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 February 16th, 2018, 12:05 PM #1 Member   Joined: Aug 2015 From: Chiddingfold, Surrey Posts: 57 Thanks: 3 Math Focus: Number theory, Applied maths A Novel attempt to find a simple proof of F.L.T A NOVEL ATTEMPT TO FIND FERMAT’S PROOF I have realized something that seems to have not been considered before, in the attempts to find a simple proof of Fermat’s “LAST” theorem and which could lead experts finding a simple proof. My attempt is based on the fact that; unlike the power two case; in every odd power, both the sum and difference of two numbers raised to any power is divisible by the sum or difference of the two numbers. Therefore, there are seven criteria that must exist in any set of three integers A B and C (A>B>C and with no common factors). These are: (A^p – B^p) is divisible by C, (A^p - C^p )is divisible by B and (B^p + C^p ) is divisible by A and A^p is divisible by (B + C) and B^p is divisible (A-C) and C^p is divisible by (A-B) and one must be even; so is it possible for all seven criteria to apply to some sets of three numbers? Trials with a QBASIC program (below) show that the only way this can happen is when A - B – C = 0, and this applies to all odd powers including unity. Obviously this applies only to unit power and although the seven criteria are met with higher powers, the result is that A^p -B^p -C^p > 0. These trials were with no common factors between any two of A B and C and with one only and only one being even. I've checked with A up to 300 in power 3. Other powers limited because of overflow: - up to 70 in power 5 , up to 15 in powers 7 and 9. I leave it to the experts to see if my results can be explained other than that Fermat’s last theorem is true. Terence Coates HNC Electrical Engineering CLS INPUT "enter power ", p# INPUT " A minimum, A maximum ", min, max FOR A# = min TO max 3 FOR B# = 2 TO A# - 1 FOR C# = 1 TO B# - 1 Mbc# = B# N# = C# ' See if there is a common factor WHILE N# <> 0 R# = Mbc# MOD N# Mbc# = N# N# = R# WEND Mab# = A# N# = B# WHILE N# <> 0 R# = Mab# MOD N# Mab# = N# N# = R# WEND Mac# = A# N# = C# WHILE N# <> 0 R# = Mac# MOD N# Mac# = N# N# = R# WEND ' Skip if there is a common factor, M# is the highest common factor ' and if A,B and C are all odd. IF Mbc# = 1 AND Mab# = 1 AND Mac# = 1 THEN U# = A# + B# + C# ' Test all relevant Modulus = 0 S# = A# ^ p# MOD (B# + C#) + C# ^ p# MOD (A# - B#) + B# ^ p# MOD (A# - C#) + U# MOD 2 + (B# ^ p# + C# ^ p#) MOD A# + (A# ^ p# - B# ^ p#) MOD C# + (A# ^ p# - C# ^ p#) MOD B# MOD C# IF S# = 0 THEN WRITE A#, B#, C#, S#, T ' SLEEP END IF END IF NEXT C# NEXT B# NEXT A#  February 16th, 2018, 01:38 PM #2 Senior Member   Joined: Aug 2012 Posts: 2,386 Thanks: 746 It's been noted that any "proof" of FLT using basic arithmetic properties of the integers must be wrong, because such a proof would also be valid in the p-adics. But FLT is false in the p-adics. See nguyen quang do's answer in this thread. https://math.stackexchange.com/quest...m-simple-proof Thanks from topsquark and SDK February 16th, 2018, 01:45 PM #3 Senior Member   Joined: Sep 2016 From: USA Posts: 642 Thanks: 406 Math Focus: Dynamical systems, analytic function theory, numerics It has been proved that a solution to FLT in this manner is impossible. Namely, because your criteria can be restated as required solutions to satisfy certain congruences modulo $A,B,C$ etc. Then this becomes equivalent to the claim that FLT should hold over finite fields but this is known to be false. In particular, for every sufficiently large prime $p$, there exists a nontrivial solution to $x^n + y^n = z^n$ where the minimum size of $p$ depends only on $n$. It follows that for every $n$, FLT fails over infinitely many finite fields. It is for this reason that elementary algebraic manipulation can never lead to a proof for FLT since such a proof would also hold modulo $p$ for any $p$ which we know is false. Please note that this is not a case of "mathematicians just don't know". It is the opposite, it is a mathematical fact that congruence/algebraic methods can never yield a proof because such a proof would "prove" infinitely many other statements for which counterexamples are known. See the following for a more detailed explanation. https://en.wikipedia.org/wiki/Hensel's_lemma Thanks from topsquark February 16th, 2018, 02:18 PM   #4
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 Originally Posted by magicterry . . . although the seven criteria are met with higher powers, the result is that A^p -B^p -C^p > 0.
Doesn't that suggest that the criteria can easily be met? You would need to find some connection between meeting them and avoiding A^p - B^p - C^p > 0, but there's no apparent reason to suppose that might be done in any simple way for arbitrary odd values of p.

