What is the validity of my father's proof for Fermat's Last Theorem?

[digiflux]

I am trying to get my father's ingenious proof of Fermat's last theorem the recognition that it deserves. Please visit the link below.

www.fermatproof.com

Thanks!

 

[matt grime]

I'm not sure who you talked to about the mathematics, but we'll assume you're serious and not a troll.

1. Let r=2ab be the decomposition into subfactors (a=p_1p_2... and b=p_3p_4... which is by the way appallingly badly chose notation)

then according to you the triples are of the form if I can decipher you slightly ambiguous notation (it is not clear what the dots should represent in your expressions)

x=2ab+b^2

y=2ab +a^2


z=2ab+a^2+2b^2


This observation is drawn from the trivial assertion that the squares are the sum of the odd numbers. And is a moderately interesting variation on the usual formula for generating primitve triples, however, I can't seem to find 5,12,13 as one such triple. wlog x=5, then a=2 b=1 is the only solution, and implies y=8.



2. Your history lesson probably ought to include the notice that 5 years after he claimed to have a general proof, he produced a sepific proof for n=4 I think, which would be surprising if Fermat actually had a general result. Something which practically no serious mathematician believes. Besides, we'd all look idiotic if there were a proof from the book. Surely one can say that given the huge number of plausible but subtley flawed proofs (presumption of unique factoization) it is likely Fermat was also wrong.

I haven't read the generalization to higher powers, give me a minute and I'll point out the first error.

 

[matt grime]

Hmmm, not as easy as usual. Will think about it.

OK, now I've got it. The detritus that must fit into the r^3 space. You never actually calculate what that is,and I'll not believe the proof until you do. I mean, you do some fiddling with it, and say it must do this that and the other, but never write down what it is. It's z^3-x^3 - x^2(z-y)-something else I can't be buggered to work out, and you claim that isn't possible because it reduces to a quadratic that can't be true. But hey, you don't do that at all, as far as I can tell.

Also even if this is true for n=3, it doesn't imply a reductio ad absurdum because you can't from it deduce that n=4 reduces to n=3

 

[Russell E. Rierson]

Hello digiflux, thanks for putting this interesting mathematics on the internet.

I agree that Fermat could have been onto something...

Here is a quote from the paper at www.fermatproof.com :

To begin, a model for squared numbers will be introduced and used to devise a method to create all Pythagorean (x^2 + y^2 = z^2) relationships. Equations will be derived from this process which indicate the existence of a Pythagorean equation in the model for squared numbers.

A model for higher powers of "n" will then be introduced. This model will be an extension of the model for squared numbers. Simple manipulations of this model will show that the "end game" packaging of quantities postulated to be x^n and y^n into spaces known to be x^n and yn requires that x, y, and z form a Pythagorean equation ! This is totally incompatible with the postulation that x^n + y^n = z^n where n >2.

The proof is thus Reductio ad Absurdum. . The recogniton of the afore-mentioned equations in the packaging process is the essence of the proof.

So far, so good. The generalized Pythagorean equation is this:

[a^2 - b^2]^2 + [2*a*b]^2 = [a^2 + b^2]^2

All odd numbers, 2x+1, can be represented as a^2 - b^2

[x+1]^2 - x^2

x^2 + 2x +1 - x^2 = 2x+1



The equation [a^2-b^2]^2 + [2*a*b]^2 = [a^2 + b^2]^2 generates all Pythagorean triples, but it is not the only one that does. There are other expressions and relations that can generate all Pythagorean triples, for example:

An infinite stack of predictable algebraic expressions to generate
positive integer solutions for the Pythagorean Theorem.


a b c

4n(n+1)-3 , 4(2n+1) , 4n(n+1)+5

4n(n+2)-5 , 4(3n+3) , 4n(n+2)+13

4n(n+3)-7 , 4(4n+6) , 4n(n+3)+25

4n(n+4)-9 , 4(5n+10) , 4n(n+4)+41

etc. etc. etc.




a^2 + b^2 = c^2


Very interesting at www.fermatproof.com
...

