Sklansky -Fermat Conjectures

[David Sklansky]

Conjecture One: A to the nth plus B to the nth (when n is an integer, five or greater) cannot equal equal C to the nth plus q, for some if not most q's.

Conjecture Two: If there are in fact q's for which the conjecture holds, some will be formally unprovable. In other words it might be true that (A to the n) + (B to the n) can never equal (C to the n) plus (lets just say) the number 846879032 (n greater than four), yet no proof of this fact is even theoretically findable.

[Roland19]

David Sklansky, you just made my night. You're my hero.

[Daliman]

[ QUOTE ]
Conjecture One: A to the nth plus B to the nth (when n is an integer, five or greater) cannot equal equal C to the nth plus q, for some if not most q's.

Conjecture Two: If there are in fact q's for which the conjecture holds, some will be formally unprovable. In other words it might be true that (A to the n) + (B to the n) can never equal (C to the n) plus (lets just say) the number 846879032 (n greater than four), yet no proof of this fact is even theoretically findable.

[/ QUOTE ]

DUUHH

[Roland19]

Honestly, though, I'd be interested to hear the relevance this has to anything. Where did this come from and what does it mean? I'm no math wizard, but sometimes I like to pretend that I am, and if I knew what the hell you were talking about, that would help further my pursuit of witchcraft and wizardry. Harry Potter out.

[nate1729]

David --

What motivates these conjectures? They "ring true" to me in the sense that they strike my mathematician's instinct as meaningful and not extremely likely (were I to lay odds) to be false; I do not, however, know enough of the technical material at the bottom of Wiles' proof to know if these conjectures would arise naturally from a careful consideration of it. (The reason I ask is that they also seem like they might have arisen from just idle thinking about Fermat's conjecture.)

And if you're trying to prove these and there's a lemma or two a guy with a BA in math from a good university (I studied mostly algebra and analysis) might be able to sink his teeth into, please PM me.

--Nate

[kurosh]

Ok? What's your point? Do you want someone to try and prove/disprove it?

[BarkingMad]

I'm no mathmetician. So, maybe some of the criteria flew over my head, but I plugged the formula into Excel, inserted some simple numbers for the A-B-C values, and used solver to find "Q".

For N = 5
A = 5
B = 4
C = 4
Q = 3125

5^5 + 4^5 = 4149

4^5 + 3125 = 4149

Again, maybe some of the constraints of the "conjecture" flew over my head. If I understood it correctly though, it seems like there are many Q's for which the conjecture would hold true.

[kurosh]

You misunderstood.

[Sponger15SB]

[censored] I was totally gonna make this exact post and you beat me to it!

[TimTimSalabim]

A, B and C have to be distinct values. If you make B = C, then they both drop out of the equation and there's no point to it.

[Stork]

[ QUOTE ]
, yet no proof of this fact is even theoretically findable.

[/ QUOTE ]

Can you prove this?

[KingMarc]

Aren't these the million dollar problems?

[BarkingMad]

[ QUOTE ]
A, B and C have to be distinct values. If you make B = C, then they both drop out of the equation and there's no point to it.

[/ QUOTE ]

OK,

Let...

A = 8
B = 7
C = 6
N = 5

and

Q = 41799

8^5 = 32768
7^5 = 16807

32768 + 16807 = 49575

49575 - 6^5 = 41799

I guess I'm working the problem backward by assigning values to A B C & N then finding for Q. It also has occured to me that (as suggested by DS) maybe there are very few q's, and I'm just finding them easily by going backward.

Of course, it is also possible that I'm messed up like a chowder sandwich, and I'm just not smart/educated enough to see what makes this conjecture intriguing, or what it is that should be difficult about it.

[brassnuts]

[ QUOTE ]
I'm just not smart/educated enough to see what makes this conjecture intriguing, or what it is that should be difficult about it.


[/ QUOTE ]

Yeah, could someone at least point us in the right direction?

[BarkingMad]

After doing some googling, i get it now... I think.

The Sklansky - Fermat conjectures look like Fermats Last Theorem with q as an added twist.

I read about this here...

Pierre de Fermat (http://scienceworld.wolfram.com/biography/Fermat.html)

According to the link, some wicked smart dude named A. Wiles proved Fermat's Last Theorem in 1995.

Can the Sklansky - Fermat conjecture be proved as well? I have no idea. I'll leave that to the Good Will Hunting's of the world.

[RocketManJames]

[ QUOTE ]
Yeah, could someone at least point us in the right direction?

[/ QUOTE ]

Let me describe how I am reading what Sklansky has said, and maybe this clears it up.

