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Welcome to NRICH.

The Fermat Game

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This discussion is unedited.

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By Barry Hale (M898) on Sunday, December 10, 2000 - 06:12 pm:

I've recently been playing with Fermat's Last Theorem, and I'd like to present the following to NRICH members in the spirit of a mathematical game.

I took up maths as a distraction from the dreariness of everyday life, and set myself a project to resolve this question:

Is it absolutely clear that the assertion of the Fermat equation is consistent at a basic level?

After a bit of struggle, I managed to cobble together a scheme that looks promising.

The cubes of odd integers are 1, 3, 5, or 7 mod 8, while the cubes of even integers are always 0 mod 8, so a sum of two cubes would be

x³ + y³ = z³

0 + 3 º 3 (mod 8)

where x is even and y is 3 mod 8.

If x is even and 2 mod 8, the binomial expansion of the sum would be give a result of

8l + 2³ + 3³ = 8l + 8 + 27 = 8l + 35

which is 3 mod 8. The cube of a single integer 3 mod 8 would expand to 8j + 27.

Now we can pose the linear equation

[integer] = 8k + r

- r + [integer] - 8k = 0

If [integer] happens to be a sum of cubes as outlined above, then one would have to resolve how

8l + 35 = 8j + 27

My addition to this simple arrangement is a structure that is provably the sum of cubes ("happens to" becomes "must equal"), which can be applied to any odd power, and cases where y > x are distinct from the cases x > y. This last point is important because it allows one to draw certain conclusions mod 8 that seem to indicate the inconsistency I was seeking.

I have this out to some maths teachers and on various Web boards and newsgroups and no one has yet replied back "Well Barry, your argument has the fundamental flaw ...", so it looks like (great sigh of relief) I might have actually posed a puzzler.

The postings to "Open Discussions" seem to indicate a playful spirit and, given the international audience, I'm sure someone will resolve my puzzle shortly.

A game of Fermat anyone?

Best Regards,
Barry A. Hale
Maryland
USA

Attached are an ASCII and a "pretty" Word 7 for Windows 95 summary of my arguments. For non-English code page users: the congruence symbol in the Word version comes from the built-in Symbol font for Windows 98 SE (SYMBOL.TTF), hopefully non-English versions will install this same font.

ASCII version yafp.txt (16 k)

Word 7 version picturea4s.doc (62 k)

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By Michael Doré (Md285) on Sunday, December 10, 2000 - 10:08 pm:

This looks highly interesting but I'm afraid I don't follow very much of your document. In the last paragraph of the first page, what is n? Also what is the definition of X_ml? And I'm afraid I don't understand the equation that follows:

X_3i(x,y) + u_3i = -n²x + 2nx² + 2(2n-1)xy + u_3i (i = 3n - 1)

or indeed anything that comes after that. If you could explain that part I'd be very grateful.

Yours,

Michael

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By Barry Hale (M898) on Monday, December 11, 2000 - 02:16 am:

That was quick. I expected to have to check in every couple of days. Thanks.

The section you're referring to is seeking to find an alternative way for expressing a sum of two cubes.

I present the formula for a sum of two cubes as

(x + y) + 3(x + y) + ... + (2x - 1)(x + y) + (2x + 1)y + ... + (2y - 1)y

the idea from that point is to find a compact way to represent a partial expansion of this series. X_ml represents l terms being expanded, starting from the term (2x - 1)(x + y). m is the power that x and y are being raised to, in this case 3. So if only 3 terms are expanded

(2x - 3)(x + y) + (2x - 1)(x + y) + (2x + 1)y
= 2x² + 2x² + 2xy + 2xy + 2xy - 3x - x - 3y - y + y
= 4x² + 6xy - 4x - 3y

i = l = 3 = 2(2) - 1 = 2n - 1

the coefficient on x² is n² = 2² = 4
the coefficient on xy is 2(2n - 1) = 2(2(2) - 1) = 2(3) = 6
the coefficient on x is - n² = - 2² = -4
the coefficient on y is - (2n - 1) = - (2(2) - 1) = -3
u₃₃ is the unexpanded balance of the series.

these are the parts of the formula that you're referring to.

34³ + 43³ (3 terms expanded) is

- 4(34) - 3(43) + 4(34²) + 6(34)(43) + u₃₃
= - 136 - 129 + 2(1156) + 6(1462) + u₃₃
= - 265 + 2312 + 8772 + u₃₃
= 10819 + u₃₃

the cubes sum to 118811 so the balance, u₃₃, is 107992

118811 = 10819 + 107992

X₃₃ is the symbol for this expansion of 3 terms.

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By Anonymous on Monday, December 11, 2000 - 09:02 am:

Hi Barry,

What does gcd stand for?
_____
/-|0\
\ = /
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By Barry Hale (M898) on Monday, December 11, 2000 - 02:08 pm:

greatest common divisor

Looks like NRICH Club is living up to it's claim: "This site is for interaction - not just browsing!"

Back to (gag) work.

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By Michael Doré (Md285) on Monday, December 11, 2000 - 05:56 pm:

Same as HCF. Barry - thanks for your explanation; I think I understand a bit better what's going on now. I am still a bit confused about why we have F_ml and X_ml if they're defined to be the same.

OK, I think I just about understand up to just before the section entitled Methods. Then as I understand it you start to write out X₃₇ + u₃₇ in base 8.

