Fermat’s Last Theorem: A one-operation proof

[moshek]

Unfortunately the "one operation" involved here appears to be hand-waving!

Please explain to me what is the real different
between hand waving to hand writing ?

It's a real question and not a joke !

Moshe

 

[Hurkyl]

Often times, when consciously skipping past significant details of an argument, a person will wave their hands around.

Thus, the term "hand waving" has been adopted to refer to an argument whose significant details have been omitted.

 

[moshek]

Often times, when consciously skipping past significant details of an argument, a person will wave their hands around.

Thus, the term "hand waving" has been adopted to refer to an argument whose significant details have been omitted.

So Fermat like Our Victor are doing hand-waving when they wrote that a^n+b^-c^n=0 have no solution ( n>2) even if they may have the same intuition.

 

[HallsofIvy]

I've always thought of "hand waving" as meaning skipping over the crucial parts of a proof. As in writing the crucial equation on the blackboard, waving your hand at it and saying "of course, this is obvious to any intelligent person"!

 

[Victor Sorokine]

Regular interesting idea

So Fermat like Our Victor are doing hand-waving when they wrote that a^n+b^-c^n=0 have no solution ( n>2) even if they may have the same intuition.

Regular interesting idea for my friends

(1°) Let a^n + b^n – c^n = 0, where for positive integers a, b, c
(2°) the number u = a + b – c > 0, where (abc)_1 =/ 0, u_(k) = 0, u_{k+1} * 0, k > 0.
(3°) Let's transform the numbers u_{k+1} and u_{k+2} into n – 1 (or "9").

(4°) Now U' = a_(k+1)^n + b_(k+1)^n – c_(k+1)^n > 0 and
(5°) U'' = a^n + b^n – c^n – U' = U – U' < 0; from here U'' = – U'.

Lemma. (6a°) If a = bn – c > 0, where c_1 =/ 0, then a_1 = n – c.
And (6b°) if d = en – c_1 > 0 and f = gn – c_1 < 0, then d_1 = n – f_1.

At last:
(7°) U'_{k+2} = (a_{k+2} + b_{k+2} – c_{k+2})_1 > 0 (cf. binominal theorem for 1°).
But some POSITIVE ending of the NEGATIVE number U'' has equal value:
(a_{k+2} + b_{k+2} – c_{k+2})_1. Therefore (cf. 6a°) U''_{k+2} =/ (– U') _{k+2}.
And therefore U_{k+2} =/ 0.

vs

 

[HallsofIvy]

Regular interesting idea for my friends

(1°) Let a^n + b^n – c^n = 0, where for positive integers a, b, c
(2°) the number u = a + b – c > 0, where (abc)_1 =/ 0, u_(k) = 0, u_{k+1} * 0, k > 0.
(3°) Let's transform the numbers u_{k+1} and u_{k+2} into n – 1 (or "9").

(4°) Now U' = a_(k+1)^n + b_(k+1)^n – c_(k+1)^n > 0 and
(5°) U'' = a^n + b^n – c^n – U' = U – U' < 0; from here U'' = – U'.

Lemma. (6a°) If a = bn – c > 0, where c_1 =/ 0, then a_1 = n – c.
And (6b°) if d = en – c_1 > 0 and f = gn – c_1 < 0, then d_1 = n – f_1.

At last:
(7°) U'_{k+2} = (a_{k+2} + b_{k+2} – c_{k+2})_1 > 0 (cf. binominal theorem for 1°).
But some POSITIVE ending of the NEGATIVE number U'' has equal value:
(a_{k+2} + b_{k+2} – c_{k+2})_1. Therefore (cf. 6a°) U''_{k+2} =/ (– U') _{k+2}.
And therefore U_{k+2} =/ 0.

vs

Wonderful example! He has not said what (abc)_1 means, he has defined u but not u_k, and he has a lemma that says "If a = bn – c > 0, where c_1 =/ 0, then a_1 = n – c." without telling us what c_1 or a_1 mean!

Handwaving at its finest!