Fermat's Theorem:

X^n + Y^n = Z^n where n > 2

Fermat says there cannot exist integers X, Y, and Z that satisfy this equation. As far as I know, there is no constraint that X, Y, and Z must be distinct. There is a constraint that n must be greater than 2, otherwise the theorem is obviously false... look at pythagorean triples.

Sklansky-Fermat Conjecture 1:

A^n + B^n = C^n + Q where n > 4

There exist some values of Q where the above equation cannot have integer solutions for A, B, and C.

Sklansky-Fermat Conjecture 2:

If you find such a Q where no integer solutions for A, B, and C exist, then it is possible that there is no way to formally prove it.

Hopefully this clears up what Sklansky is trying to say. And, Mr. Sklansky, if I misunderstood, please correct my understanding of your problem.

-RMJ

[AgentBishop]

[ QUOTE ]
Conjecture One: A to the nth plus B to the nth (when n is an integer, five or greater) cannot equal equal C to the nth plus q, for some if not most q's.

Conjecture Two: If there are in fact q's for which the conjecture holds, some will be formally unprovable. In other words it might be true that (A to the n) + (B to the n) can never equal (C to the n) plus (lets just say) the number 846879032 (n greater than four), yet no proof of this fact is even theoretically findable.

[/ QUOTE ]

Poker must bore the piss out of you.

[Hiding]

this one wants me to log into lagrange.edu too??? Other than that q=pi+ln2 /images/graemlins/grin.gif

[Hiding]

[ QUOTE ]
findable.

[/ QUOTE ] Secondly, is that a word? (I don't know if secondly is either though)

[AlienCorpse]

[ QUOTE ]
Conjecture One: A to the nth plus B to the nth (when n is an integer, five or greater) cannot equal equal C to the nth plus q, for some if not most q's.

Conjecture Two: If there are in fact q's for which the conjecture holds, some will be formally unprovable. In other words it might be true that (A to the n) + (B to the n) can never equal (C to the n) plus (lets just say) the number 846879032 (n greater than four), yet no proof of this fact is even theoretically findable.

[/ QUOTE ]

my head just asplode

[Zeno]

Submit a paper.

Two suggested journals:

Mathematica Journal (http://www.mathematica-journal.com/submissions/)

Journal of Interger Sequences (http://www.cs.uwaterloo.ca/journals/JIS/)

There are many others to choose from.

Teaser article Newton, Fermat, and Exactly Realizable Sequences (http://www.cs.uwaterloo.ca/journals/JIS/VOL8/Huang/huang11.pdf)


-Zeno

[pzhon]

[ QUOTE ]
Conjecture One: A to the nth plus B to the nth (when n is an integer, five or greater) cannot equal equal C to the nth plus q, for some if not most q's.

[/ QUOTE ]
I have a few comments about this conjecture.

/images/graemlins/diamond.gif This is easy to prove for many values of q and n. For example, it is easy to prove that a^6 + b^6 = c^6 + 3 has no solutions, since every 6th power is of the form 7k or 7k+1, so when divided by 7, the left hand side would leave remainder 0, 1, or 2, while the right hand side would leave remainder 3 or 4.

/images/graemlins/diamond.gif This is a generalization of Fermat's Last Theorem, but not in a direction that looks particulaly promising from the perspective of modern algebraic and analytic number theory. If you are interested in generalizations of Fermat's Last Theorem with more connections to deep parts of mathematics, see the ABC Conjecture (http://www.math.unicaen.fr/~nitaj/abc.html):

For every epsilon greater than 0, there exists some k(epsilon)>0 so that for any positive integers A, B, C satisfying gcd(A,B)=1 and A+B=C,

C < k(epsilon) squarefree(A,B,C)^(1+epsilon),

where squarefree(A,B,C) is the product of the prime factors of A, B, and C (with repetition removed).

/images/graemlins/diamond.gif Rather than conjecture specifics about particular values of q, perhaps it would be more interesting to conjecture that for any q and n, there are at most finitely many triples (a,b,c) satisfying a^n+b^n=c^n+q.

I'm not a number theorist, and it could be that both the specific cases and the general finiteness conjecture are settled.

[ QUOTE ]
Conjecture Two: If there are in fact q's for which the conjecture holds, some will be formally unprovable.

[/ QUOTE ]
That doesn't look like a conjecture. That looks like a guess. I'd bet $1k against it, even money.

[David Sklansky]

Conjecture Two: If there are in fact q's for which the conjecture holds, some will be formally unprovable.


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That doesn't look like a conjecture. That looks like a guess. I'd bet $1k against it, even money.


My point is that I think there are qs for which the conjecture holds for no logical "reason" other than "sparseness".