Then I certainly agree that X₃₇ + u₃₇ can be written as 8*integer + another integer in more than one way, and so I can see that there should be solutions to the equations with -35 and -27 in bold. However I don't then understand the sentence: "the working equation consists of a binomial part and a sum of cubes" part, and I'm not sure how the sum of cubes part is linked to the binomial part by t₁ = r₈. I'd be grateful for any explanation.

Thanks,

Michael

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By Barry Hale (M898) on Monday, December 11, 2000 - 07:42 pm:

Hold that thought. Still at work. Short answer: F and X are formal distinctions.

The binomial v. sum... parts come from a lack of formal names, but two eq. at play, one held const., the other variable. Sorry about the bold, overemphasis.

The link: not looking, but eight variables, 7 parametric eq. for each variable, 7 parameters, but parameters set across eq--set t[1] you set it for all eq. at once, set t[2] you set across 6 eq., etc. Change t[1] and the other t must change, thus the linkage.

Too verbose, perhaps?

Will post better this eve. (prob. after 12a London time).

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By Barry Hale (M898) on Tuesday, December 12, 2000 - 06:59 am:

Why have F_ml and X_ml?

Well, … you’re right. The original idea was to make a (too fine?) distinction

[partial sum of cubes] + w_ml = z³
w_ml = z³ - [partial sum of cubes]

[partial sum of cubes] + u_ml = x³ + y³
u_ml = (x³ + y³) - [partial sum of cubes]

I guess this was about how and what symbols are used. The symbol F_ml, using z³, was created and held in reserve just in case there was some future fine point made that would actually require this distinction.

But then how is the partial expansion sythesized except by using the sum of cubes? And we're arguing that the sum and the power are identical anyway, so perhaps a pointless distinction.

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By Barry Hale (M898) on Tuesday, December 12, 2000 - 12:17 pm:

Answers pending.
Didn't have time to finish last night. Will do today.

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By Barry Hale (M898) on Tuesday, December 12, 2000 - 06:44 pm:

Question:

I know I haven't finished the last answer, but here's a thought: how would I synthesize X₃₇ from Z₃₇? How would I derive the terms of my sum-of-cube series from the series for the single-integer cube?

Confusing things more and I haven't the slightest idea how to proceed, but I just wanted to pose that thought.

Actually enjoying your methodical approach. I scoured the globe for illumination and you are providing it. Excellent.

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By Michael Doré (Md285) on Wednesday, December 13, 2000 - 01:50 am:

Thanks for your explanation, Barry.

I think I now nearly understand the first three pages. On the fourth page I'm slightly confused as to why we have 7 parameters if there are 7 variables. For instance if we are to write:

x² + y² = 1

in parametric form, we could do so x = cos t y = sin t, i.e. use only one parameter though there are two variables. So shouldn't the number of parameters be one fewer than the number of variables, hence 6?

I think the answer to this question lies in your statement "If only the first 6 parameters are set (t₇ = 0), the parametric solution for k₁₀ happens to equal k₉." but I still don't fully understand.

Of course there are loads and loads of solutions to the linear equation you have written. By Euclid if you take any two of the terms with coprime coefficients then this can be set to any integer. I guess the problem is that the variables are not any old value but have conditions imposed on them (e.g. x₁ is a square).

I am not entirely sure what you mean about synthesizing X₃₇ from Z₃₇ in your last post. Do you mean can we calculate Z₃₇ if we know the value of X₃₇?

Yours,

Michael

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By Barry Hale (M898) on Wednesday, December 13, 2000 - 06:19 am:

Parts

[integer] = 8k + r

[sum of cubes] = [binomial product]

[sum of cubes] = [binomial part 1] + [binomial part 2]

- [binomial part 2] + [sum of cubes] - [binomial part 1] = 0

- [part 2: two particular values] + [maintained at a value] - [what happens when part 2 changes?] = 0

- r + [integer] - 8k = 0

Links

The parametric solution is a coordinated system of expressions. In this particular case, the equation being worked with has 8 variables, requiring 7 parameters. The coefficients on the parameters used for the example

	t1	t2	t3	t4	t5	t6	t7
r8	1
x1	-1	1
x2	15	-16	1
x3	90	-96	7	1
x4	-630	672	-49	-8	1
u37	-57330	61152	-4459	-728	98	8
r9	57330	-61152	4459	728	-98	-7	-8
k10	-8190	8736	-637	-104	14	1	1

In the example, t₇ was set to 0, and the solution for k₁₀ equals k₉

r₈,[part 2: two particular values]
x₁,[maintained at a value]
x₂,[maintained at a value]
x₃,[maintained at a value]
x₄,[maintained at a value]
u₃₇,[maintained at a value]
r₉,[not used]
k₉,[what happens when part 2 changes?]

The maintenance of the central part at a certain value depends on the value of the first parameter t₁, which equals r₈. This is a coordinated system: if the value of t₁ changes, adjustments have to be made to the entire chain of expressions in terms of the parameter values.

The value of k₉ is the only value that "floats", but its parametric solution depends on the same parameters as u₃₇, and since we're adjusting the parametric expression for u₃₇ to maintain a value, k₉ has to follow suit.

[sum of cubes] is linked to [binomial part 1] and [binomial part 2] by changes in r₈ = t₁.

----------------------------------------------

8 variables, 7 parameters. The last two variables are r₉ and k₁₀, set by the same 7 parameters. Look again at how these are defined, how the equation is set up, and at the parametric solution above and you will find that if the final parameter is set to 0 then the parametric solution for k₁₀ should equal k₉.

