# Fermat's last theorem: a simple general proof ?

by Vladimir F. Tamari, 4-2-8-C26 Komazawa, Setagaya-ku, Tokyo, Japan 154.

Aug 4, 1995

Abstract: A simple general proof is presented to the theorem that the sum of two integers raised to a power larger than 2 cannot be equated to another integer raised to the same power. Using the binomial theorem, it is shown that any number c raised to a power n can be represented as a 1 added to a sum of descending powers of C = (c-l). It is assumed that (a), (b) and (c) are integers, and that B=(b-l) and A=(a-l). It is shown that after equating the expansions of (an+bn) = (1+1+(integral multiples of A and B) = cn = (1+integral multiple of C), an ‘extra’ 1 remains. When the A, B, or C terms ‘absorb’ the 1, they inflate fractionally, rendering them non-integers. This contradiction proves Fermat's last theorem for any n, and the exception when n=2 is explained. How non-integral solutions result from ‘absorbing’ the 1 into the C terms is detailed in Appendix I. Variations on the proof are suggested in Appendix II.  A sequence of powers is presented as a table of differences in Appendix III. A new conjecture, a generalization of Fermat’s theorem to include any case where the number of terms on both sides of the equation is unequal is presented in Appendix IV.

Key Words: Fermat, Last theorem. Lamé ovals. Asymmetrical operations. Differences table.

Pierre de Fermat's 1637 marginalia [1] , [2] stated that

1)      xn + yn = zn

has no integral solutions if n >2. This has been proven in general by T. Wiles [3] using the results of modern number theory. An elementary proof, using the mathematics of Fermat’s own time, is still required. Let (a), (b) and (c) be integers, with A=(a-1),

B=(b-l), C=(c-l):

2) cn = (C+l)n         =    l + {(n 1) C    +(n 2)C2    + .. +  (nn)Cn   }

3) = an + bn = (A+l)n + (B+l)n = 1+ 1 + (n 1)(A+B) + (n 2) (A2+B2) + …+ (n n) An+Bn)

where (n r) =   are the binomial coefficients, the integers of Pascal's triangle with 0! =1. All the A, B and C terms are gathered, and Eqs. (2) and (3) equated. One 1 in Eq.(3) is printed here in bold to emphasize its role in preventing integral solutions.

4a)    an + bn =   1 +    1 + {qA}+{rB}

||                 ||

4b)     = cn   =    1  +         {sC}

(q),(r)and (s) are integers, the result of multiplying (n r) integers with multiples of A , B and C respectively, and adding up the terms. For example when n=3, sC=∑{(3 r)Cr =  C3+3C2+3C. Vertical double lines equate 1=1, and

5)    {qA}+{rB}= {sC} - 1

Eq. (5) can be reduced using an asymmetrical balanced operation: one or more set of the A, B and C variables ‘absorb’ the 1, and each of the absorbing variables is fractionally changed,  becoming non-integers. (Appendix I details this process). But if the A, B or C variables are non-integers. This contradicts the original assumption that (a), (b), and (c) are integers, proving Fermat's last theorem for any (n). For the exception n =2, however, x2 = z2 - y2 = (z+y)(z-y).

This is a simple expression that can be solved without resorting to any binomial methods, allowing an infinity of solutions.Q.E.D.

### Appendix I: Algebraic and numerical demonstration of the proof.

No numerical example of an impossible integral solution for Eq. (l) with n >2 can be given. Instead, a slightly different equation will be used at first, and the exact process of how ‘absorbing’ the 1 into one or more of the variables, say C, inevitably leads to non-integral solutions of Eq. (l) will be analyzed in detail. Using Eq. (4a), but adding 1 to both sides of Eq. (4b) creates a balance between the four 1 terms:

 6a)   4a) an + bn = 1 + 1+ {qA}+{rB} 6b) =1 + cn = 1 + 1+{sC} 7) an + bn = 1 + cn

Since {qA} + {rB} = {sC}, Eq.(7) can have integral solutions. Let the 1 be divided into g increments ∆ and added to each (c), inflating its value to g = ( c + ∆ ), This asymmetrical operation does not affect the balance of Eq. (7), merely simplifies it to:

8)     an + bn = gn

To validate this process, plot Eqs.(7), (8), in the a-b plane, keeping (c) and (g) constant and putting (cn +1) = gn = (c+∆)n : both curves represent an identical Lamé oval [4] (but the length of its half-axes is expressed differently, as g= (cn +1)(1/n) The value of Δ will be a fraction 0<Δ ={( cn +1) (1/n)  - c} < l. And since (c) is an integer and Δ a fraction, (g) is a non-integer. Similar results are obtained when any other number replaces the 1 in Eq. (7). In the great majority of cases even Eq.(7) does not have integral solutions. But it was just shown that when it does, (g) in Eq.(8) always ends up being a non-integer. Starting from a known integral solution to Eq.(7), a=64, b=94, c=103, so that A=63, B=93, C=102:

 a3 + b3 = 1 + c3 643 + 943 = 1 + 1033 262144 + 830584 = 1 + 1092727 (A+1)3 + (B+1)3 = 1 + 1+ 3(A+B) +3(A2+B2) +(A3+B3) 1092728 = 1 + 1+ 468 37854 1054404 = 1 + 1+ {3A+3A2+A3}+{3B+3B2+B3} = 1 + 1+ 262143 + ( 830583 1+(C+l)3 = gn = 1 + 1+ {C3+3C2+3C} 1+ (103)3 = (103.0000314198..)3 = 1 + 1+ 1092726

The process of dividing the 1 into increments Δ and absorbing them into each C is:

Δ= (g - c ) = 0.0000314198...this is confirmed by Δ ={(1033 + l)-3- 103}=0.0000314198..

