[b]Fermat's last theorem[/b] (event

[AndrewMozley]

Fermat's last theorem (eventually proved) states that :

[b]No three positive integers [b]a, b, and c[/b] satisfy the equation :
     a[sup]n[/sup] + b[sup]n[/sup] = c[sup]n[/sup] 
for any integer value of n greater than 2[/b]

Consider the equation a[sup]k[/sup] + b[sup]l[/sup] = b[sup]m[/sup]
. .. where a, b, c, k, l, m are all positive integers and k, l, m (may be the same) are all greater then 2

Is there a solution to this equation where a, b, c have no common divisor (greater than 1)?

 

[mikrom]

There are known solutions, when (a, b, c) have a common divisor - e.g:
2[sup]n[/sup] + 2[sup]n[/sup] = 2[sup]n+1[/sup]

There are known solutions, when one of the exponents is equal to 2 - e.g.:
1[sup]n[/sup] + 2[sup]3[/sup] = 3[sup]2[/sup]
2[sup]5[/sup] + 7[sup]2[/sup] = 3[sup]4[/sup]

But, when (a, b, c) have no common divisor and all exponents (k, l, m) are equal or greater 3, then there is not a solution. There is a mathematically based conjecture about this - see:

https://en.wikipedia.org/wiki/Beal_conjecture

 

[AndrewMozley]

Thank you very much, Mikrom, for your helpful and polite reply. And there I was, thinking that I was the first person to consider this conjecture . . . .

Andrew

 

[mikrom]

Here is very interesting article by Peter Norwig about trying to find counterexamples to Beal's Conjecture:

http://norvig.com/beal.html

 

[karluk]

Interesting. Unfortunately I don't think the conjecture is amenable to the efforts of amateur mathematicians like me. My guess is that the conjecture is true, so it probably won't be productive to search for a counterexample, and attempts to prove the conjecture involve techniques that are way beyond my level of competence. Take a look, for example, at the abstract of the paper that found all of the primitive solutions of x[sup]2[/sup] + y[sup]3[/sup] = z[sup]7[/sup].

We find the primitive integer solutions to x[sup]2[/sup] + y[sup]3[/sup] = z[sup]7[/sup]. A nonabelian descent
argument involving the simple group of order 168 reduces the problem to the determination
of the set of rational points on a finite set of twists of the Klein quartic curve X. To restrict
the set of relevant twists, we exploit the isomorphism between X and the modular curve
X(7), and use modularity of elliptic curves and level lowering. This leaves 10 genus-3 curves,
whose rational points are found by a combination of methods.

By way of reference, I personally was aware that the alternating group of degree five is the smallest nonabelian simple group, but I had no idea that the second smallest has order 168, nor how to use it to prove the result in the paper.

 

[mikrom]

The question is, how it is exactly formulated .. i.e.
either
1) a, b, c have no common divisor
or
2) a, b, c have pairwise no common divisor

For the 1. case there seems to be counterexamples - see here:

https://www.scipress.com/BMSA.12.39.pdf

 

[karluk]

I hope I'm not misunderstanding your link, but counterexample 1 is

[2[sup]88[/sup] + 9999999999999[sup]3[/sup]] = 10[sup]39[/sup]

The left hand side of this equation is the sum of an even number plus an odd number, hence odd. The right hand side is even, so I fail to see how equality holds. Is the author, S. Saravanan, using brackets [] on the left hand side to mean something other than parenthesis?

 

[mikrom]

It looks like the result was achieved when calculating with floating point numbers, for example calculator on Linux calculates it the same way - see the screenshot.
But that is of course not a correct result and not a counter-example for the conjecture.

 

[mikrom]

probably in the article mentioned, the brackets [] would mean approximately.

computed with python:

>>> 2**88
309485009821345068724781056

>>> 9999999999999**3
999999999999700000000000029999999999999

>>> 2**88 + 9999999999999**3
1000000000000009485009821375068724781055

>>> (2**88 + 9999999999999**3) / 10**39
1.0000000000000095

 

[AndrewMozley]

I see that scipress.com claims to review articles (such as Mr Saravanan’s article) before publication. But all four counter-examples seem to start with an untrue arithmetic statement; so unless we accept that square brackets imply that the following ‘=’ sign means ‘ is approximately equal to ‘, the review process seems to be pretty minimal !

But I hope I am not moving into murky waters. - Andrew

 

[mikrom]

But, nonetheless, I find the approximations interesting. I would be particularly interested in the procedure, how did the author come up with the approximate solutions. Whether he wrote a computer program for this, or came up with the estimation of the results in some other way.

 

[karluk]

I see that scipress.com claims to review articles (such as Mr Saravanan’s article) before publication.

No reasonable review process would have passed this article. Either the arithmetic is simply wrong, or there is some sort of implied approximation going on. But if the math is approximate, then the examples are not counterexamples to the Beal conjecture and the article should have been rejected for not showing what it claimed.

I doubt that the author has any math skills at all that go beyond the ability to punch numbers into a calculator. All of the calculations go up to and beyond the limits of a calculator's precision, and so may have incorrectly convinced S. Saravanan that his numbers were showing equality and he was about to collect the $1 million prize for finding a counterexample to the Beal conjecture.

 

[mikrom]

But if the math is approximate, then the examples are not counterexamples to the Beal conjecture and the article should have been rejected for not showing what it claimed.

The headline "BEAL'S CONJECTURE-COUNTER EXAMPLES" is like most of these days marketing - to fool people.

 

[Chris Miller]

There's an unrelated conjecture. Not with a million dollar pledge.

The Collatz conjecture. It's also a pita for maths professors getting proposed solutions they are asked to check and confirm.

The conjecture itself goes like that:
Given any natural number n you have two rules to determine a successor:
a) if n is even, the next number in the series will be n/2
b) if n is odd, the next number in the series is 3n + 1

The actual conjecture is that starting with any natural number n you arrive at 1 or continuing to apply the rule you end in a repeating loop of 1->4->2->1.

Nobody could prove whether either all numbers eventually end there or there is a counter-example of a loop repeating somewhere on a high number never getting down to 1.

What's interesting about it is, that it's a delicate definition, if you vary the rule for odds to 3n-1, it has a simple counter-example: 7->20->10->5->14->7... never reaching 1. Generalization to an+b also didn't help find patterns and solve the conjecture more generally.

It's checked for very high starting numbers and it's easy to program it, you have an easy way of cutting calculations down, if you just keep in mind (or a variable) that you proved it for 1 to n, it's sufficient to only calculate successors of n+1 until you reach something lower than n+1, as that's already checked.

The simplicity and still yet unpredictability of this number series makes it so fascinating, on the other side it's not important to prove or disprove it, nothing is conjectured based on the assumption of this conjecture being true or false, it's just playing with number theory.

Chriss