There are an infinite number of solutions, but there are an infinite number of solutions to x³ + y³. In my case values are being constrained, but isn't the symbol "y³" shorthand for the constraint that y be multiplied by itself 3 times? Is this a fair comparison?

The idea is to impose a certain structure that can be associated with a sum of cubes.

So now that the concept has been presented, I'm calling on your expertise. The setup section is completed and you seem to accept the ideas so far, but "understand" ¹ "agree".

Should we proceed? Have I imposed a sufficient structure to start figuring out how 8l + 35 = 8j + 27 using this setup?

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By Barry Hale (M898) on Thursday, December 14, 2000 - 05:56 pm:

Pending a response to the last post, I'll start answering my own question with the idea that my construction X_ml is too general in some ways and not specific enough in others. More to follow.

To Michael: thanks for your input.

Happy Holidays,
Barry

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By Barry Hale (M898) on Thursday, December 21, 2000 - 10:26 pm:

Progress?

Looks like the second half of my essay can stand on its own--a question of "if it can be determined which of x and y is greater" versus "it must be the case one of x and y is greater".

Thought I'd figured this thing out and could finally stop thinking about it, but more mysteries. It seems that my conclusions may actually hold up for odd n > 5, even n > 4, but n = 3,5 (that have been proven for ages) are open questions using my system (sigh).

More to follow
[but if anyone able to help is still following this thread, please feel free to put me out my misery ;-) ]

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By Michael Doré (Md285) on Thursday, December 21, 2000 - 11:03 pm:

Barry, sorry for the delay - I haven't been able to reply on this over the week. I'm still not really following you, so (starting tomorrow) I'm going to try and summarise your document bit by bit, then you can point out where all my misunderstandings are. I think that the way it is written out does need a bit of modification - it is often quite hard to tell whether an equation is a definition of a variable, or an independent assertion.

I wouldn't worry too much about n = 3,5 - if your method works for all other odd n then that will be quite some achievement!! I can give a reference for the proof of the n = 3 case if you want. I'm not sure about n = 5. But that is unimportant - it is n > 5 that is the hard part, and if your method works it will probably turn the world of number theory upside down, and certainly make national press. The 1995 proof of FLT is far, far from elementary. But I'm afraid I'm lost well, well before the end of your document, so I can't comment on whether it works or not.

Yours,

Michael

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By Barry Hale (M898) on Friday, December 22, 2000 - 03:41 pm:

I am redoing my essay, but please feel free to comment on the original draft. Your patience is remarkable, but I hope you're having fun (this is only a Game).

The current rewriting was inspired by your comments. In a nutshell: seek clarity of expression (just the facts, ma'am).

As soon as this query is cleared up I intend to take up another problem to plague NRICH with. This is an amazingly calming diversion: it makes other (serious) problems much more tolerable.

A question: do binomial expansions uniquely express powers of binomials? For instance, can

(n + a)³ = n³ + 3n²a + 3na² + a³

also be written as the incomplete expansion of the cube of another binomial (m + b)

3m²b + 3mb² + b³

where the term m³ is not present (a,b,n,m all integers)? This likely has something to do with factorization theories. Web links welcome.

Best Regards,
Barry

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By Michael Doré (Md285) on Friday, December 22, 2000 - 04:11 pm:

Hello Barry,

I will post the beginning of my summary shortly. Just to answer your last question, can:

n³ + 3n²a + 3na² + a³ = 3m²b + 3mb² + b³

where m,n,a,b are all integers?

Well we can rewrite it as:

(n + a)³ = (m + b)³ - m³

or

m³ + (n + a)³ = (m + b)³

By Fermat, we know the only solution to this is when one of the cubes is zero. So the only solutions are:

m = 0, n + a = b

or

n = -a, b = 0 (now m is free to be anything)

or

n + a = b = -m

Yours,

Michael

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By Barry Hale (M898) on Friday, December 22, 2000 - 07:30 pm:

You say by Fermat--Fermat's Little Theorem? I have this in a text, but the way it is presented doesn't let me see the solution you just gave (or I didn't understand it).

Can you amplify or point to a Web reference? My searches on binomials so far have only lead to highly abstract applications.

Thanks,
Barry

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By Michael Doré (Md285) on Friday, December 22, 2000 - 08:06 pm:

No, sorry I meant Fermat's Last Theorem!!

We can prove that:

x³ + y³ = z³

has no solution unless xyz = 0. This is a special case of FLT, but it has a far more elementary proof than the general case.

Therefore:

m³ + (n + a)³ = (m + b)³

only has solutions where m = 0, or n + a = 0, or m + b = 0.

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By Barry Hale (M898) on Friday, December 22, 2000 - 09:13 pm:

Sorry to keep posting like this (I promise this will be the last today), but if I knew that

z³ = (y + 8k)³

could I then say

x³ + y³ = y³ + 3y²(8k) + 3y(8k)² + (8k)³

x³ + y³ - y³ = 3y²(8k) + 3y(8k)² + (8k)³

x³ = 3y²(8k) + 3y(8k)² + (8k)³

with x³ equal to the incomplete expansion I queried about? If I understand you correctly, this means that y = 0. True?

I haven't finished digesting what you sent (or think the last through), I probably shouldn't post yet, but I want to prepare for "homework" this weekend. So don't laugh too hard when you read this.

One informative-looking link with info about polynomials and factoring (almost too informative--are mathematicians taught to forget how to speak English?)

