Re: Fermat's last theorem and a counter example
- From: quasi <quasi@xxxxxxxx>
- Date: Tue, 07 Mar 2006 08:16:04 -0500
On Tue, 07 Mar 2006 08:05:16 -0500, quasi <quasi@xxxxxxxx> wrote:
On Tue, 07 Mar 2006 03:04:11 EST, bassam king karzeddin
<bassam@xxxxxxxxxx> wrote:
On Tue, 21 Feb 2006 11:43:56 EST, bassam kingYes Sir
karzeddin
<bassam@xxxxxxxxxx> wrote:
Dear Allnumber system for (x, y, z, n), where (n>2)
FLM states simply that the following equation:
x^n + y^n = z^n
Doesn't have solution in the whole positive integer
following conditions must obey:
Iff, exists a counter example to FLM, then the
What is FLM? Do you mean FLT?
I wish to know the difference as I'm a Civil Engineer and not a profissional mathematician
I'll assume you mean FLT.
Anyway, since FLT has presumably been established, it
seems
meaningless to talk about counterexamples.
But let's assume, hypothetically, that FLT has not
yet been proven.
Then I think what you are trying to say is this:
Any counterexample to FLT must satisfy the conditions
below:
All prime factors of (x+y) must be some primefactors of (z)
All prime factors of (z-x) must be some primefactors of (y)
All prime factors of (z-y) must be some primefactors of (x)
2*z > x+y > z
However the only clear implication is one way:
If a counterexample to FLT exists, then your
conditions can be
satisfied.
Thus, I think you should replace the "iff" in your
claim by "if".
The statement "if P then Q" means: if P is true, then Q is true. An
alternate way of saying this is: P => Q (P implies Q).
The statement "if Q then P" means: if Q is true, then P is true. An
alternate way of saying this is: Q => P (Q implies P).
If both the implications P => Q and Q => P hold, then we say:
P iff Q (P if and only if Q)
An alternate way of saying this is P <=> Q (P is equivalent to Q).
However, when only one of the 2 implication holds, the order is key.
Example:
Let P be the statement "x = a^2 for some integer a".
Let Q be the statement "x = a^n for some integer a and some integer
n>1."
Then the statement P => Q is true but the statement Q => P is false.
Now, assuming (hypothetically) that FLT was not yet known to be true,
consider the 2 statements:
P: FLT is true.
Q: There do not exist positive integers x, y, z satisfying the
following conditions:
(1) x < y < z < x + y
(2) x, y, z are pairwise coprime
(3) all prime factors of x + y divide z
all prime factors of z - y divide y
all prime factors of z - x divide x
Typo -- condition (3) should have been:
(3) all prime factors of x + y divide z
all prime factors of z - x divide y
all prime factors of z - y divide x
Clearly (not P) => (not Q). Equivalently, Q => P..
But it's an "if then", not an "iff".
Thus, if you could show Q is true, then you get an alternate proof of
FLT, however showing Q false does not show that P is false.
In fact, the counterexample x=35, y=46, z=51 shows that Q actually is
false.
You then added a further conditions to Q:
(4) x + y, z - x, z - y each have at least 2 prime factors
With the new condition, a counterexample to Q has not yet been shown,
but now the tie to FLT is less clear. In other words, assuming FLT is
false, it's clear that you can find positive integers satisfying (1),
(2), (3) but why would condition (4) have to hold?
quasi
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- From: quasi
- Re: Fermat's last theorem and a counter example
- From: bassam king karzeddin
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