2[5*1] + 1^2 = 11

2[5*1] + 2[5]^2 = 60

2[5*1] + 2[5*1]^2 + 1 = 61

A + B = C

2A + 2B + 1 = 2C + 1

11^2 + 60^2 = 61^2


3 = 2^2 - 1^2

3^2 = 5^2 - 4^2

3^3 = 14^2 - 13^2

3^4 = 41^2 - 40^2 = 9^2

3^p = A^2 - B^2


[2x+1]^p = A^2 - B^2

 

[digiflux]

Thanks for looking at my father's proof. I have absolute confidence in his abilities. He was a genius who worked at Vought developing rocket missile systems. I on the other hand have a degree in Fine Arts...lol. Oh well. Please forward my father's link to others who are interested in a Fermat proof. Oh and sign the guestbook.

Thanks Again...

 

[matt grime]

But your confidence appears to be misplaced. As I said it contains two obviously false assertions about generating pythagorean triples and claiming a reductio ad absurdum proof. The proof in case n=3 is half reasoned and not convincing and appears to make some unjustified conclusions about possible cases. That isn't to say the n=3 argument is wrong, just that it needs more examing. If you are genuine and not just a troll you should look at it and think about it - it uses nothing more than high school maths.

 

[Russell E. Rierson]

Originally posted by matt grime
But your confidence appears to be misplaced. As I said it contains two obviously false assertions about generating pythagorean triples and claiming a reductio ad absurdum proof. The proof in case n=3 is half reasoned and not convincing and appears to make some unjustified conclusions about possible cases. That isn't to say the n=3 argument is wrong, just that it needs more examing. If you are genuine and not just a troll you should look at it and think about it - it uses nothing more than high school maths.

[zz)] [zz)] [zz)]




Actually the proof does generate all Pythagorean triples.

Look at it again.

http://hometown.aol.com/parkerdr/math/pythagor.htm

http://www.fermatproof.com/index.html

As the number of prime factors in "r" increases, the number of P. triplets doubles with each additional prime as follows:

"r" in prime factors Numerical value of "r" No. of P. triplets generated
2(1*3*5*7*11) 2310 16
2(1*3*5*7*11*13) 30030 32
2(1*3*5*7*11*13*17) 510510 64
2(1*3*5*7*11*13*17*19) 9699690 128

This method generates all valid Pythagorean triplets. Since any given P. triplet can be shown to have an even root number, "r", as shown in Figures 2 and 3, the P. triplet will eventually be generated as sequential even root numbers are selected and processed. The only limitations will be the capacity of the computational devices used.

 

[matt grime]

please read the first reply i gave on this where i asked if the generation was of the form I rewrote without the ambibuigty in the decompositions into prime factors (P_1P_2... is used as is p_1p_2p_3p_4... and p_3p_4... which is ambiguous as it is not clear what the ...means) where I prove that 5,12,13 is not of the form specified. or post a correction to my interpetation (which is just an interpretation; I'm prepared to accept i misread it).

There is no dimensionless argument so it is not correct to argue reductio ad absurdum, do you dispute that? Given your own bizarre postings on FLT I'll require some direct refutation like posting the correct generating function (there are I believe many, and i could well accept that this is very close to one and it is just my pedantic reading of mathematics by a non-mathematician) - the spirit of its argument may be correct but the object he writes down at the end doesn't look correct.

 

[Russell E. Rierson]

Originally posted by matt grime
please read the first reply i gave on this where i asked if the generation was of the form I rewrote without the ambibuigty in the decompositions into prime factors (P_1P_2... is used as is p_1p_2p_3p_4... and p_3p_4... which is ambiguous as it is not clear what the ...means) where I prove that 5,12,13 is not of the form specified.

x = r + odd

y = r + even

z = r + even + odd

Let a = 2

Let b = 1

r = 4 = 2*(2*1)

x = 2ab + b^2

y = 2ab + 2a^2

z = 2ab + 2a^2 + b^2

x = 2*(2*1) + (1)^2 = 5

y = 2*(2*1) + 2*(2)^2 = 12

z = 2*(2*1) + 2*(2)^2 + (1)^2 = 13



I still am not sure about the reducto ad absurdum though [?]

Very intersting still...

 

[matt grime]

yes, i figured out shortly after i wrote it that it should be 2ab+b^2 and 2ab+2a^2, which works.It isn't relevant to the rest of the proof that it generates all triples anyway.

The proof for n=3 ought to be rewritable in a purely algebraic form omitting the diagrams, where it ought to become clear if it is correct (my gut feeling is that it isn't).

There is no obvious way from geometrical arguments to induct.

