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**Explanations**

"Your logic is flawed. Statement 2 does not follow from statement 1 since u_{k+1} can be any number less than n."ramsey2879 said:So much for a concise proof which does not require study of other documents which you did not refer to until now.

Anyways. Your logic is flawed. Statement 2 does not follow from statement 1 since u_{k+1} can be any number less than n.

Also while a,b,and c can each be multiplied by d to get u_{k+1} to equal n-1, how does this lead to statement 3 that u_{k+2}=2? On the contrary, u_{k+2} may equal n-1.

Similarly: statement 5 does not follow from statements 1-4

Statement 7 does not follow from statements 1-6.

About the counterexample. It still stands even though U'+U'' =/0. If the logic of (20°) in doc: Revista Foaie Matematică: www.fmatem.moldnet.md/1_(v_sor_05).htm

in pdf: http://fox.ivlim.ru/docs/sorokine/flt.pdf is correct it would make no difference whether or not U' + U'' = 0. The formula for U' and U'' should work for any a,b and c or your statement is not logical.

1) u > 0, hence

(AFTER multiplication Fermat's equation by such d^n that (ud)_1 = n – 1, or "9")

2) u_{k+1} = "9" > 0.

" Also while a,b,and c can each be multiplied by d to get u_{k+1} to equal n-1, how does this lead to statement 3 that u_{k+2}=2?"

Let (u/u^k)_(2) = gn + (n – 1). Then there is such number f = 1 + xn that in the product [gn + (n – 1)](1 + xn) second digit is equal 2. Or [gn + (n – 1) – xn]_2 = 2, from here x = g – 2.

"The formula for U' and U'' should work for any a,b and c…"

Yes, but only in the Fermat's equation U' + U'' = 0, from here U'_{k+2} = U'' _{k+2} = 0 (since U'' _(k+2) = 0!).

V.S.