Applied Abstract Algebra
[This site appears to have moved. - The Editor]

I know it's getting late over there, so once again my apologies.

Thanks for your time,

Barry

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By Michael Doré (Md285) on Friday, December 22, 2000 - 11:06 pm:

Oh, I see, you were still trying to prove FLT for n = 3. Sorry, I thought it was a totally different problem.

All I did was assume the result of FLT for n = 3, and then go on to show that m = 0, n + a = 0 or m + b = 0. You agree with this right? Obviously we can't use this to prove FLT in n = 3, as this would be circular.

In that case I don't have any immediate ideas of where you can go from:

x³ = 3y²(8k) + 3y(8k)² + (8k)³

But anyway, if you believe you've proved it for odd n > 5 then that is certainly worth investigating!

By the way, do you know the standard proofs for n = 4 and n = 3? The method for n = 4 is beautifully simple - see x⁴ + y⁴ = z², the last message. The idea quite simply, is to use the fact that x²,y²,z are a Pythagorean triple, apply the generation formula, and then it turns out you can find positive integers x',y',z' so that x'⁴ + y'⁴ = z'² and z' < z. Therefore, by strong induction on z, there is no solution.

For n = 3 someone pointed out to me a magazine which contained the proof. The magazine is called Crux Mathematicorum, and is published by the Canadian Mathematical Society. The proof is not exactly straight-forward to figure out, but easy to understand. It basically revolves around polynomial arithmetic modulo x² + x + 1.

I don't know about n = 5; I suspect it can be done without high-level techniques, but I'm not sure.

Yours,

Michael

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By Barry Hale (M898) on Friday, December 22, 2000 - 11:38 pm:

Breaking my promise. So binomial expansions are not unique representations of powers of binomials?

An incomplete expansion of one binomial can indeed equal the normal expansion of another binomial?

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By Michael Doré (Md285) on Saturday, December 23, 2000 - 09:29 pm:

In general yes.

For example the equation:

a³ + 3a²b = c³ + 3c²d + 3cd² + d³

has an incomplete binomial on the LHS, and a complete binomial on the RHS. It is equivalent to:

a³ + 3a²b = (c + d)³

When a = 3, b = 7, c = 3, d = 3 (for example) we get equality.

It is true that the equation:

a³ + 3a²b + 3ab² = (c + d)³

(so the LHS now is only missing 1 term)

has no solution. But the only way I know of proving this is to use FLT for n = 3 (then the proof of the non-existence of integral solutions for the equation is trivial).

OK then, let's make a start on your document.

Preconditions

If there exist natural x,y,z such that

xⁿ + yⁿ = zⁿ

(n natural, >2) then we can choose x,y,z to be pairwise coprime (simply pick the solution with minimal z; then if two of x,y,z have a common prime factor p then as the Fermat equation is satisfied, the third variable is also divisible by p, hence x/p, y/p, z/p are another solution, contradicting the minimality of z).

Hence we have a solution x,y,z in which no more than two of x,y,z are even and as they can't all be odd (this fails mod 2) we conclude that exactly one of x,y,z is even.

Now we set n = 3. If x is even then we have:

y³ = z³ (mod 8)

If y is even, it is similar; just replace y with x. If x,y are odd then z is even, so either:

x = 1, y = 7 (mod 8)
x = 3, y = 5 (mod 8)
x = 5, y = 3 (mod 8)
x = 7, y = 1 (mod 8)

OK, I have no problems so far.

Next, you say: the possibility of x and y both being odd fails to yield integer solutions in the procedure that follows. So are you saying that by the end of the argument we will show that there are no solutions where x,y are both odd?

The other case is where one of x,y is even; the other odd. Without loss of generality, we can call the even one x, so y is odd. But then you say, "with the added condition y > x". How do you know this? (I missed this when I read through earlier.) Are you taking only a subcase? If so does this mean that we aren’t actually going to prove FLT – only that a subset of the possible (x,y,z) values don’t have a solution?

Well let’s continue anyway.

Concept

We know that:

n²(1 + 3 + 5 + … + (2n-1)) = n³
So x³ + y³ = z³ can be written:

x(1 + 3 + … + (2x-1)) + y(1 + 3 + … + (2y-1)) = z(1 + 3 + 5 + … + (2z-1))

If we take y > x:

(x + y)(1 + 3 + … + (2x-1)) + y((2x+1) + (2x+3) + … + (2y-1)) = z(1 + 3 + … + (2z-1))

Let A_r = (x + y)(2r – 1) for r <= x, and A_r = y(2r – 1) for r > x. Let B_r = z(2r – 1). We can now write x³ + y³ = z³ as:

A₁ + A₂ + … + A_x + A_x+1 + A_x+2 + … + A_y = B₁ + B₂ + … + B_z

We are going to define three polynomial quantities F_ml, X_ml, Z_ml. The first two are polynomials in x and y; the third is a polynomial in z. We define X_ml = F_ml always. F_ml is going to be l of the A terms of the left side. Z_ml is going to be l of the B terms of the right hand side. For the time being we’re only defining F_ml, X_ml, Z_ml for m = 3, correct?

For the case where m = 3, we define:

F_m1 = A_x
F_m2 = A_x + A_{x + 1}
F_m3 = A_x-1 + A_x + A_x+1
….

In other words when r is even,

F_mr = F_m(r-1) + A_{x + r/2}

When r is odd:

F_mr = F_m(r-1) + A_{x – (r-1)/2}

Throughout this definition we’re keeping x fixed.