 

[Victor Sorokine]

For recent disputants of the forum

Wonderful example! He has not said what (abc)_1 means, he has defined u but not u_k, and he has a lemma that says "If a = bn – c > 0, where c_1 =/ 0, then a_1 = n – c." without telling us what c_1 or a_1 mean!

Handwaving at its finest!

For recent disputants of the forum:

a_k, or a_{k} – 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.

 

[Victor Sorokine]

Shorter, simpler, more clearly, more complete

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.

 

[ramsey2879]

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.

In other words neither a,b,or c in the formula a^n+b^n=c^n is divisible by n. In this proof the numbers are in base n, therefore, any number not divisible by n can not end in 0. x_a or x_{a} means the digit of the number a positions from the right. x_1 = the last digit of the number x

(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.

"u_{k+1} * 0" makes no sense but from past writings, it is clear that Victor is saying that u mod n^k = 0, while u mod n^(k+1) =/ 0 for some k> 0.

(3°) If the digit u_{k+1} = 1 then we multiply the equality 1° by a 2n.

Victor is "forgetting" again to write this proof in a manner such that those unfamiliar to the thread can understand it. If u_{k+1} =/ 1 then we can multiply a,b and c by some d such that u_{k+1} = 1, and yet the new a,b and c still satisfy the original assumptions. Thus there is no limitation on the proof to assume that u_{k+1} = 1. By the same reasoning if a,b and c were multiplied by 2d then u_{k+1} would equal 2. But Victor is "wrong" to suggest that we should multipled a,b and c by 2n, since then u_{k+1} would equal 0, not 2. In this fashion, Victor obtains new a,b and c such that u_{k+1} = 2

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.

It is a fact that the sum of the digits a_{k+1} + b_{k+1} minus the digit c_{k+1} must equal 1 or 2 modulus n.

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',

For k=1, n^(k+1)=100 and n^(k+2)=1000. Let a=11, b=12 and c = 10 such that a+b-c = 20. K = 1 then 1^3+2^3-0^3 = U' = 100 and a^n+b^n-c^n = 20000 = U''+U' = U''+100. However, 20000 is not equal to 100 + 100*(1+1-1)+1000*P. Thus this part of Victor's proof is clearly incorrect.

(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.

Although U'_2 = U''_2= 0 it is not correct to state that they should for any reason equal 1 or 2. Victor simply has not establish any contradiction.

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.

U' must equal - U'' for a^n+b^n-c^n =0 but Victor has not shown that U' = an+d and U'' = bn+d must hold, hence the Lemma simply does not apply.

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.

As shown above the proof for case 1 fails in more than one way, thus the "analogous proof" whatever that is must also fail.

 

[moshek]

I've always thought of "hand waving" as meaning skipping over the crucial parts of a proof. As in writing the crucial equation on the blackboard, waving your hand at it and saying "of course, this is obvious to any intelligent person"!

HallsofIvy:

I feel that Victor improve now his hand waving about FLT. but still he need to write the missing points in his work.

Will you be kind and explain for me because of my bad English what the meaning is of: "Euclid alone has looked on beauty bare"

Thank you
Moshe

 

[moshek]

In other words neither a,b,or c in the formula a^n+b^n=c^n is divisible by n. In this proof the numbers are in base n, therefore, any number not divisible by n can not end in 0. x_a or x_{a} means the digit of the number a positions from the right. x_1 = the last digit of the number x

"u_{k+1} * 0" makes no sense but from past writings, it is clear that Victor is saying that u mod n^k = 0, while u mod n^(k+1) =/ 0 for some k> 0.

.

dear ramsey2879

thank you very much for doing the translation to victor work. he really miss many points and I hope he will complete them a.s.a possible. I have to admit that I am still very far to understand his prove (?) but I hope I can give my final opinion at October.

As shown above the proof for case 1 fails in more than one way, thus the "analogous proof" whatever that is must also fail

Let's hope for Victor and bless his great effort and kindness to share here his Organic vision [ the base n idea with a_{k+1} observing ] obout Fermat work.