[mybutthurts]

A+b is what n gd conjecture?
im a run of the mill player can someone tell me what this is in very simple idiot english

[Rotating Rabbit]

[ QUOTE ]
Conjecture One: A to the nth plus B to the nth (when n is an integer, five or greater) cannot equal equal C to the nth plus q, for some if not most q's.

Conjecture Two: If there are in fact q's for which the conjecture holds, some will be formally unprovable. In other words it might be true that (A to the n) + (B to the n) can never equal (C to the n) plus (lets just say) the number 846879032 (n greater than four), yet no proof of this fact is even theoretically findable.

[/ QUOTE ]

Have you ever had any formal mathematical training because it doesnt look like it; this is complete nonsense.

[drudman]

Is the basic idea: pick an integer q, and if it does satisfy (a^n)+(b^n)=(c^n)+q, there's no way to prove it?.

??

[EliteNinja]

What is the practical purpose of discovering q?
(I'm an engineering student)

Can you build anything useful from this concept?

[PairTheBoard]

[ QUOTE ]
Conjecture Two: If there are in fact q's for which the conjecture holds, some will be formally unprovable.


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


That doesn't look like a conjecture. That looks like a guess. I'd bet $1k against it, even money.


My point is that I think there are qs for which the conjecture holds for no logical "reason" other than "sparseness".

[/ QUOTE ]

It might be much easier to show that such q's exist than to actually find them.

PairTheBoard

[ToneLoc]

Please fix your title to "Sklansky conjecture".
This is still a bit pompous, but shows a bit more respect to a great, dead mathematician, who might object to people using his name just for adding a twist to the original conjecture.

J.

[wmspringer]

Just thought I'd throw in - if anyone's interested in the general form (Fermat's Last Theorem), Simon Singh's book Fermat's Enigma is an excellent, relatively nonmathematical writeup.

[David Sklansky]

I was trying to introduce my idea about why I believe some number theory conjectures are true for no "reason".

[drudman]

Isn't this just a consequence of what Godel showed?

I honestly have no idea, just wondering. Could be totally off.

[Mano]

Wasn't this proven by Andrew Wiles?

[pzhon]

[ QUOTE ]
Isn't this just a consequence of what Godel showed?


[/ QUOTE ]
http://www.math.hawaii.edu/~dale/godel/godel.html has a nice exposition of what Godel showed in his incompleteness theorems.

Godel proved that in any sufficiently robust logical system with a computable set of axioms, there are either statements with no counterexample that are not provable, or everything is provable, including false statements. Basic logic/sets + arithmetic is an example of a sufficiently complicated system, so either everything is inconsistent, or there are statements in arithmetic that are true, but can't be proven.

There are statements one might care about that are examples of Godel's incompleteness theorem. See this relatively accessible collection of slides from a lecture: http://www.maths.bris.ac.uk/~maadb/research/seminars/bur1999sem02/online/ . That gives two examples of true, interesting statements that are not provable in "Peano Arithmetic," a partial axiomatization of number theory.

Now Sklansky suggests that if a^n+b^n=c^n+q has no solutions, this might be unprovable. I don't buy it at all. It's conceivable, but I would bet against it.

There are many algebraic techniques which apply to this type of equation, such as extending the rationals by adjoining an nth root of q, and factoring both sides. There are numerous results on sums of nth powers such as Waring's problem. There are many similar problems where you can typically prove that the number of solutions is finite, then get an effective bound on the size of the solutions, then exhaustively check the possibilities up to that bound, finding a few sporadic solutions which seem to exist for no particular reason. (Which perfect powers of positive integers differ by 1? 2^3 and 3^2.) In other situations, like the solutions to y^2=x^3+k, k>0, it is known that there are finitely many integer solutions, but there is a rich struction on the hierarchy of rational solutions, and solutions in extensions of the rationals. The sporadic-seeming integer solutions are meaningful.

[pzhon]

Oops, I deleted my exposition of Hibert's 10th, but forgot to change the title back.

In short, Hilbert's 10th problem was to find an algorithm for determining whether a Diophantine equation has solutions in integers. This was solved in the negative direction: There are families of Diophantine equations for which no algorithm determines whether they have solutions. However, I doubt a^n+b^n=c^n+q is one of those.

[jediael]

1)

Proof:
Choose C s.t. C^n < A^n + B^n
Put q = A^n + B^n - C^n

Then A^n + B^n = C^n + q

So there exists A, B, C, q for which the assertion is true.

Conversely, fix any C^n. Pick q = C*C^n. Then for any A^n, B^n we have A^n + B^n = C^n + q i.e. A^n + B^n = C^(n+1) i.e. A^n + B^n = (C²)^n which cannot have a solution due to Fermat's Theorem.

So for each C, there exists a q for which the assertion is false.