We define Z_mr as B₁ + B₂ + … + B_z where z is some fixed integer?

OK, so how is F_mr defined when r is between x and y – x? I believe this isn’t covered by the definition I’ve used.

I don’t follow what you mean by: “sum of cubes is expanded starting at the disjunction between terms (2x - n)(x + y) and terms (2x + n)y” What exactly is n here?

I don’t really see how F_ml and Z_ml approximate x³ + y³, or z³. If you set l to be the right value, perhaps, but if this is the case, I don’t see why we are working with 7th level expansions. Sure we can define:

u_3r = w_3r as x³ + y³ – F_3r (so again it is also dependent on x,y).

Likewise:

v_3r = z³ – Z_3r

(And I guess if we actually get round to defining X_mr, Z_mr for m not equal to 3 then we could define v_mr as z³ – Z_mr and similarly for u,w_mr.)

But I really don’t see where we can go with the equation:

F₃₇ + w₃₇ = Z₃₇ + v₃₇ (*)

F₃₇ is probably not even close to x³ + y³. Sure the error term means that F₃₇ + w₃₇ can equal x³ + y³, but as w₃₇ and v₃₇ can be anything, I don’t see how (*) can be useful.

Finally you write down a linear equation (which I agree with) but then say: "a failure to find linear solutions would also result in a failure for the non-linear system". Well yes, but since two of the components of the linear equation are u₃₇ and v₃₇, and these can be anything a failure to find linear solutions does not sound likely. I am obviously missing something.

Perhaps you could elaborate slightly on the "game" aspect of this. I think it may shine light on the points about the proof that I've been misunderstanding. Are there any holes in your proof known to you? (I'm assuming you've checked it over a few times, and believe it is a legimate proof of FLT - am I right?)

Thanks,

Michael

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By Michael Doré (Md285) on Sunday, December 24, 2000 - 01:36 am:

Mistakes above:

n²(1 + 3 + 5 + ... + (2n-1)) = n³

should be

n(1 + 3 + 5 + ... + (2n-1)) = n³.

The definition of Z_mr for m = 3 should have been:

B₁ + B₂ + ... + B_r

not

B₁ + ... + B_z

Any more?

Yours,

Michael

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By Barry Hale (M898) on Monday, December 25, 2000 - 02:06 am:

Thanks for the input. I've have found some problems myself. Revisions shortly (I hope).

The "Game" reference:

Yes, I think I might have something. Not trying to waste your time or take up space on NRICH. I've been working on this all day, for instance. Total rebuild.

Yes, I've checked it quite a few times (the original report was 30 pages). I'm asking for help because I felt I had at least reached a point where an extra set of (informed) eyeballs was called for.

However

The idea here is to gain insight--why is this thing so darn hard? Only one way to find out (did I ever).

But think about this for a sec. This problem has baffled folks for centuries and the only solution that anyone arrived at was a "Well, I've got an answer, but it's going to take a while to explain ..." (sound of 200-page single-spaced report hitting desk) type of thing. So you would have to be loony to take this too seriously.

This is a game. I really am enjoying this (between the headaches). No harm is done in attempting it. If you make even a minor amount of progress, consider yourself fortunate. If you're really fortunate and actually prove something, Wow!

I'm not alone: at least two teachers I've written to either are playing with this or know of someone who is, and there are few simple proofs floating around the Web. As long as a direct, simple proof is open game I think this will be a diverting temptation.

Your insights and time are appreciated.

Merry Christmas,

Barry

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By Barry Hale (M898) on Tuesday, January 2, 2001 - 10:56 pm:

Version 2.0

New! Improved!

Cleaned things up and tried to be as clear as possible. Equations lined up like soldiers (ready to be mowed down). Checked my ideas against Pythagoras this time (see the first and last pages).

Came across an interesting possibility. It would be kind of neat if it works out.

Ver. 2.0 testfermat1.doc (124 k)

Happy New Year,

Barry

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By Michael Doré (Md285) on Wednesday, January 3, 2001 - 12:09 am:

Barry - thanks for taking time to clarify your original document. I thought I'd have a quick look at it now (the start at least):

"8W + 8 + 27 = 8Z + 27"

You say this intuitively seems to be incorrect. I don't quite understand why not. It implies W + 1 = Z, and I'm not sure why this is counterintuitive. Of course I'm not looking for you to give a rigorous answer (that, after all, is what the rest of the document is aimed to do) but I'd be interested in how it goes against intuition.

Next, in the Preconditions section, I don't quite see how you've proved that with the equation:

xⁿ + yⁿ = zⁿ

you can ignore the case where x,y are odd.

You say: "xⁿ + yⁿ = 8Z requires a combination of non-integers, xⁿ/8 + yⁿ/8 to derive the integer Z." Well I agree that Z can be written as the sum of two non-integers, but that doesn't imply that x,y are not integral.

Thanks,

Michael

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By Barry Hale (M898) on Thursday, January 4, 2001 - 02:18 am:

Intuition: Trying to find a way to say that there is something about the thing that just doesn't seem right. Can't explain any better.

Both odd: Kept on thinking this one over. Doesn't seem that it should be a valid combination. When I used both odd with the parametric system I was using earlier, t₆ always came out non-integer. Will rethink.


Thanks,

Barry

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By Michael Doré (Md285) on Saturday, January 6, 2001 - 09:58 pm:

Thanks Barry. Sorry if I sounded pedantic about the counter-intuitive bit, but I really couldn't see how the equations appears to be inconsistent. It's not important anyway.