Moshe

 

[Victor Sorokine]

For Moshe and ramsey2879

Dear Moshe and ramsey2879,
I am glad to hear you. Thanks.

One more precise definition:
The digit u_{k+1} = 2 does not work, since a_(k)^n + b_(k)^n – c_(k)^n < 0.
Correctly: u_{k+1} = n – 1 (or "9"), u_{k+2} > 1. For example u_{k+2} = 2; then v = a_{k+2} + b_{k+2} – c_{k+2} = (1 or 2) =/ 0.
Then U' = a_(k+1)^n + b_(k+1)^n – c_(k+1)^n > 0, where U'_{k+3} > 0,
and U'' < 0, where U''*_{k+3} = v > 0. Therefore U'_{k+3} = – v. But in the NEGATIVE number U'' = – qn + v U''_{k+3} = – v.
And therefore U_{k+3} = – 2v =/0!
The END?

Victor

P.S. Where are hurkyl and Robert Ihnot?

PPS. Numerical example in base 3: k = 2 and u_3 = 2.
In the worst case a_(3) = 211, b_(3) = 211, c_(3) = 222,
or in base 10: a_(3) = 22, b_(3) = 22, c_(3) = 28;
U' = 22^3 + 22^3 – 26^3 = 21296 – 17576 > 0! U" < 0, v = (1 or 2) > 0. And therefore U_{k+3} = (n – 2v or 2n – 2v) =/ 0.

 

[ramsey2879]

Dear Moshe and ramsey2879,
I am glad to hear you. Thanks.

One more precise definition:
The digit u_{k+1} = 2 does not work, since a_(k)^n + b_(k)^n – c_(k)^n < 0.
Correctly: u_{k+1} = n – 1 (or "9"), u_{k+2} > 1. For example u_{k+2} = 2; then v = a_{k+2} + b_{k+2} – c_{k+2} = (1 or 2) =/ 0.
Then U' = a_(k+1)^n + b_(k+1)^n – c_(k+1)^n > 0, where U'_{k+3} > 0,
and U'' < 0, where U''*_{k+3} = v > 0. Therefore U'_{k+3} = – v. But in the NEGATIVE number U'' = – qn + v U''_{k+3} = – v.
And therefore U_{k+3} = – 2v =/0!
The END?

Victor

P.S. Where are hurkyl and Robert Ihnot?

PPS. Numerical example in base 3: k = 2 and u_3 = 2.
In the worst case a_(3) = 211, b_(3) = 211, c_(3) = 222,
or in base 10: a_(3) = 22, b_(3) = 22, c_(3) = 28;
U' = 22^3 + 22^3 – 26^3 = 21296 – 17576 > 0! U" < 0, v = (1 or 2) > 0. And therefore U_{k+3} = (n – 2v or 2n – 2v) =/ 0.

This is a proof? Where is the logic to this?

 

[moshek]

This is a proof? Where is the logic to this?

Dear ramsey2879

Since you understand victor work 10 times better then me I want to ask you a very simple question . Can you see the picture that Victor have in his mind about FLT. can you help him to find it's right logic ?

 

[HallsofIvy]

Dear ramsey2879

Since you understand victor work 10 times better then me I want to ask you a very simple question . Can you see the picture that Victor have in his mind about FLT. can you help him to find it's right logic ?

Are you saying that you believe that Victor does not even understand the logic of his own proof? ramsey2879 was at least just asking him to clarify his logic!

 

[moshek]

Are you saying that you believe that Victor does not even understand the logic of his own proof? ramsey2879 was at least just asking him to clarify his logic!

I believe today that Victor have a good vision why Fermat was right, but he may need help with the logical tool and mayby he can work with ramsey on that.

 

[Victor Sorokine]

The logic of the proof

This is a proof? Where is the logic to this?

The logic of the proof:

a^n + b^n – c^n = U = U' + U" = 0, where U'_(k+2) = U"_(k+2) = 0, U'_{k+3} = U''_{k+3} = v > 0, BUT U' > 0 and U" < 0. Therefore or U'_{k+3} =/ U''_{k+3} either U' =/ – U", and therefore u =/ 0.