 

[Sariaht]

o = odd, e = even

o_a^g +/- o_b^g is not o_c^g

e_a^g +/- e_b^g = e_c^g can be divided with 2^g.

e_a^g +/- e_b^g is not o_c^g.


o_a^g +/- o_b^g = e_c^g (this must be proven wrong for g>2).

e_c^g is devideable with 2^nx.

(o_a/2^x)^g +/- (o_b/2^x)^g = o_c^g

Exchange o_a with 2n - 1, and o_b with 2s - 1.

((2n-1)_a/2^x)^g +/- ((2s-1)_b/2^x)^g = o_c^g

Try to prove it with "binomials".

It actually works.

Try with g = 3 first.

Since no 2^gx higher than 4 can work, it's actually a proof.

4² = 5² - 3² still works.

1² = (5/4)² - (3/4)²

1² = ((2s - 1)/4)² - ((2n - 1)/4)²

( n = 2, s = 3 then x = 2 and g = 2 )

etc.

I sweare this can be used to solve Fermat's last theorem.

 

[Sariaht]

Please, reply for me?

 

[matt grime]

it's unclear what you are attempting to say, but it appears, that you've divided odd numbers by two so you're immediately out of the realm of FLT, and there ie nothing special about n=2 so you've proved that there are no pythagorean triples despite writing one down

 

[Sariaht]

Look.

o_aⁿ +/- o_bⁿ = e_cⁿ. This must be proven wrong for n>2.

all the other cases are impossible.

e_cⁿ is divideable with 2ⁿ.

e = even, o = odd

 

[matt grime]

yes, that is rather obviously what you must prove. but you've not done so.

 

[suyver]

Originally posted by Sariaht
(o_a/2^x)ⁿ +/- (o_b/2^x)ⁿ = o_cⁿ

Here you are deviding an odd number by a power of two. As a result, you are no longer dealing with whole numbers.

 

[Sariaht]

Originally posted by matt grime
yes, that is rather obviously what you must prove. but you've not done so.

well two odd numbers cannot become a third odd number.

And two even numbers cannot become an odd.

ofcourse two even numbers can become a third, then you can divide both sides with 2^nx

 

[Sariaht]

Originally posted by suyver
Here you are deviding an odd number by a power of two. As a result, you are no longer dealing with whole numbers.

No, but now we are dealing with binomials instead!

Ain't that so?

 

[suyver]

Originally posted by Sariaht
No, but now we are dealing with binomials instead!

I guess I am just dumb. Binomial coefficients are of the form

$$\frac{a!}{b!\; (a-b)!} \equiv \left(\begin{array}[c] a a \\ b \end{array}\right)$$

What has that to do with Fermat or with your division by 2^x ?

 

[matt grime]

what are you blathering about, sariaht?

if p^n+q^n=r^n then we may assuming that after dividing out by the largest power of 2 as necessary that at most one of these numbers is even. yes we know this. that is why we often look for p,q,r coprime - if any two of p,q and r are divisible by a number so is the third, therefore we may divide through and reduce to the case of coprime numbers, in particular at most one can be even, at most one is a multiple of 3 etc.

AND?

your reduction in the complexity of the problems doesn't reduce to anything at all that we didn't know already.so why is it that expanding

(x/2+y/2)^n tells you anything?bearing in mind you're now in the ring Z[1/2]

 

[Sariaht]

Originally posted by matt grime
what are you blathering about, sariaht?

if p^n+q^n=r^n then we may assuming that after dividing out by the largest power of 2 as necessary that at most one of these numbers is even. yes we know this. that is why we often look for p,q,r coprime - if any two of p,q and r are divisible by a number so is the third, therefore we may divide through and reduce to the case of coprime numbers, in particular at most one can be even, at most one is a multiple of 3 etc.

AND?

your reduction in the complexity of the problems doesn't reduce to anything at all that we didn't know already.


so why is it that expanding

(x/2+y/2)^n tells you anything?


bearing in mind you're now in the ring Z[1/2]

because x and y are uneven, and uneven numbers can be written 2n - 1.
the quote in ((2n - 1)/2^x)^g that has the lowest value is 1/2^xg.
Sorry for making you waiting, I just got home from school

 

[matt grime]

you have n meaning two different things. and i still fail to see why that tells you anything. in particular you've n ot excluded the case n=2 (n the exponenet) by any genuine means (ie said something non-trivial that is only true for every n>2)

there is no reason why 2^x should divide the 2m in 2m-1 either and even if it did, so what?