 1 +c3 = (c+Δ)3 =( C+1 +Δ)3 = 1  +(C+Δ)3 +3(C+Δ)2 +3(C+Δ) = gn 1+1033= (103.0000314198.. )3 = 1+(1061208.9800..) +(31212.01922..) +(306.0000942..) = (103.0000314198..)3

Eqs.(7),(8) can also be put in a form resembling Eq.(5), showing how the variable and its multiplicant both change to non-integers when the 1 is absorbed.

 Qa +Rb = 1  +    Sc = Tg 4096 x 64 +8836 x 94 = 1+10609 x 103 = 10609.0064724964  x  103.0000314198 =1092728

When n=2 and a=21, b=28, c==35, a2=(c+b)(c-b) gives 441=(35+28)(35-28)=63x7=441.

### Appendix II: Variations on the general proof

The following proofs 1. and 2. only differ in the details of the initial approach, so just the changes in the first steps will be given.The final steps are more or less the same as those of the general proof. Proof 3. is incomplete, and is given in more detail as it differs somewhat from the others.

1. Putting cn-bn =(1-1)+ 1+ 1 + {(n 1)(C+B) + (n 2) (C2-B2) + …+ (n n) Cn-Bn)} and comparing the format with the an expansion. A digit 1 has to be added to replace the (1-1). As before, its Δ increments are added to C. Alternatively , either in this case or in Eq.(5), the 1 can be distributed in increments among one or more of the (A), (B) terms (+ΔA and+ΔB) and (C) terms (- Δc).

2. Expanding the integers by setting a=(A-l) , b=(B-l), etc., (rather than a=(A+l), b==(B+l) etc.). This yields a slightly more complicated binomial expansion (with alternating + and - signs).

3. Suppose from the start that an integral solution does exist. Then,

9)      an+bn = cn

10)   ja + kb = mc with (j,k,m) integral multiples of (a,b,c) respectively. Expanding Eq.(9) as before, cancelling the m terms and gathering the variables:

11)  1+JA+KB=MC

With J,K,M integers, integral multiples of (A,B,C) multiplied by the integers (n r) of Pascal's triangle. The three (a), (b) and (c) terms of Eq.(9) have generated the three A, B and C terms of Eq. (11). The 1, an anomaly, must have come from either (a),(b) or (c), and hence must be absorbed by the A and/or B terms. Say the B is chosen.Putting 1 +KB == K(B+Δ) with Δ==1/K:

12)   JA + K(B+A)-==LC

Let β= (B+Δ);  β is a non-integer, since it is the sum of an integer B and a fraction Δ:

13)   JA + Kβ = LC

Equations (10) and (13) have both been derived from Eq.(9) using rational processes. Each equation consist of two terms added to equal a third, and each term consists of a coefficient multiplied by a variable. Each term of one equation is related uniquely to its opposite term in the other equation. Yet in two pairs of terms (the a,A and c,C terms) all the coefficients and variables are integers. But in the third pair (b,β) one of the variables is always a non-integer.Intuitively, this lack of symmetry is a contradiction, showing that the original supposition that all (a,b,c) variables were integers is false. The specialized mathematician is invited to prove this contradiction, thereby completing this version of the proof.

4.  Various combinations of the above methods.

### Appendix III: Tables of Differences

A power of an integer is regarded as a sum of the differences of powers of a descending sequence of overlapping pairs of integers. This sequence is more complicated than the binomial methods, and no proof is claimed using this method. The analysis is included here because it provided the initial insight into the concept of the missing 1. For example, when n=3:

Starting from the first tier with powers n of a series of integers, (a), this trapezoid of differences of powers has n+1 tiers. From the second tier and above, each number gives the difference between the two numbers directly below it For example 18=37-19 and 37==64 -27.The uppermost tier is always composed of just one integer, n!, showing the regularity found in a sequence of integral powers. The sum of a sequence of second tier numbers gives the differences between the powers of the integers at the beginning and end of the chosen series. For example 53 = 1 + 7 + 19 + 37 + 61 = 125 , so that 83 - 33 =37 + 61 + 91 + 127 + 169 = 485. In general,

an        ={(a-a+l)n-(a-a)n}+{(a-a+2)n-(a-a+l)n}+ . . .      +{(a-1)n – (a-2)n} + {an-(a-1)n}

=                          +            7             +19+37  +   .  .  .                 +{an-(a-l)n}

=cn-bn  = {(b+1)n - b n {(b+2) n - (b + 1) n } +   ...          + {(c-1) n -(c-2) n }+ {c n -(c-l n }

the expansion of a n involves a   l  + 7 + 19... as the first terms, but the c n -b n side of the equation results in a series that would start and end with terms including b n and c n without a 1.

### Appendix IV : Fermatesque Theorems

It was shown in the proof that the reason Eq.(l) has no integral solution is that an un-cancelled 1 appears in the terms of its expanded form. Conversely, Eq.(7) can have integral solutions because the four 1 terms in its expanded form (Eqs.(6a) and (6b)) cancel out leaving only A,B,and C.terms. If this analysis is correct, then Fermat's last theorem can be extended to cover any similar equations where variables are raised to the same power n, but where the number of terms on either side of the equation differ by one or more. For example two terms equaling three terms:

14)    a n + b n = cn + dn + e n

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[1]  E.Bell, The Last Problem, (ODudley, Ed.), Math. Assoc, America, 1990.

[2] .N. Adachi, Feruma No Dai Teiri . Sugaku seminar books. Nippon Hyoron-sha 1984, ["Fermat's Great Theorem", in Japanese].

[3] A. Wiles, R. Taylor, Annals of Mathematics May 1995.

[4] N. Gridgeman, Math. Gazette, 54 (1970) 31-37.