P.S. For n = 5 and k = 2 in base 5 min(U') = min[a_(k+1)^n + b_(k+1)^n – c_(k+1)^n] = 422^5 + 422^5 – 4444^5 = (in base 10) (3.5 – 2.9)10^10 > 0.

V.S.

 

[ramsey2879]

The logic of the proof:

a^n + b^n – c^n = U = U' + U" = 0, where U'_(k+2) = U"_(k+2) = 0, U'_{k+3} = U''_{k+3} = v > 0, BUT U' > 0 and U" < 0. Therefore or U'_{k+3} =/ U''_{k+3} either U' =/ – U", and therefore u =/ 0.

P.S. For n = 5 and k = 2 in base 5 min(U') = min[a_(k+1)^n + b_(k+1)^n – c_(k+1)^n] = 422^5 + 422^5 – 4444^5 = (in base 10) (3.5 – 2.9)10^10 > 0.

V.S.

Wait a moment please. How did you get U'_(k+2)=U''_(k+2)? How did you get U'_{k+3}=U''_{k+3}=v>0? What about my counter example in base 3?
"For k=1, n^(k+1)=100 and n^(k+2)=1000. Let a=11, b=12 and c = 10 such that a+b-c = 20. K = 1 then 1^3+2^3-0^3 = U' = 100 and a^n+b^n-c^n = 20000 = U''+U' = U''+100. However, 20000 is not equal to 100 + 100*(1+1-1)+1000*P. Thus this part of Victor's proof is clearly incorrect." Here U'_{k+3} = U'_4 = 0 but U''_4 = 2. Also, U'_(k+2) = 100 =/ 0. P.S. Have you actually computed U' and U'' in your example where a=b=422 and c=4444 in base 5? If so, what are they?

 

[Victor Sorokine]

Answers

Wait a moment please. How did you get U'_(k+2)=U''_(k+2)? How did you get U'_{k+3}=U''_{k+3}=v>0? What about my counter example in base 3?
"For k=1, n^(k+1)=100 and n^(k+2)=1000. Let a=11, b=12 and c = 10 such that a+b-c = 20. K = 1 then 1^3+2^3-0^3 = U' = 100 and a^n+b^n-c^n = 20000 = U''+U' = U''+100. However, 20000 is not equal to 100 + 100*(1+1-1)+1000*P. Thus this part of Victor's proof is clearly incorrect." Here U'_{k+3} = U'_4 = 0 but U''_4 = 2. Also, U'_(k+2) = 100 =/ 0. P.S. Have you actually computed U' and U'' in your example where a=b=422 and c=4444 in base 5? If so, what are they?

"How did you get U'_(k+2)=U''_(k+2)? "
Cf. (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

"How did you get U'_{k+3}=U''_{k+3}= v > 0?"
1) u > 0, hence:
2) u_{k+1} = "9" > 0, hence:
3) u_{k+2} = 2 > 0, hence:
4) v = (a_{k+2} + b_{k+2} – c_{k+2})_1 = (1; 2) > 0; hence:
5) U''_{k+3} = (a_{k+2} + b_{k+2} – c_{k+2})_1 = v > 0.
6) U' > 0 since u_{k+1} = "9" = n – 1; hence:
7) U'_{k+3} > 0.
8) From U'_{k+3} + U''_{k+3} = 0 [where U'_{k+3} and U''_{k+3} are last non-zero digits] we have: U'_{k+3} = n – v > 0.
9) From U' + U'' = 0 we have: U'' < 0 [but U''_{k+3} > 0]; hence:
10) The positive number (–U'') has negative last digit (–v); hence [cf. Lemma]:
11) U' =/ (–U''), hence:
12) U = U' + U" = a^n + b^n – c^n =/ 0.