I'm still having trouble following your argument. In the bit about xⁿ/8 + yⁿ/8 = Z then in general there are solutions when x,y,Z are integers. Simply take x = 1 (mod 8), y = 7 (mod 8). So I think I'm missing something fundamental here.

I don't really understand the table on page 3. I may be being a bit slow, but I don't understand where the constant column of 2s are coming from.

Thanking you for any further explanation,

Michael

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By Barry Hale (M898) on Sunday, January 7, 2001 - 08:34 pm:

Sorry not to reply sooner. Been fighting a bug of some kind.

Both Odd
The only way to tackle the special cases one 1 mod 8, the other 7; one 3 mod 8, the other 5 is to do a separate report. I just might be able to knock these cases out. Will let you know. So the report can only address one of x,y even, the other odd.

Page 3
Given that one of x,y is even, the other odd, there are 16 combinations mod 8. The table illustrates these combinations.

So in the cases where x is even and 2 mod 8, y can be 1,3,5,7 mod 8. This choice is, relatively speaking, constant to the varying choices 1,3,5,7 for y. Just an illustration.

But I think you're trying to make a point. Why not pick out the column of zeroes, for instance? In the case n = 3, x³ would be 0 mod 8 for x º 2 mod 8, perhaps allowing the elimination the column of twos. The table is choices mod 8 for x and y not x³ and y³.

I think I see what you're getting at: the next sentence has "each choice of s³ ..."

OK, amend this with "for each choice of s, s³ is an even, constant C ..."


Regards,

Barry

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By Michael Doré (Md285) on Monday, January 8, 2001 - 01:05 am:

No, no I wasn't trying to make a point; I just didn't understand. I think I do understand that bit now; thanks. More feedback to come in the morning...

Yours,

Michael

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By Michael Doré (Md285) on Tuesday, January 9, 2001 - 12:26 am:

Sorry, I got distracted. This certainly looks very interesting. But I'm having problems with page 5. I kind of understand the gist of the first four pages though there are still lots of details I haven't fully grasped. (I would recommend saying explicitly for each equation whether it is an assertion or a definition of a variable; and in the latter case to specify which variable is being defined.) On the table in page 5, I'm not sure of the significance of the estimate (1 + 24/z) column.

And then I'm totally lost in the next two paragraphs. You say, "the ratio between a candidate and a base value needs to be <= 2". If I understand your terminology, I think you've shown that two consecutive candidates must have ratio less than 2. But why does this mean that any candidate has ratio less than two with the base value? (I think I see what you mean by base value here, but I'm not totally sure.)

Thanks,

Michael

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By Barry Hale (M898) on Tuesday, January 9, 2001 - 04:19 am:

You should understand that I am coming to NRICH for academic/professional assistance with my idea, as I haven't studied math in a while (don't ask how long) except for my recent efforts.

Fresh Start
You have a point: I am not defining things as I should. I actually am aware of this.

But on the other hand, I really do think I might have something here. So should I not express myself because I'm perhaps a bit (OK, maybe quite) sloppy? Seems the spirit of "Open Discussions" is to invite such expressions.

So (same wavelength) please point out where it seems an assertion is being made (or should be) and where a variable has not been defined properly.

--------------------------

Current Comments

The estimates and illustrations in this section are an effort to get the reader to see what I'm seeing.

The first estimate is a rough estimate of the ratio between adjacent cubes that are congruent r mod 8:

43³, 51³, 59³, ... (r=3)

the estimate is in the table to show that it actually is a reasonable estimate, since I then use that estimate to get another estimated value--a safe estimate because it's actually an underestimate.

< 2y³

One case: y > x, y = 43. With x³ + 43³ I'm saying every case where x < 43. Since x is at most 42, the sum 42³ + 43³ < 2(43³).

So Fermat is saying that one or more of the cases x³ + 43\+(3} might sum to yet another cube. Which one? It should be one congruent with 43 º 3, i.e., 51³, 59³, ... a set of candidate, prospective values for z³.

Starting from 43, the base value, we begin counting up 3 mod 8 looking for this cube, but we can't use a cube > 2(43³), so we can only count to 51, since the next cube, 59³, is 2.583y³

New case: y > x, y = 27, a new base. But we can't pose the cases x³ + 27³, because the first candidate, 35³, is 2.178(27³).

Has all of this been long-winded enough?

Apologies,

Barry

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By Michael Doré (Md285) on Thursday, January 11, 2001 - 11:41 pm:

Sorry for the delay. I'm just putting together a summary of the first few pages of your second document so you can comment on how well I've understood it.

Regarding your last message:

> So should I not express myself because I'm
> perhaps a bit (OK, maybe quite) sloppy?

Of course not. Having ideas is the most important thing. However, I'm having trouble following some of the concepts in their current form, and I expect that some other people might have the same difficulty. So I'm merely suggesting ways of altering the text so that more people have access to your ideas without being confused by notation.

Let's take an example on the first page:

nis^2m + njr^2m + c º nkr^2m, n º 2m + 1 = 1,3,5,7 mod 8

Where exactly have i,j,k come from? In the first of the two equations, under what modulo does the congruence apply?

Please don't think I'm being critical - I personally have a lot of trouble expressing myself clearly at the best of times. I simply bring it up because I'm having a bit of difficulty following it (and this may be more my fault) but I think you need to assume the person reading it has little intelligence. It is just for this reason I suggest making it clear exactly whether a variable is being defined by the equation, or whether we're already supposed to know about the variable and the equation is a new assertion.

