2017-02-18 16:28:26 UTC
assume there is at least one solution for,
(1.) a^p + b^p + c^p = 0,
where a, b, and c are non-zero pairwise co-prime
integers and p >= 3 is an odd integer.
proof (by contradiction).
(2.1.1) (a + b) | (a^p + b^p).
(2.1.2) (a + b) | (-c^p).
(2.2.1) (a + c) | (a^p + c^p).
(2.2.2) (a + c) | (-b^p).
(2.3.1) (b + c) | (b^p + c^p).
(2.3.2) (b + c) | (-a^p).
(2.1.2), (2.2.2), and (2.3.2) yield,
(2.4) (a + b)(a + c)(b + c) | (abc)^p.
definition of Y:
(3.1) Y = (b + a)(b + c)(a + c).
(2.4 and 3.1->3.2) Y | (abc)^p.
< snip: I don't understand why the following section (section 8) is unnecessary in this proof.>
Edit: there may be a certain set of general equations (undefined by me) that fall in a catory that make this method of proof effective.
expansion of Y:
(8.1) Y = (b^2 + bc + ab + ac)(a+ c).
(8.2) Y = b^2a + b^2c + abc + bc^2 + ba^2 + abc + a^2c + ac^2.
(8.3) Y = ab(a +b) + ac(c + a) + bc(c + b) + 2abc.
(8.4) Y = ab(a + b + c - c) + ac(a + c + b - b) + bc(b + c + a - a) + 2abc.
(8.5) Y = (a + b + c)(ab + ac +bc) - abc.
(8.6) Y = (ab + ac + bc)(a + b + c) - abc.
end of section 8.
definition of X:
(5.1) X = Y + abc.
(3.2 and 5.1->5.2) (X - abc) | (abc)^p.
let B be such that,
(6.1) B = gcd(X, abc).
let F be such that,
(6.2) BF = X.
(5.3 and 6.1->6.3) (X/B - abc/B) | (abc)^p.
(6.3 and 6.2->6.4) (F - abc/B) | (abc)^p.
(6.1 and 6.2->6.5) 1 =gcd(F, (abc)/B) .
(6.5->6.6) 1 = gcd((F - (abc)/B), (abc)/B).
(6.6->6.7) 1 = gcd((F - (abc)/B), (abc)^p/B).
(6.8) (abc)^p/B | (abc)^p/B,
(6.9) (abc)^p/B = (abc)^p/B.
This may seem as if the givens have little (are independent) of the proof. I conjecture instep, that the general class of equations yield similar if not identical proofs in which (or whereby),
(U - V) | W,
GCD(U,V) = 1
except THAT, V | W, U | W (and also UV | W),
where |UV| > W. (oops, contradiction).
There may exist a given set of equations that can be proved false under this general method of proof.
(6.7, 6.8 and 6.4 ->6.10)
[(F - abc/B)(abc)^p/B] | (abc)^p/B.
(6.10 -> 6.11) [(F - abc/B)(abc)^p/B] > (abc)^p/B.
(recognizing Y=/= 0 yields a contradiction)
End of proof.
Simon C. Roberts.