This proves Conjecture One, i.e. that for some but not for all q, the equation A^n + B^n = C^n + q does not have a solution.

[pzhon]

[ QUOTE ]

This proves Conjecture One, i.e. that for some but not for all q, the equation A^n + B^n = C^n + q does not have a solution.

[/ QUOTE ]
Just to be clear, jediael was trolling. He gave no proof, only gibberish including shuffling quantifiers (every person has a mother, so some woman is the mother of every person) and asserting that C^(n+1)=(C^2)^n.

[David Sklansky]

Please don't do a lot of work on this subject. I had a deeper point and was trying to introduce the subject with these "conjectures". I'll do it differently and better when I'm not as busy.

[Orpheus]

Point of clarification:

You say
[ QUOTE ]
Conjecture One: A to the nth plus B to the nth (when n is an integer, five or greater) cannot equal equal C to the nth plus q, for some if not most q's.

Conjecture Two: If there are in fact q's for which the conjecture holds, some will be formally unprovable. In other words it might be true that (A to the n) + (B to the n) can never equal (C to the n) plus (lets just say) the number 846879032 (n greater than four), yet no proof of this fact is even theoretically findable.


[/ QUOTE ]
Do you mean "for some integer Q and ALL integers N>5" or "SOME integers q,N where N>5"?

Many of the posters thus far seem to have read the problem differently than I did and are attacking specific N,Q values. A single counterexample would have disproved Fermat's Last Theorem, because he stated that NO solutions exist for integer A,B,C,N>2. Your Conjecture #1, however seems to be saying that some Q DOES exist (for all integers A, B C, N>5), which can be proved by a single example of Q, but cannot be disproved by elimination (except, through, say induction or elimination by class of all candidates)

It seems to me (in my sleep deprived state) that there may be a simple inductive proof that no single Q>I (where I is a small integer) can work for ALL N>5. I'll see if my logic still makes sense to me this weekend (How many millions of people have thought they had inductive proofs of FLT at 5 am in the morning?), but it'd really help to know I had the problem right before I attempted it.

Obviously, a disproof of #1, or even a partial disproof for Q>I and a sufficiently small I, might defeat Conjecture #2 before it gets out of the gate

Dammit, Jim, I'm a doctor (among other things), not a mathematician.

[DMBFan23]

we really need the morgan o malley gimmick accout to show up here

[AlphaWice]

[ QUOTE ]
1)

Proof:
Choose C s.t. C^n < A^n + B^n
Put q = A^n + B^n - C^n

Then A^n + B^n = C^n + q

So there exists A, B, C, q for which the assertion is true.

Conversely, fix any C^n. Pick q = C*C^n. Then for any A^n, B^n we have A^n + B^n = C^n + q i.e. A^n + B^n = C^(n+1) i.e. A^n + B^n = (C²)^n which cannot have a solution due to Fermat's Theorem.

So for each C, there exists a q for which the assertion is false.

This proves Conjecture One, i.e. that for some but not for all q, the equation A^n + B^n = C^n + q does not have a solution.

[/ QUOTE ]

(c^2)^n does not equal c^(n+1).

[Abednego]

You're just trying to find out if I am really Andrew Wiles aren't you?

[gaming_mouse]

[ QUOTE ]
In other words it might be true that (A to the n) + (B to the n) can never equal (C to the n) plus (lets just say) the number 846879032 (n greater than four), yet no proof of this fact is even theoretically findable.


[/ QUOTE ]

This idea of truth without proof has always seemed like a contradiction to me.

I thought the whole point of axiomatic mathematics was that you replace the notion of truth with the notion of provability. That is, by definition, what is true is what can be proved.

[SheetWise]

Alice laughed. "There's no use in trying," she said: "one can't believe impossible things."

"I daresay you haven't had much practice," said the Queen. "When I was your age, I always did it for half-an-hour a day. Why, sometimes I've believed as many as six impossible things before breakfast."

SCOTUS

[gaming_mouse]

[ QUOTE ]

This idea of truth without proof has always seemed like a contradiction to me.

I thought the whole point of axiomatic mathematics was that you replace the notion of truth with the notion of provability. That is, by definition, what is true is what can be proved.

[/ QUOTE ]

Btw, to clarify this, I don't mean that this is my own personal notion of truth. It is just my understanding of the term "truth" as it used in mathemetical contexts.

[jason1990]

I'm no logician, but my understanding is that the notion of "truth" transcends provability. In my possibly misguided perception of things, I imagine it having something to do with the fact that proofs must end. So, for instance, it may be true that statement P(n) holds for all n, but the only "proof" would be to check P(1),P(2), and so on. So although you could never prove it (in finite time), it is still true in the sense that you can never find a counterexample.