More to come,

Michael

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By Barry Hale (M898) on Friday, January 12, 2001 - 03:17 am:

OK. I'm starting to see.

Actually ran into this in a text on Diophantine equations that would have been quite useful except the professor never defined all of his notation (the way most texts do), so I couldn't follow it easily.

You will see the use of i, j, k at the beginning of "Definitions"

x = 8i + s, y = 8j + r, z = 8k + r (s = 0,2,4,6; r = 1,3,5,7)

This definition is maintained throughout. Everything in the report is mod 8, but in the summary I guess I should have said

"ns^2mi + nr^2mj + c º nr^2mk (mod 8) (n = 2m + 1; s = 0,2,4,6; r = 1,3,5,7)

where

x = 8i + s, y = 8j + r, z = 8k + r"

Also, I have since revised this part to say "if the equality the above congruence is based on is rearranged ..." rather than "if the congruence above is restated as an equality ..." and should change the b in 8b - c to something else, since I use b later on in a different context.

You make an important point: if the summary is daft-looking, then readers won't look any further.

Any and all other comments welcome.

Thanks,

Barry

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By Michael Doré (Md285) on Friday, January 12, 2001 - 05:59 pm:

OK, thanks that clears it up. It's probably worth putting the definitions of i,j,k at the start if they are used in the introduction.

I think I'm starting to understand the table at the start of P.5 a bit better now. If I understand correctly the accumulated estimate is the product of each individual estimates prior to and including that line (previously I'd assumed it was the sum). I'm still not entirely sure about the derivation of the inequalities that follow on the same page, but I think I'm getting there.

Before I post my summary (which may have to wait till after I'm back at college tomorrow) I was wondering: a lot of the text seems to be based around the equation x³ + y³ = z³. Is this a special case to let the reader get familiar with the ideas, which can later be generalised to other n? Or is the document designed to prove FLT for n = 3 only?

Yours,

Michael

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By Barry Hale (M898) on Saturday, January 13, 2001 - 04:47 am:

The inequality

Say y > x. z is the cube of one of the numbers greater than and congruent with y. But you have the limitation z³ < 2y³. The problem is infeasible until the ratio between consecutive cubes becomes small enough to conform to this limitation.

The idea is that you'd be wasting your time asking anything about 2³ + 5³, for instance, because the next number 5 mod 8 is 13, whose cube is 17.57 times 5³.

So when does the problem start making practical sense? The formula is an estimate of this lower boundary. The table is this boundary where z can feasibly be within one increment of the base value I mentioned in the earlier post.

But what if z had to lie more than one increment away? What about 4 inccrements away? In order for z³ to fall under twice the base value, the ratio between adjacent values would be around the fourth root of 2.

Confused? Me too, but I've tested this (maybe not as much as I should) and it seems to work.

The inequality comes from the work started on pg. 4. Looking at it again I could have been a little clearer how I arrived at it.

-------------

The case n = 3 is used as a first case. Other powers are also considered. The ideas with n = 3 seem to carry over to higher n. See pg. 7.

Kind of concentrated on n = 3 because my Excel atarts going simple with larger powers.

Do you know of any Excel add-ons or other software for dealing with very large integers?

Thanks,

Barry

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By Michael Doré (Md285) on Saturday, January 13, 2001 - 10:27 pm:

No, sorry. But how far did you go with Excel?

I see what you mean now about the inequality. Certainly the idea is a very good one in principle. Is this basically the same approach as in Version 1 or is it entirely different?

I must admit I still am not quite entirely sure on P.5. Isn't the accumulated estimate going to go to infinity as k-> infinity? So presumably this is where the lower bound table comes in (next page), where we show that no only do low values not work but high ones also don't work.

What I really don't understand about this part I think are the lines of inequalities in the middle of P5. We have 2^1/I >= 1 + nb/z. What is I here?

Thanks,

Michael

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By Barry Hale (M898) on Sunday, January 14, 2001 - 08:00 pm:

Replying before I have digested again, but wanted to try to catch you before it gets too late there.

Different approach. This work is an extension of that idea with new observations.

k goes to infinity, but any results have to conform to the limitation z³ < 2y³.

The "I" is the increment from whatever base value you're considering. So in the example problem I use the example x³ + 43³ = z³ to stand for all the cases even x < 43. The only feasible answer for z is 51, since its cube is < 2(43³).

43 = 8(5) + 3
51 = 8(5 + 1) + 3
59 = 8(5 + 2) + 3
.
.
.

I = 0, 1, 2 for the first 3 entries. The next section after this uses a congruence to show that for n > 5, I º 0 and so = 8t for some t.

So for the example problem where n = 7, we would need z to be a number 3 mod 8 and at least 8(5 + 8) + 3 = 107, 107 ⁷ = 1.6 E 14, 590 times 43⁷.

Our lower bound for feasiblility has moved up because of the requirement k - j º 0. to find the new lower bound, we need adjacent congruent 7th powers to have a ratio of about 2^1/8 so that if we go 8 increments up the prospective z is < 2 times any given base value. This is what the inequality should tell us.

We calculate (7(8))/(2^1/8 - 1) = 618 is around when this occurs. 619 º 3 mod 8, so use the example x⁷ + 619⁷ = z⁷, where z is some integer 3 mod 8, therefore at least 627.

627⁷ is 1.0941 times 619⁷, that is, about the 8th root of 2, so a requirement of 8 increments away is feasible.

