Victor Sorokine

Shorter, simpler, more clearly, more complete

I don't more participate in the discussion about the previous versions of the proof.
My final choice is last (September) proof. Here is:

Lemma: In prime base n, if whole numbers a = pn + d > 0 (< 0) and b = qn + d < 0 (> 0), where whole d > 0, then a =/ – b (– a =/ b ) by any p and q.
Example in base 7: 50 + 3 =/ – (– 50 + 3), 50 + 3 =/ – (– 60 + 3)…

PROOF of FLT

Case 1: The last digit of the number abc is not equal to zero, or (abc)_1 =/ 0.

(1°) Let a^n + b^n – c^n = 0, where for positive integers a, b, c
(2°) the number u = a + b – c > 0, u_(k) = 0, u_{k+1} * 0, k > 0.
(3°) If the digit u_{k+1} = 1 then we multiply the equality 1° by a 2n.
Now u_{k+1} = 2 and the digit (a_{k+1} + b_{k+1} – c_{k+1})_1 = v =/ 0 since v = or 1 ether 2.

a^n = a_(k)^n + (n^(k+1))a_{k+1} + (n^(k+2))P_a, b^n = …, c^n = …, and:
a^n + b^n – c^n = [a_(k)^n + b_(k)^n – c_(k)^n] +
(n^(k+1))[a_{k+1} + b_{k+1} – c_{k+1}] + (n^(k+2))P, where
(4°) [a_(k)^n + b_(k)^n – c_(k)^n = U',
(5°) (n^(k+1))[a_{k+1} + b_{k+1} – c_{k+1}] + (n^(k+2))P = U",
and U'_(k+1) = U"_(k+1) = 0, U'_{k+1} == U"_{k+1} == v > 0.

BUT the number U' is positive/negative and number U" is positive/positive. Therefore (cf. Lemma) U' =/ –U". And therefore U' + U" = a^n + b^n – c^n =/ 0.

Case 2: (ac)_1 =/ 0, b_(t) = 0, b_{t+1} =/ 0, [or (ab)_1 =/ 0 and c_(t) = 0, c_{t+1} =/ 0]

In this case u = a + bn^(nt – 1) – c [or u = a + b – c n^(nt – 1)]. The proof is analogous.

The proof is done.

P.S. For recent disputants of the forum:

a_k, or a_{k} (only for the forums) – the digit at the place k from the end, in the number a (thus a_1 is the last digit);
a_(k) – is the k digits’ ending (it is a number) of the number a (a_(1) = a_1)
[cf. Revista Foaie Matematică: www.fmatem.moldnet.md/1_(v_sor_05).htm].

V.S.

the number u = a + b – c > 0, u_(k) = 0, u_{k+1} * 0, k > 0.