 

[Sariaht]

Originally posted by matt grime
you have n meaning two different things. and i still fail to see why that tells you anything. in particular you've n ot excluded the case n=2 (n the exponenet) by any genuine means (ie said something non-trivial that is only true for every n>2)

there is no reason why 2^x should divide the 2m in 2m-1 either and even if it did, so what?

I didn't say it should. if g (sorry about the n) is high, then 1/2^xg won't transform into a hole number whith the rest of the terms in the binomial funktion(s) ( if I'm not misstakeing ).

look at the numbers; ((2n - 1)/2)^xg + ((2s - 1)/2)^xg. the two lowest therms cannot become a whole number together with the other numbers for g > 2. That would be weird.

 

[matt grime]

it's perfectly possible for numbers of the form x/2 and y/2 to have their n'th powers sum to an integer, even if x and y are odd. Just saying it would be strange is not mathematiclaly sound nor accurate. THe key is that the sum must also be an n'th power of some integer. And there is no reason to presume that can't happen, unless one assumes FLT is true.

example, 9/2 and 7/2. the sum of 7 cubed and 9 cubed is divisible by 8 by elementary modulo arithmetic.

 

[Sariaht]

What I am trying to say is two eights (2/8 = 1/4) cannot become a whole number together with two f fourths (2f/4 = f/2).

((2n + 1)/2^x)^g + ((2s + 1)/2^x)^g is not o^g for g>2.

Sorry there.

for g = 3, x = 1

(s³ + n³) + 1/4 + 3/2(s² + n²) + 3/4(n + s) is not o³.

 

[matt grime]

Did you actually work out the numbers in my last post or trust that you must be correct?

set n=4, s=3, x=1, g=3 in you last notation and the answer is 67, which looks a lot like an odd number to me.

And I didn't generate that at random; I can make many more for other g and other powers of 2.

 

[Sariaht]

Sorry then!

 

[matt grime]

your edited post is now correct, but i don't see any proof for it that does not already require FLT.

 

[matt grime]

I just found a pm referring to this. You want x=2 then? Stick with n=3,

we need to find odd numbers r and s with r/4 and s/4 having their third powers sum to a number divisible by 4^3 = 64.

how about 63 and 65, two numbers congruent to 1 and -1 mod 64, so their thrir third powers are still 1 and minus 1 mod 64, thus their sum is divisible by 64. You see how to generalize that for any x? n= any odd number is then doable, n even a little harder, but not much if you allow differences rather than sums, which you did earlier.

 

[Russell E. Rierson]

3^2 = 5^2 - 4^2

5^2 = 13^2 - 12^2

7^2 = 25^2 - 24^2

9^2 = 41^2 - 40^2



[x+1]^2 - x^2 = 2x+1

[x+1]^3 - x^3 = 3x*[x+1] + 1

[x+1]^4 - x^4 = 2x*[x+1]*[2x+1] + 1

[x+1]^5 - x^5 = 5x*[x+1]*[x^2+x+1] + 1

[x+1]^7 - x^7 = 7x*[x+1]*[x^2+x+1]^2 + 1

The answer to the riddle is hidden within ...?

 

[matt grime]

Would you mind helping out the lesser mortals round here like me who don't see what you're getting at? What riddle, what answer, what ellipsis?

 

[Russell E. Rierson]

[x+y]^2 - x^2 = 2xy + y^2

3^2 = 2*4*1 + 1^2

5^2 = 2*12*1 + 1^2

7^2 = 2*24*1 + 1^2

[x+y]^3 - x^3 = 3yx^2 + 3xy^2 + y^3


y[3x^2 +3xy + y^2]

cannot be a cube.

y[2x+y] is a square.

y[3x^2 +3xy + y^2] is not a cube

y[4x^3 + 6yx^2 + 4xy^2 + y^3] is not a 4th power

 

[matt grime]

Yes. And? I think it was archimedes who proved there were infinitely many pythagorean triples because there are infinitely many odd squares (plus some coprimality). If you happen to have a good test lying around to check every single case there then you've got a proof of FLT.

 

[Russell E. Rierson]

The proof boils down to the fact that

[binomial expansion]^n - [first term]^n

cannot be an integer nth power, for n > 2

reducto, and Wiles is

 

[matt grime]

So you've got a proof lying around that none of those expressions can possibly the n'th power of an integer? Cor, it's amazing no one thought of this line of attack!