find a minimal
counterexample, it appears to be one that can be used
to restrict the search space efficiently, at least
for fixed prime p > 2. That is, given a double outer
loop on x,y, an inner loop on z strictly between y and
x+y can have "stride" equal to p.

regards, chip
.