Similar reasoning works for x > y, except we would also have to atate what y,z are mod 8. So the cases y³ + 40³ = z³, y,z º 1, odd y < x, the base value is 33, the highest number 1 mod 8 < 40.

The first candidate for z is 41, 1.85 times 40 cubed. A feasible candidate. But go two more increments up to 57 and we get an infeasible result that is 2.89 times 40 cubed.

Now don't laugh, but I didn't see the implication that high values don't work either. Love to see that ;-)

-------------------------

Excel

I easily encounter #NUM! errors with the higher powers unless I limit myself to relatively small values. The MOD function fails if numbers get too large. Frustrating. Have to start trying to break numbers up to get results.

--------------------------

Question

If I divide binomial coefficients n!/(k!(n - k)!) by n, it seems I get integers except for the first coefficient n!/n!

True?

Another long post. Sorry.

Regards,

Barry

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By James Lingard (Jchl2) on Sunday, January 14, 2001 - 08:11 pm:

4!/(2!(4 - 2)!) = 6 and is not divisible by 4.

When n is prime your statement is true.

Since n!/k!(n-k)! is an integer we know that k! | n!/(n-k)!, so k! | n(n-1)(n-2)...(n-k+1), but since k! and n are coprime we have k! | (n-1)(n-2)...(n-k+1) and so n!/k!(n-k)! = n(n-1)(n-2)...(n-k+1)/k! is divisible by n.

Hope that's comprehensible,
James.

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By Barry Hale (M898) on Sunday, January 14, 2001 - 10:08 pm:

To Mr. Lingard:

Got the gist scanning, will review further. That actually helps a great deal. Thank you very much.

Barry

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By Michael Doré (Md285) on Tuesday, January 16, 2001 - 12:58 am:

I'm afraid I'm more confused than ever before. I've been looking over the document a lot today and it seems that the summary I was about to post totally misses most of the important points. Sorry for being slow in responding but I need time to get my head round it.

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By Michael Doré (Md285) on Tuesday, January 16, 2001 - 01:22 am:

By the way, the statement that p choose r (shortened to C(p,r)) is a multiple of p (for 0 < r < p) also follows from Fermat's Little Theorem, which states.

n^p = n (mod p)

Set n = x + 1 and:

(x + 1)^p = x + 1 (mod p)
x^p + C(p,1)x^p-1 + C(p,2)x^p-2 + ... + C(p,p-1)x + 1 = x + 1

The x^p and the x cancel, and the 1s cancel:

C(p,1)x^p-1 + C(p,2)x^p-2 + ... + C(p,p-1)x = 0 (mod p)

is an identity in x, and it follows that C(p,1) = C(p,2) = ... C(p,p-1) = 0 (mod p).

A more general result than Fermat's Little Theorem is that:

n^phi(r)+1 = n (mod r)

where phi is Euler's phi function and r (unlike p) does not have to be prime.

So if you set n = x + 1 and equate coefficents on both sides you get:

C(phi(r) + 1,a) = 0 (mod r)

where 0 < a < phi(r) + 1. Don't know if that's useful...

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By Barry Hale (M898) on Tuesday, January 16, 2001 - 01:50 pm:

If I understand you, the whole idea of a lower bound on the feasibility of the problem is bugging you (amongst others, it seems).

Maybe just focus on the cases y > x.

The way I've been visualizing this, the interval (yⁿ,2yⁿ) represents a "window" of feasibility. I've graphed this on a logarithmic scale and I get something like


yⁿ [feasible zⁿ] ... 2yⁿ [infeasible zⁿ ... [infeasible zⁿ] ... infinity

Put this graph into motion. As y increases, the interval (yⁿ,2yⁿ) widens and "captures" an increasing number of feasible zⁿ.

Move y backwards, the window shrinks, and every potential zⁿ falls outside. y needs to meet minimal values.

-------------------

Point taken about the divisibility of binomial coefficients. Thanks again.

-------------------

I'm sure your comments will be insightful.


Barry

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By Barry Hale (M898) on Tuesday, January 16, 2001 - 07:04 pm:

I think I see what might be bugging you. But best if you give your take in case you see something different/additional.

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By Barry Hale (M898) on Tuesday, February 27, 2001 - 07:28 am:

Version 2.5

A New Look, with Great New Featues!

V2.5 v2_5.zip (49 k)

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By Brad Rodgers (P1930) on Saturday, March 3, 2001 - 08:59 pm:

I'm afraid I don't have the number-theory background to understand most of this paper, but have you proved Fermat for only 3, or for all n?

Thanks,

Brad

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By Barry Hale (M898) on Monday, March 5, 2001 - 01:38 am:

Are you referring to my last post? If so, the very last post should be referring to prime values of n.

No proof. This is more of an exploration, thus the various versions (and fragments thereof) posted. The latest version is looking interesting (well, to me anyway :-). Still playing with it.

Thanks for the interest,

Barry

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By Barry Hale (M898) on Saturday, March 17, 2001 - 04:06 am:

Vesion 2.5, Amendment 1

Pythagoras is behaving himself, but Fermat is starting to act silly.

Amendment 1 2_5a1.zip (31 k)

Probably seeing stuff, but this looks complete.

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By Barry Hale (M898) on Tuesday, March 20, 2001 - 12:13 am:

Vesion 2.6

Consolidated, cleaned up, corrected

Version 2.6 testfermat2_6s.doc (65 k)

Question:
What is the best way to check these results?

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