As a first step, you might look up the simplest proof of the theorem for p = 3 and judge for yourself whether you would call the methods used "simple". February 16th, 2018, 06:48 PM #5 Senior Member   Joined: May 2016 From: USA Posts: 1,310 Thanks: 551 I must have misread this. Did someone just allege that something about the infinitude of numbers has been proved by looking at the first 300 of that infinity? A statistician would say that 300 out of infinity is a sample size that is infinitesimal, particularly given that the sample is NOT random. Reminds me of the old joke about engineers: 3 is prime, 5 is prime, and 7 is prime: therefore all odd numbers greater than 1 are prime. (There is a joke both more bitter and more realistic about primes and accountants.) February 16th, 2018, 08:47 PM   #6
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 Originally Posted by JeffM1 Reminds me of the old joke about engineers: 3 is prime, 5 is prime, and 7 is prime: therefore all odd numbers greater than 1 are prime. (There is a joke both more bitter and more realistic about primes and accountants.)
There's a whole collection of such jokes: https://rationalwiki.org/wiki/Fun:Pr...bers_are_prime February 16th, 2018, 09:56 PM #7 Senior Member   Joined: May 2013 Posts: 118 Thanks: 10 Magicterry Your idea isn't new.Some members in this forum have unsuccesfully attempted to show A-B-C=0. See,for example,the thread started the 6th of Septempber 2015 under the name "An elementary proof of the Fermat's last theorem" I've been trying this for a while,but each time a wall rose before me February 17th, 2018, 08:45 AM #8 Math Team   Joined: May 2013 From: The Astral plane Posts: 2,266 Thanks: 934 Math Focus: Wibbly wobbly timey-wimey stuff. I just got back from a (rather long) break. Another one of these? @magicterry, if you use the search system (under FLT) you will find a rather large number of other posts addressing this issue. Perhaps reading them will guide you in a more profitable direction. -Dan Thanks from greg1313 March 5th, 2018, 08:20 AM   #9
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Math Focus: Number theory, Applied maths Further research I have added another condition to be excluded. i.e, B + C must be larger than A.
To enable higher powers to be included I have changed A^p MOD (B +C) to (A MOD (B + C) )^p.

The number of sets of A,B,C out of a possible 4455100 with A up to 300 that meet all the conditions is only 52. That is 20 up to power 3 with another 16 up to power 5, another 8 up to power 7, another 8 up to power 9 and no more to a higher power. It is noted that most sets have A - B = 1, and the A value generally only occurs once, twice occasionally. Attached is a spread sheet showing my results up to A = 300.
I notice that with any pair of A and B, the form of (A^p -B^p)^(1/p) plotted against p resembles a logarithmic curve.

Since this subject involves three integers it is interesting to relate these to a triangle. With power 2 of course it is a right angle triangle where the cosine of 90 is zero, and the other corners have rational cosines but irrational angles.
With power unity the triangle collapses to a straight line with cosines of +/-1.
For higher powers and all three sides as integers, then all three corners have rational cosines , but all with irrational angles.
So is this why higher powers have more than three integers?
Perhaps with say power 4 and 158^4 +59^4 = 134^4+133^4 we could represent this with a four sided shape. We could then make one corner have a rational angle of +/-90 or +/ 30 or +/- 150 with rational cosines. But we would have no choice of angles for the other three corners.
There seems to be a similarity between the minimum number of integers to satisfy each power, and several other mathematical subjects.
1. The number of terms in a binomial expansion
2. The amount of required to determine the coefficients in a curve of a given
order.
3. The number of equally spaced ordinates required to calculate the area
under a curve of a given order.
Attached Images scan0003.jpg (95.0 KB, 3 views) March 5th, 2018, 04:26 PM   #10
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Quote:
 Originally Posted by magicterry  I have added another condition to be excluded. i.e, B + C must be larger than A. To enable higher powers to be included I have changed A^p MOD (B +C) to (A MOD (B + C) )^p. The number of sets of A,B,C out of a possible 4455100 with A up to 300 that meet all the conditions is only 52. That is 20 up to power 3 with another 16 up to power 5, another 8 up to power 7, another 8 up to power 9 and no more to a higher power. It is noted that most sets have A - B = 1, and the A value generally only occurs once, twice occasionally. Attached is a spread sheet showing my results up to A = 300. I notice that with any pair of A and B, the form of (A^p -B^p)^(1/p) plotted against p resembles a logarithmic curve. Since this subject involves three integers it is interesting to relate these to a triangle. With power 2 of course it is a right angle triangle where the cosine of 90 is zero, and the other corners have rational cosines but irrational angles. With power unity the triangle collapses to a straight line with cosines of +/-1. For higher powers and all three sides as integers, then all three corners have rational cosines , but all with irrational angles. So is this why higher powers have more than three integers? Perhaps with say power 4 and 158^4 +59^4 = 134^4+133^4 we could represent this with a four sided shape. We could then make one corner have a rational angle of +/-90 or +/ 30 or +/- 150 with rational cosines. But we would have no choice of angles for the other three corners. There seems to be a similarity between the minimum number of integers to satisfy each power, and several other mathematical subjects. 1. The number of terms in a binomial expansion 2. The amount of required to determine the coefficients in a curve of a given order. 3. The number of equally spaced ordinates required to calculate the area under a curve of a given order.
Let me reiterate in the hopes it penetrates the crazy. It is a mathematical THEOREM that no proof using the methods you are describing is possible. Tags attempt, find, flt, proof, simple Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post relaxursoul Number Theory 0 November 14th, 2017 12:20 PM tejolson Real Analysis 1 March 17th, 2013 07:54 PM Math4dummy Calculus 3 February 22nd, 2012 08:56 PM Riazy Calculus 2 January 21st, 2011 08:45 AM Riazy Calculus 1 January 15th, 2011 05:23 PM

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