"What about my counter example in base 3?
For k=1, n^(k+1)=100 and n^(k+2)=1000. Let a=11, b=12 and c = 10 such that a+b-c = 20. K = 1 then 1^3+2^3-0^3 = U' = 100 and a^n+b^n-c^n = 20000 = U''+U' = U''+100. However, 20000 is not equal to 100 + 100*(1+1-1)+1000*P. Thus this part of Victor's proof is clearly incorrect." Here U'_{k+3} = U'_4 = 0 but U''_4 = 2. Also, U'_(k+2) = 100 =/ 0."
In your example do not fail a condition: U''+U' = 0.

"P.S. Have you actually computed U' and U'' in your example where a=b=422 and c=444 in base 5? If so, what are they?"
422^5 + 422^5 – 444^5 = minimum = d [if a=b=422 and c=444; here 22 + 22 = 44, u_3 = 4 + 4 – 4 = 4, or "9"].
Really, 421^5 + 422^5 – 443^5 > d, 423^5 + 421^5 – 444^5 > d, etc.
And 444^5 + 400^5 – 444^5 = maximum.

V.S.

 

[ramsey2879]

"How did you get U'_(k+2)=U''_(k+2)? "
Cf. (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

"How did you get U'_{k+3}=U''_{k+3}= v > 0?"
1) u > 0, hence:
2) u_{k+1} = "9" > 0, hence:
3) u_{k+2} = 2 > 0, hence:
4) v = (a_{k+2} + b_{k+2} – c_{k+2})_1 = (1; 2) > 0; hence:
5) U''_{k+3} = (a_{k+2} + b_{k+2} – c_{k+2})_1 = v > 0.
6) U' > 0 since u_{k+1} = "9" = n – 1; hence:
7) U'_{k+3} > 0.
8) From U'_{k+3} + U''_{k+3} = 0 [where U'_{k+3} and U''_{k+3} are last non-zero digits] we have: U'_{k+3} = n – v > 0.
9) From U' + U'' = 0 we have: U'' < 0 [but U''_{k+3} > 0]; hence:
10) The positive number (–U'') has negative last digit (–v); hence [cf. Lemma]:
11) U' =/ (–U''), hence:
12) U = U' + U" = a^n + b^n – c^n =/ 0.

"What about my counter example in base 3?
For k=1, n^(k+1)=100 and n^(k+2)=1000. Let a=11, b=12 and c = 10 such that a+b-c = 20. K = 1 then 1^3+2^3-0^3 = U' = 100 and a^n+b^n-c^n = 20000 = U''+U' = U''+100. However, 20000 is not equal to 100 + 100*(1+1-1)+1000*P. Thus this part of Victor's proof is clearly incorrect." Here U'_{k+3} = U'_4 = 0 but U''_4 = 2. Also, U'_(k+2) = 100 =/ 0."
In your example do not fail a condition: U''+U' = 0.

"P.S. Have you actually computed U' and U'' in your example where a=b=422 and c=444 in base 5? If so, what are they?"
422^5 + 422^5 – 444^5 = minimum = d [if a=b=422 and c=444; here 22 + 22 = 44, u_3 = 4 + 4 – 4 = 4, or "9"].
Really, 421^5 + 422^5 – 443^5 > d, 423^5 + 421^5 – 444^5 > d, etc.
And 444^5 + 400^5 – 444^5 = maximum.

V.S.

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.

 

[Victor Sorokine]

Explanations

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.

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

 

[ramsey2879]

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

As to the first two paragraphs these statements belong in the proof to enable others to follow your logic. As to the counterexample, the counterexample is contrary to your admission that the formula for U' and U'' should work for any a,b,or c (whether or not a^n+b^n = c^n. I studied the text where you arrived at
"(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'', including
Cf. (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
but the text does not support the conclusion since these lines do not follow logically one line from another. As seen by my counterexample, U'' = U-U' =/ (n^(k+1))[a_{k+1} +b_{k+1}-c_{k+1}] +(n^(k+2))P as stated in (5°). If you still maintain that it is so, then you should start out with U'' = a^n + b^n -c^n - U' and line by line show logically how it can be so when my counterexample shows it not to be so.

