Fermat’s Last Theorem states: There are no three positive integers A, B, and C satisfy the equation Aⁿ + Bⁿ = Cⁿ for any integer value of n greater than 2.

ABSTRACT

Fermat’s assertion that Cⁿ≠Aⁿ+Bⁿ is proven by the application of Pascal’s Triangle.

Fermat’s Last Theorem (FTL) can be proven through examination of the coefficients resulting from (A+B)ⁿ, and then apply the principals of Pascal’s Triangle where the sum of coefficients must equal 2ⁿ. Through the examination of the coefficients of binomial fragments resulting in (A+B)ⁿ-Aⁿ=Bⁿ or (A+B)ⁿ-Bⁿ=Aⁿ and applying the defined in properties of Pascal’s triangle to that fragment, it is proof that Fermat was correct in his assertion. The sum of the coefficients from (A+B)ⁿ-Aⁿ=x, or (A+B)ⁿ-Bⁿ=y, x and y can be proven to always have 2ⁿ-1 as the sum of the coefficients. 

Pascal’s triangle demonstrates the sum of the binomial coefficients of any (A+B)ⁿ will always be a member of the set {2ⁿ}. When Aⁿ or Bⁿ, each with a coefficient of 1, is subtracted from the resulting expansion of (A+B)ⁿ, the resulting difference of the coefficients will equal {2}ⁿ-1 and no longer a member of Pascal’s assertion. The sum of the coefficients of this fragment is now always an odd numbers and can never be a member of {2ⁿ}.

Fermat is proven correct in his theory.

Key Words: 1; Fermat; 2, FTL; 3, Pascal

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1        Introduction

FTL has been the catalyst for some of the most important math discoveries. In 1996, Sir Andrew Wiles published his proof of Fermat’s Last Theorem. This was a marvel to mathematics and defined new principles to math, as did many previous, but failed attempts. This paper will show a simple proof which existed in Fermat’s time. It would have been a great loss to human understanding of theoretical math had this simple proof been known before Sir Wiles made his discovery. Yet, a simple proof does exist, exactly as Fermat said is should.

2        Fermat and Pascal

To truly appreciate this proof of FTL, a history of  Pierre de Fermat (1607-1665) and Blaise Pascal (1623-1662) is best known. They were both mathematicians living in France in the same times. It is well known they corresponded and met. In a series of letters^[1] between the two from 1654 to 1650, their work together resulted in the Probability Theory.  

It should be no surprise where Fermat would have intimate knowledge of Pascal’s other works as Pascal had sought Fermat’s council in other matters. This would have included Pascal’s works on coefficients, found in the Treatise on the Arithmetical Triangle^[2], resulting in what we now call Pascal’s Triangle. I believe it possible FTL was a spinoff from this work.

3        Chasing A & B

The historical approach to FTL has only been through the manipulation of A and B. Andrew Wiles finally succeeded with this in 1996 and opened a whole new universe of theoretical math. For that reason, Fermat’s Last Theorem has been a landmark in mathematics. The energy spent in the pursuit of FTL’s proof has launched great learning. 

 If what I suspect is true, Fermat has been amused for almost 400+ years. Fermat perhaps expected Pascal to also chase A an B down the rabbit hole. Why not? Mathematicians ever since have done so as it is the most obvious approach. This is the approach all theories and attempts at proofs available to me have used. Even the acknowledged proof offered by Sir Andrew Wiles used this approach.  All the while, the proof lies in Pascal’s works. It is the coefficients and their properties which are the simple proof to FTL. Fermat said his proof was simple. The mathematics for this proof existed in Fermat’s days. To prove this proof wrong would require the two proofs of Pascal’s triangle to be wrong^[3].

4        Exploring binomials where C=A+B

There are some basic truths which will be built on. For any given positive variable greater than 1, there exists two positive numbers which can be added together an equal that variable. The larger the number, the more combinations, or better known as order pair. 

Consider the variable C where C=A+B. There follows C²=(A+B)², C³=(A+B)³, C⁴=(A+B)⁴, Cⁿ=(A+B)ⁿ. 

Next consider the resulting binomial formula which results from the expansion of (A+B)², C³=(A+B) , C⁴=(A+B)⁴, Cⁿ=(A+B)ⁿ. 

(A+B)²= A²+2AB+B²

(A+B)³= A³+3A²B+3AB²+B³

(A+B)⁴= A⁴+4A³B+6A²B²+4AB³+B⁴

5        The obvious path

When looking at the binomial expansion of (A+B)ⁿ as demonstrated above, it lends to the desire to deal with the values of A and B when defining any further functions. In the exploration of FLT, all have followed this path.  This has been the historic path. But there is a less obvious representation which needs to be examined.

6        Examination of FLT, Focus on Coefficients

Further, 1Aⁿ and Bⁿ  are constants in every expansion of (A+B)ⁿ.  And identifying constants in any function is key to solving any proof.  

(A+B)³= 1A³+3A²B+3AB²+1B³

(A+B)⁴=1A⁴+4A³B+6A²B²+4AB³+1B⁴

(A+B)²= 1A²+2AB+1B², the coefficients total to 4, 2².

(A+B)³= 1A³+3A²B+3AB²+1B³, the coefficients total to 8, 2³.

