Phinnaeus Gage
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January 02, 2012, 06:42:41 PM 

I believe that Pierre de Fermat's Conjecture, known as Fermat's Last Theorem (FLT), was loosely based on Numerology in general, and Digital Root in particular, when he penned that he found a truly amazing proof. I further believe that he did this in his mind via Mathematical Induction just prior to him writing in the margin of Diophantus' Arithmetica, hence being too long, coupled with the fact that he offered little or no proof of his theorems. Take a look at the following matrix. Notice a pattern? 1 1 1 1 1 1 8 7 5 1 2 4 9 9 9 9 9 9 1 4 7 1 4 7 8 4 2 1 5 7 9 9 9 9 9 9 1 7 4 1 7 4 8 1 8 1 8 1 9 9 9 9 9 9 If not, how about now? 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 8 7 5 1 2 4 8 7 5 1 2 4 8 7 5 1 2 4 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 1 4 7 1 4 7 1 4 7 1 4 7 1 4 7 1 4 7 8 4 2 1 5 7 8 4 2 1 5 7 8 4 2 1 5 7 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 1 7 4 1 7 4 1 7 4 1 7 4 1 7 4 1 7 4 8 1 8 1 8 1 8 1 8 1 8 1 8 1 8 1 8 1 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 8 7 5 1 2 4 8 7 5 1 2 4 8 7 5 1 2 4 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 1 4 7 1 4 7 1 4 7 1 4 7 1 4 7 1 4 7 8 4 2 1 5 7 8 4 2 1 5 7 8 4 2 1 5 7 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 1 7 4 1 7 4 1 7 4 1 7 4 1 7 4 1 7 4 8 1 8 1 8 1 8 1 8 1 8 1 8 1 8 1 8 1 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 8 7 5 1 2 4 8 7 5 1 2 4 8 7 5 1 2 4 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 1 4 7 1 4 7 1 4 7 1 4 7 1 4 7 1 4 7 8 4 2 1 5 7 8 4 2 1 5 7 8 4 2 1 5 7 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 1 7 4 1 7 4 1 7 4 1 7 4 1 7 4 1 7 4 8 1 8 1 8 1 8 1 8 1 8 1 8 1 8 1 8 1 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 I believe that the original proof to FLT lies within this matrix and not based on Modular Elliptic Curves of which Fermat did not have at his disposal (I'm aware of the fact that this has been stipulated many times in the past). All that needs to be done now is penning the proof via Mathematical Induction or converting the same into a formal proof. Questions? ~Bruno~









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BurtW
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January 02, 2012, 08:53:06 PM 

Interesting. I have reproduced your matrix and think I can see what you might be talking about.
Let's use S(x) to mean the digital sum of x. The matrix shows the various solutions to S(i^(n+2)) where i is the row numbers [1..] and n is the column number [1..]
If we look at each column we can see that for all values of i:
S(i^3) = [1, 8, 9] S(i^4) = [1, 4, 7, 9] S(i^5) = [1, 2, 4, 5, 7, 8, 9] or NOT[3, 6] S(i^6) = [1, 9] S(i^7) = [1, 2, 4, 5, 7, 8, 9] or NOT[3, 6] S(i^8) = [1, 4, 7, 9] S(i^9) = [1, 8, 9]
and this pattern repeats
This is all interesting but taking the simplest case above of S(i^6) = [1, 9] and the fact that a^6 + b^6 = c^6 iff S(a^6) + S(b^6) = S(c^6)
The only three possiblities are:
S(a^6) + S(b^6) = 1 + 1 = 2 S(a^6) + S(b^6) = 1 + 9 = 1 S(a^6) + S(b^6) = 9 + 9 = 9
But the last two show that there may be a solution for n=6, when we wanted to show that there was no possible solution.

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Meni Rosenfeld
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January 02, 2012, 09:06:33 PM 

Fermat is turning in his grave hearing your suggestion that the FLT is derived from some mod 9 calculations.




ineededausername


January 02, 2012, 09:21:56 PM 

Fermat is turning in his grave hearing your suggestion that the FLT is derived from some mod 9 calculations.
+1 hahaha. Phin: you just rediscovered Euler's Theorem, congratulations

(BFL)^2 < 0



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January 02, 2012, 09:47:42 PM 

Another way to look at it is to look at the column for S(i^2), which you do not show but it is identical to the column for S(i^8).
We know that there are solutions for a^2 + b^2 = c^2 and we know that there are NO solutions for a^8 + b^8 = c^8
Yet the columns are identical.
I do find it interesting that it appears that if you prove something (some certain things related to the function S) for i^n for n=[2..7] then you should be able to show it is true for all n>1 since the patterns (columns) you have found do repeat after that.