 

[Victor Sorokine]

Ok!

but the text does not support the conclusion since these lines do not follow logically one line from another. As seen by my counterexample, U'' = U-U' =/ (n^(k+1))[a_{k+1} +b_{k+1}-c_{k+1}] +(n^(k+2))P as stated in (5°). If you still maintain that it is so, then you should start out with U'' = a^n + b^n -c^n - U' and line by line show logically how it can be so when my counterexample shows it not to be so.

OK:
1) U'' = a^n + b^n -c^n – U', or U'' + U' = = 0, where
2) U''_(k+2) = 0 (cf. 20°),
3) therefore U'_(k+2) = 0 (cf. 20°),
4) therefore (U'_{k+3} + U''_{k+3})_1 = 0 (cf. 20°).
5) therefore U'_{k+3} = – U''_{k+3} since U'_(k+2) = U''_(k+2) = 0 and U'' + U' = = 0.
6) Therefore U'' =/ - U' and a^n + b^n - c^n =/ 0.

V.S.

 

[ramsey2879]

OK:
1) U'' = a^n + b^n -c^n – U', or U'' + U' = = 0, where
2) U''_(k+2) = 0 (cf. 20°),
3) therefore U'_(k+2) = 0 (cf. 20°),
4) therefore (U'_{k+3} + U''_{k+3})_1 = 0 (cf. 20°).
5) therefore U'_{k+3} = – U''_{k+3} since U'_(k+2) = U''_(k+2) = 0 and U'' + U' = = 0.
6) Therefore U'' =/ - U' and a^n + b^n - c^n =/ 0.

V.S.

If a=5, b=4, and c=3, n=3 U''_(k+2)=200 =/ 0 so this means that 20° is not a true statement. What I meant by line by line is with the logic supporting each line so evident that no further study has to be made. Forget a reference to other work. Please show all work so it can be followed. I repeat here that so far you have not shown that U''_(k+2) must equal 0.

 

[Victor Sorokine]

Answer

I repeat here that so far you have not shown that U''_(k+2) must equal 0.

- Cf. binominal theorem for PRIME n:
a^n = [a_(k+1) + n^(k+1)a_{k+2}]^n = a_(k+1)^n + n[a_(k+1)^(n – 1)][n^(k+1)a_{k+2} + n^(k+2)P] = a_(k+1)^n + n^(k+2)[1][a_{k+2}) + n^(k+3)P.
Therefore a^n_{k+2} is not a function from a_{k+2}
(and therefore U_{k+2} is not a function from v).

 

[moshek]

Dear Victor,

Maybe you know how to prove FLT in a very simple way by looking on base מ and the k+3 digit of a^n+b^n-c^n.

Your attitude is really beautiful !

But it look that there are many holes in the writing of the prove.
Ramsey work really hard for you you sould thank him , he discover mistakes and then you correct it - that's fine.

I want to ask you were is your responsibility to preset here the most update version of your work. If you can't do that I am afraid you will never receive the recognition that you want.

So please do arrange everyting together like you write it for the first time but with the new insight you got from us , and share your paper with us

Your sincerely
Moshe

 

[Victor Sorokine]

two holes

But it look that there are many holes in the writing of the prove.

Dear Moshe,
You are right: there are 2 holes:

After transformation u_{k+1} into n – 1 [with one-digit-multiplier g]
and u_{k+2} into 2 [with two-digit-multiplier 1 + ng]
we transform also
c_{k+1} into 0 [with multiplier 1 + (n^k)g; the digits u_{k+1} and u_{k+2} do not change!]
and c_{k+2} into 0 [with multiplier 1 + (n^(k+1))g; the digits u_{k+1} and u_{k+2} do not change!]
Now U' >> 0, U'' < 0, n > v > 0 [since c_{k+2} = 0].
And therefore U_{k+3} = (n - 2v)_1 =/0.

V.S.

 

[moshek]

Last theorem of Fermat..

Ok Victor thank you !

Still you have to write the all prove together.
Now Ramsey , do you think that we can declare
That FLT can be settle in a very simple and Organic way
as was done [ until now ] nicely by Victor ?