(A+B)⁴= 1A⁴+4A³B+6B²B²+4AB³+1B⁴, the coefficients total to 16, 2⁴.

(A+B)ⁿ= 1Aⁿ+ [Pascal’s center function for each n] +1Bⁿ, the coefficients total to 2ⁿ. 

1Aⁿ and 1Bⁿ are constants for every n and always have a coefficient of 1.

• Pascal asserted “The sum of the coefficients in the nth row of Pascal’s triangle is 2ⁿ.”

There are two published proofs confirming this to be true to this assersion. The coefficients for Pascal’s Triangle are derived from (A+B)ⁿ. This represents every Cⁿ when C=(A+B). In the case for n=6, any two integers can be fed into the formula for A and B.

Case for n=6

C⁶=(A+B)⁶=1A⁶+6A⁵B+15A⁴B²+20A³B³+15A²B⁴+6AB⁵+1B⁶ 

Pascal’s triangle evaluates  the sum of coefficients of every Cⁿ for Cⁿ=(A+B)ⁿ to always be 2ⁿ. 

Pascal’s triangle demonstrates, when reading across any row for n, the coefficients will always sum to a value of 2ⁿ.  The bottom row in the below illustration represents the coefficients for the binomial, (A+B)⁵. The sum of these is 32 or 2⁵. And when vertically, the constant of 1 is present on both the Aⁿ and the Bⁿ side.

n depicts the expansion of (A+B)ⁿ when n=6 and the extraction of the coefficients. It must also be noted the constants 1Aⁿ and 1Bⁿ have the same coefficient, 1, for all values of n. This provides both horizontal and vertical constants for which makes the evaluation of Pascal’s Triangle apply easily to Fermat.

For n=6, {1, 6, 15, 20, 15, 6 and 1} are the coefficients for (A+B)⁶. and they sum to 64 or 2⁶

Cⁿ=(A+B)ⁿ. Below example demonstrates (A+B)ⁿ⁼⁶ when Fermat is (A+B)ⁿ-Aⁿ=Bⁿ.

Consider (A+B)ⁿ-Aⁿ=Bⁿ when (A+B)ⁿ=1A⁶+6A⁵B+15A⁴B²+20A³B³+15A²B⁴+6AB⁵+1B⁶. This is a true statement and will be true for every value of A and B. Now evaluate the coefficients from the resulting fragment on the right side of the equals sign when 1A⁶ is subtracted. 6A⁵B+15A⁴B²+20A³B³+15A²B⁴+6AB⁵+1B⁶. The coefficients are {6 + 15 + 20 + 15 + 6 + 1}, sum to 63, are an odd number and can never be a member of {2ⁿ}. This fragment does not pass Pascal’s test to be a member of { i ⁿ}. 

6 + 15 + 20 + 15 + 6 + 1 = 63, 63≠2⁶

Above are the coefficients after 1B⁶ is subtracted from the expansion of (A+B)⁶.

Pascal proves C⁶-A⁶=B⁶ can not exist.  

Cⁿ=(A+B)ⁿ. Below example demonstrates this reciprocal equation for (A+B)n=⁶. Consider

Consider (A+B)ⁿ-Bⁿ=Aⁿ when (A+B)ⁿ=1A⁶+6A⁵B+15A⁴B²+20A³B³+15A²B⁴+6AB⁵+1B⁶. This is a true statement and will be true for every value of A and B. Now evaluate the coefficients from the resulting fragment on the right side of the equals sign when 1B⁶ is subtracted. 1B6+16A5B+15A⁴B²+20A³B³+15A²B⁴+6AB⁵. The coefficients are {1+6 + 15 + 20 + 15 + 6}, sum to 63, are an odd number and can never be a member of {2ⁿ}. This fragment does not pass Pascal’s test to be a member of { i ⁿ}. 

Above are the coefficients after 1A⁶ is subtracted from the expansion of (A+B)⁶.

Pascal proves C⁶-B⁶=A⁶ can not exist.

7        The Proof Is In The the Coefficients

It is not the pursuit of the values of (A+B)ⁿ as used in  Andrew Wiles proof and in all other attempts I could find. Rather, the proof is simply found in by applying the rules of Pascal to the coefficients of (A+B)ⁿ. The sum of the coefficients of any (A+B)ⁿ must equal 2ⁿ.

• For A or C to be member of {iⁿ}, the sum of the coefficients of the fragments for Cⁿ–^An or Cn-Bⁿ must sum to 2ⁿ. Aⁿ or Bⁿ, when subtracted from Cn each have a coefficient of 1 leaving the resulting fragment a value of 1 leaving an odd number and short of meeting Pascal’s requirement by 1.

[1] https://www.york.ac.uk/depts/maths/histstat/pascal.pdf.  All but the last two letters, from which the Theory of Probability is found, were translated from the French by Professor Vera Sanford, Western Reserve University, Cleveland, Ohio, and appear in A Source Book in Mathematics (ed. D E Smith). The last two were translated by by Maxine Merrington and appear in Games, Gods and Gambling by F N David.

[2] Blaise Pascal’s Treatise on Arithmetical Triangle was written in 1653 and appeared posthumously in 1665.

[3] Two proof exist for Pascal’s Triangle can be found at Canada/USA Mathcamp site: https://www.mathcamp.org/files/yearly/2017/quiz/pascal.pdf.   N