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Phinnaeus Gage
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January 02, 2012, 11:20:35 PM 

Fermat is turning in his grave hearing your suggestion that the FLT is derived from some mod 9 calculations.
+1 hahaha. Phin: you just rediscovered Euler's Theorem, congratulations That's my next trick. I'm going to use Euler's Theorem to prove Zeno's Paradox is correct. I do find it interesting that it appears that if you prove something (some certain things related to the function S) for i^n for n=[2..7] then you should be able to show it is true for all n>1 since the patterns (columns) you have found do repeat after that. I believe we are on the same page here. Whatever is proved in the fist column, should be able to prove the same for the n+6x columns. Fermat is turning in his grave hearing your suggestion that the FLT is derived from some mod 9 calculations.
I like to look at it as him smiling in his grave for seeing that somebody is finally removing mod x from the calculations.




Phinnaeus Gage
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January 02, 2012, 11:27:09 PM 

Fermat is turning in his grave hearing your suggestion that the FLT is derived from some mod 9 calculations.
Something just clicked! Perhaps you're right. Fermat could have used mod 9 calculations, though he didn't know what a mod was. Does that make sense? S(i^3) = [1, 8, 9] S(i^4) = [1, 4, 7, 9] S(i^5) = [1, 2, 4, 5, 7, 8, 9] S(i^6) = [1, 9] S(i^7) = [1, 2, 4, 5, 7, 8, 9] S(i^8) = [1, 4, 7, 9] S(i^9) = [1, 8, 9] That's an interesting pattern, in and of itself. I had to reread your first post, Burt, to make sure we're on the same page and didn't misinterpret your notations. I feel you do see what I'm getting at. However it's proved for the cube root column, should hold for the 10th power column. Now, how do we go about proving this thing, short of by inductive reasoning?




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January 02, 2012, 11:43:16 PM 

The only three possiblities are:
S(a^6) + S(b^6) = 1 + 1 = 2 S(a^6) + S(b^6) = 1 + 9 = 1 S(a^6) + S(b^6) = 9 + 9 = 9
But the last one shows that there may be a solution for n=6, when we wanted to show that there was no possible solution. And for S(a^6) + S(b^6) = 1 + 9 = 1, but not S(a^6) + S(b^6) = 1 + 1 = 2.




ineededausername


January 02, 2012, 11:56:29 PM 

Phin: I don't think you understand. If you don't prove 2..7 with the mod 9 method, you can't prove >7 with the mod 9 method, due to the way the proof is structured. It's gotta be all or nothing. You can't prove 2..7 with other methods and then conclude the rest works. And, sir, if mod 9 could be used, someone would've found that out by now

(BFL)^2 < 0



TT


January 03, 2012, 12:06:36 AM 

I hope you guys are trolling...but in case you're actually thinking about carrying on with this line of reasoning, I thought I'd point out that you only need to consider prime n since if n is composite, say n = pq, you have x^(pq) = (x^p)^q = (x^q)^p, so if a solution existed for n, it would also exist for p and for q.
And no, you cannot prove FLT generally using simple congruences...but you can definitely prove it for a bunch of small primes.




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January 03, 2012, 12:34:11 AM 

No trolls here. We are having a fun discussion with our buddy Bruno. He has (re)discovered a few interesting patterns and we are talking to him about it. Your point about primes is well taken.

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January 03, 2012, 12:48:09 AM 

Let me put this another way. In my original post I tried to use your proposed method to prove the assertion:
There are no integer solutions to the formula a^6 + b^6 = c^6.
I attempted a proof and I failed. My point was that since I could not use this method to prove the above assertion then the entire induction will not work.
However you were kind of on to something. If I could have proved the assertion:
There are no integer solutions for a^n + b^n = c^n for all values of n where n is between 3 and 8 using the digit root (mod 9) method then we could have easily gone on to prove this is true for all values of n > 2.
But as I showed, I cannot even prove it for the simplest case (n=6) using this method.
I hope that makes sense.