Thank you
Moshe


p/s : My own interest in mathematics is about the hidden connection
between mathematics an physics as you can read in my paper ]

Intelligent life

Moshe Klein
‏2005–03–19

Planet Mars was associated twice during human history with the search for intelligent life. First was the effort to understand the retrograde movement of Mars on the background of the night sky. Second were the marks on Mars face that were at some point confused to be canals created by other cultures. Perhaps there is connection between Mars and the research of life in the area of mathematics?
Thousands of years ago people observed that Mars is moving sometimes in the opposite direction relative to the other stars. Mars is moving slowly on the background of the night sky so that you can distinguish the difference every day. From time to time it suddenly stops and then starts to move in the opposite direction. Then it stops again and continues its movement in the regular direction.
To explain this unusual movement the astronomers invented complex systems of wheels in which planet Earth was always in the center. The most complicated system was invented by Ptolemy in the second century AD, and included 12 wheels. But 1500 years later (about 400 years ago) Copernicus invented revolutionary idea to put the sun in the center of the world and not the planet Earth. In this model the explanation to the movement of Mars became very simple, bases on some interaction of movements between Mars and Earth. This is how the science revolution started, and it came with many hard struggles. The followers of Copernicus, Galileo and Newton, established the scientific way of thinking by developing new suitable mathematics.
The peak of this way of thinking brings in the 20 century the development of two central theories: Relativity and Quantum theory. These two theories changed the way man understood his place in the world completely. We are not passive observer of phenomena, but rather we have active part that is based on the interaction between us and the world.
Is there a way to establish similar principles in the rational field of mathematics? At first glance it seems there is no way to do it. The world of mathematics seems to be absolute and to exist independently of man. But on the other hand we know today that there is a need to develop a new view of mathematics that will bring better understanding of the relations between mathematics and the world of phenomena. This understanding was expresses in a lecture by Alain Connes, who developed 20 years ago a Non commutative Geometry, and is considered one of the leading mathematicians in the world today. "… We need today a new understanding of mathematics that is not necessarily founded on logic but rather on geometry." This observation ended the last lecture in the conference "100 to Hilbert" (free transcript from the speech). This conference was held in commemoration of another famous lecture – the lecture of Hilbert in Paris in 1900. Hilbert ended his lecture a hundred years ago with a vision of the discovery of an organic unity of mathematics.
A simple observation through these two important conferences as two eyes, one took place in Paris and the other in Los-angels with a difference of 100 years between them, raises the question if and how mathematics is moving, like Mars, on the background of the culture of mankind. This also raises the problem that has not been solved yet, of the inherent hidden connection between mathematics and the real world. But like the observation of Copernicus that allowed us to put the Sun in the center of the world and by doing this to understand much more simply the interactions between Earth and Mars, we can point to a third eye that will allow us to solve the problem of mathematics. The new common center of mathematics and physics is the discovery of the Organic unity of mathematics – Intelligent Life.

 

[ramsey2879]

- Cf. binominal theorem for PRIME n:
a^n = [a_(k+1) + n^(k+1)a_{k+2}]^n = a_(k+1)^n + n[a_(k+1)^(n – 1)][n^(k+1)a_{k+2} + n^(k+2)P] = a_(k+1)^n + n^(k+2)[1][a_{k+2}) + n^(k+3)P.
Therefore a^n_{k+2} is not a function from a_{k+2}
(and therefore U_{k+2} is not a function from v).

Standing alone these lines make no sense. For instance, in the counter example of Hurkyl, n=13, k = 2 and a=CCCCCCC so a^n = [CCC + n^(k+1)*C +N^(k+2)P]^n where P>0 =/ [a_(k+1) + n^(k+1)a_{k+2}]^n

 

[ramsey2879]

Dear Moshe,
You are right: there are 2 holes:

After transformation u_{k+1} into n – 1 [with one-digit-multiplier g]
and u_{k+2} into 2 [with two-digit-multiplier 1 + ng]
we transform also
c_{k+1} into 0 [with multiplier 1 + (n^k)g; the digits u_{k+1} and u_{k+2} do not change!]
and c_{k+2} into 0 [with multiplier 1 + (n^(k+1))g; the digits u_{k+1} and u_{k+2} do not change!]
Now U' >> 0, U'' < 0, n > v > 0 [since c_{k+2} = 0].
And therefore U_{k+3} = (n - 2v)_1 =/0.

