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

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

(OP)

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

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No three positive integersa, b, and csatisfy the equation : a^{n}+ b^{n}= c^{n}for any integer value of n greater than 2

Consider the equation a

^{k}+ b

^{l}= b

^{m}

. .. 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)?

## RE: [b]Fermat's last theorem[/b] (event

2

^{n}+ 2^{n}= 2^{n+1}There are known solutions, when one of the exponents is equal to 2 - e.g.:

1

^{n}+ 2^{3}= 3^{2}2

^{5}+ 7^{2}= 3^{4}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

## RE: [b]Fermat's last theorem[/b] (event

Andrew

## RE: [b]Fermat's last theorem[/b] (event

http://norvig.com/beal.html

## RE: [b]Fermat's last theorem[/b] (event

^{2}+ y^{3}= z^{7}.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.

## RE: [b]Fermat's last theorem[/b] (event

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

## RE: [b]Fermat's last theorem[/b] (event

[2

^{88}+ 9999999999999^{3}] = 10^{39}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?

## RE: [b]Fermat's last theorem[/b] (event

But that is of course not a correct result and not a counter-example for the conjecture.

## RE: [b]Fermat's last theorem[/b] (event

approximately.computed with python:

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## RE: [b]Fermat's last theorem[/b] (event

scipress.comclaims 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

## RE: [b]Fermat's last theorem[/b] (event

## RE: [b]Fermat's last theorem[/b] (event

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.

## RE: [b]Fermat's last theorem[/b] (event

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

## RE: [b]Fermat's last theorem[/b] (event

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