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Phinnaeus Gage
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January 03, 2012, 12:51:06 AM 

Phin: I don't think you understand. If you don't prove 2..7 with the mod 9 method, you can't prove >7 with the mod 9 method, due to the way the proof is structured. It's gotta be all or nothing. You can't prove 2..7 with other methods and then conclude the rest works. And, sir, if mod 9 could be used, someone would've found that out by now For a second, I though I knew what mod 9 was. Guess I was mistaken. I did just find out something about prime numbers. With the exception of the number 3, no other numbers factor down to 3 or 6, just like what Burt shown here: S(i^3) = [1, 8, 9] S(i^4) = [1, 4, 7, 9] S(i^5) = [1, 2, 4, 5, 7, 8, 9] or NOT[3, 6] S(i^6) = [1, 9] S(i^7) = [1, 2, 4, 5, 7, 8, 9] or NOT[3, 6] S(i^8) = [1, 4, 7, 9] S(i^9) = [1, 8, 9] No trolls here. We are having a fun discussion with our buddy Bruno. He has (re)discovered a few interesting patterns and we are talking to him about it. Your point about primes is well taken.
Do you feel that FLT has been approached this way before?




Phinnaeus Gage
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January 03, 2012, 01:03:08 AM 

Let me put this another way. In my original post I tried to use your proposed method to prove the assertion:
There are no integer solutions to the formula a^6 + b^6 = c^6.
I attempted a proof and I failed. My point was that since I could not use this method to prove the above assertion then the entire induction will not work.
However you were kind of on to something. If I could have proved the assertion:
There are no integer solutions for a^n + b^n = c^n for all values of n where n is between 3 and 8 using the digit root (mod 9) method then we could have easily gone on to prove this is true for all values of n > 2.
But as I showed, I cannot even prove it for the simplest case (n=6) using this method.
I hope that makes sense.
For starters, as long as n is not divisible by 3, we know that n ^{6} + (n + 1) ^{6} and n ^{6} + 3x ^{6} and n ^{6} + 4x ^{6} would have no solutions. Can we agree on this so far?




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January 03, 2012, 01:03:50 AM 

Let me clear up one thing for you. The digital sum/digital root function, where you add up the digits and then take the results and then add up those digits and then keep going until you get a number between 1 and 9 is pretty much identical to just taking the number and dividing it by 9 and then taking the remainder. This is called the mod operation or remainder operation.
In one case you get numbers between 1 and 9 in the other case you get results between 0 and 8 (o replaces 9).
So the digital sum/digital root function is pretty much the same as the mod 9 function.
It is very common to use mod operations in number theory (including the math behind Bitcoin) so I am certain this approach has been tried.

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ineededausername


January 03, 2012, 01:10:18 AM 

And, Phin, when you take the exponent of things in mod 9 (or any mod) it's the same as taking the exponent, then performing the mod operation. Also, Euler's Theorem states that a^(phi(n)) = 1 (mod n) for gcd(a,n) = 1, and phi(9) = 6.

(BFL)^2 < 0



Phinnaeus Gage
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January 03, 2012, 01:25:06 AM 

Let me clear up one thing for you. The digital sum/digital root function, where you add up the digits and then take the results and then add up those digits and then keep going until you get a number between 1 and 9 is pretty much identical to just taking the number and dividing it by 9 and then taking the remainder. This is called the mod operation or remainder operation.
In one case you get numbers between 1 and 9 in the other case you get results between 0 and 8 (o replaces 9).
So the digital sum/digital root function is pretty much the same as the mod 9 function.
It is very common to use mod operations in number theory (including the math behind Bitcoin) so I am certain this approach has been tried.
So, what you're telling me is that for the past 30+ years, I've been carrying this in my head for no reason? At least I have one more idea that'll be pretty hard for you all to poke holes in:




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January 03, 2012, 01:39:01 AM 

And for S(a^6) + S(b^6) = 1 + 9 = 1, but not S(a^6) + S(b^6) = 1 + 1 = 2.
Sorry, just figured out what you were saying here. My op has been updated.

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Phinnaeus Gage
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January 03, 2012, 06:44:02 AM 

And for S(a^6) + S(b^6) = 1 + 9 = 1, but not S(a^6) + S(b^6) = 1 + 1 = 2.
Sorry, just figured out what you were saying here. My op has been updated. All jokes aside, is my approach truly a dead end? ~Bruno~




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January 03, 2012, 12:54:01 PM 

I am sorry to say it, but I think it s a dead end.

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