V.S.

I refuse to consider anymore a proof that is presented only piecemeal. Each time a hole in the proof is found you come up with additional detail that should have been presented originally. As requested by Moshe please submit a complete proof, line by line, with no detail missing, and which has been checked for typing errors. As a preliminary note, however, you should clearly state that each of a,b and c are multiplied by these multipliers, or your proof fails. However, it doesn't seem logical that the digit u_{k+2} can not change when a,b,c are multiplied by 1+(n^k)g to make c_{k+1} into 0.

 

[arildno]

ramsey:
I've followed this for a while, and it is quite clear that Victor Sorokine is making up his proof as he goes along.
He doesn't have any proof, as he has claimed, only some muddled ideas as to how a proof might look like.

 

[Victor Sorokine]

+ physics

p/s : My own interest in mathematics is about the hidden connection
between mathematics an physics as you can read in my paper ]

Thanks +
Another step to the perfection

Algorithm of the Proof (Case 1):
1) Transformation of u_{k+2} into 2 [or > 1, but < n] with multiplication by 1 + ng.
2) Transformation of c_3, c_4, … c_k, c_{k+1}, c_{k+2} into 0 with multiplication by 1 + gn^2, 1 + gn^3,… 1 + gn^(k+1).
Then:
3) v = (a_{k+2} + b_{k+2} – c_{k+2})_1 = (1; 2) > 0.
4) U' = a_(k+1)^n + b_(k+1)^n – c_(2)^n >> 0.
5) U" = U – U' < 0.
6) (– U")_{k+3} = v; U"_{k+3} = – v.
7) U'_{k+3} = – v.
8) U_{k+3}= (U'_{k+3} + U"_{k+3})_1 = – 2v =/0.
The proof is done.

Next topic: Case 2.

Victor

 

[ramsey2879]

ramsey:
I've followed this for a while, and it is quite clear that Victor Sorokine is making up his proof as he goes along.
He doesn't have any proof, as he has claimed, only some muddled ideas as to how a proof might look like.

All the more reason to insist that Victor present a complete proof in a single paper that is not lacking detail, i.e., such that each line is clearly correct from what has been just stated above. At this time, I don't think anyone can make out a proof from what Victor has given us. Does anyone disagree?

 

[arildno]

Frankly, I can't make head or tails out of this.
The least we should demand, is that V.S. shapes up his, at times, extremely confusing notation and presents a properly formatted proof for perusal (preferably LATEX).

As it is now, it is utterly impenetrable, at least to me.

 

[Victor Sorokine]

One step back: return to "Fantastic idea for my friends"

Fantastic idea for my friends
Right contradiction: the number u is infinite

One step back: return to "Fantastic idea for my friends" (cf. Forum, 08.30.2005)

Heart of the proof:
After transformation of u_{k+1} into 1 the equation U = a_(k+1)^n + b_(k+1)^n – c_(k+1)^n = 0 has no solution since the number u = a_(k+1) + b_(k+1) – c_(k+1) is odd.

Attempt to correct this situation with help of the insertion in the numbers a_(k+1), b_(k+1), c_(k+1) following digits from a, b, c is doomed to failure:
transformation of U_s [=/ 0] into 0 requires to add an odd/even number to U_s and add en even/odd number to u. Therefore after this operation U = a_(s)^n + b_(s)^n – c_(s)^n =/ 0 and therefore there is other digit U_r =/ 0.

Attempt to correct this situation with help of the insertion in the numbers a_(s), b_(s), c_(s) following digits from a, b, c is doomed to failure…

AND SO AD INFINITUM!

Think yourself about that!

V.S.