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FERMAT’S LAST THEOREM: THE COMPRHEHENSIVE PROOF RESULTING IN CALCULATING ALGORITHM.

 

by Ivan M. Obushenko, Ph.D.

(Ukraine)

E-mail: obush@estart.com

  Printed on March 8-10, 2005

 

 

Summary

The new approach for consideration of the legendary Fermat’s Last Theorem has been developed. One advantage has been received from proposed consideration: the calculating algorithm for PC – “ Obushenko-Fermat MachineÔ ” making the calculation process by fully automated and expendable unlimitedly however it had been shown that it is unnecessary, which was unexpexted. The “byproduct” of calculation is a finding of all integer triads fulfilling to the Pythagoras Theorem.

 

Introduction

 

On February 25, 2005 I heard from the Ukrainian radio broadcast a presentation by Dr.Phys.-Math.Sci.  on subject of the famous Fermat’s Last Theorem. She (well, what a difference, if it would be He?J) mentioned about enormous efforts made by Robert Wiles whose results were not recognized and accepted as the final proof (the same conclusion I met at reference [2]) that surprised me a lot because the acclaim made in the media about 10 or more years ago produced the impression of success on naïve public (well, on myself too). I remembered the theorem from my studies at the Senior High School (well, I had the really good teacher despite she lost to me a few “good” problems; so what? As a teacher she was very resourceful: I realized that when we’ve got a substitute man for the whole trimester…) when like anybody else I tried my “wings” on that prestigious subject too, which doesn’t seemed to be complicated on the first look and now – many years later –  started again immediately on a fresh sight. Dr. Sci. summarized her lovely presentation by words about lying beyond the real world when you have deal with such thing as Fermat’s Theorem. Very scientifically and encouraging doesn’t it? It seemed to me I got some idea from air how to do it although it was not quite clear. Anyway I wrote some notes, which gave me impression that I am on right way and next day I told to my son that I did it. One more day later I’ve met the dead ends as before. But I already gave my word to my son and so I had nothing to do as go forward.

 

Fermat’s Last Theorem

 

For integers a, b, c, and  n>2  an + bn ¹ cn .

 

The PROOF

 

The current consideration is substantially based on following Lemma expressing my basic idea led me forward.

 

Lemma. For any three numbers a, b, c, for which  a + b >  c some triangle can be constructed.

 

That Lemma arisen from the everyday experience and can’t be proved in principle on my opinion. So it rather Axiom than Lemma . Indeed, let’s make a drawing for a, b <  c :

 

   a          m               b             

Drawing 1.

                                                           

                 c

 

Case#1. What, if    a + b <  c

 

Let’s put for clarity |a|<|b|<|c|, where |a|=mod a .

For a + b <  c and  a + b +m = c (according to the Drawing 1.) and assuming all numbers to be positive one can conclude:

a2 + b2 <  c2 =[(a + b)+m]2 =  a2 + b2 + 2ab + 2(a + b)m +  m2.                                       (1)

By multiplying that inequity (1) by  c we will obtain

            a2c + b2c <  c3                                                                                                                                                                                  (2)

and by observing that a3< a2c and b3< b2c we can conclude that

a3 + b3 <  c3 .

By applying the Mathematic Induction Method we now can easily prove that for any

a + b <  c from fulfillment of inequity an-1 + bn-1 <  cn-1  the following one will be true:

 

      an + bn <  cn                                                                                     (3)

 

like it was for (1) and (2).   

 

What if one or two or all three integers will be negative ?

 

If the smallest is negative.  Then for even power n=2k  we’ll get obviously an + bn <  cn  because (-a)2k=a2k for k≥1 and we are getting back to the already proved expression (3) above.

For the odd power n=2k+1

                                                                 -an + bn <  cn                                                                                (4)

because we set from beginning |a|<|b|<|c|, where |a|=mod a.

 

If the intermediate is negative (b<0).   Then for even power n=2k  we’ll get obviously

an + bn <  cn  because (-b)2k=b2k for k≥1 and we are getting back to the expression (1).

For the odd power n=2k+1

                                                                  an - bn <  cn                                                                                 (5)

because we set from beginning |a|<|b|<|c|, where |a|=mod a.

 

 

If the biggest is negative.  Then again for even power n=2k we’ll get obviously an + bn <  cn because (- c)2k= c2k.

For the odd power n=2k+1

 

                                                           b2k+1 + a2k+1 > - c2k+1                                                                        (6)

on the left we have the positive number, while on the right side of inequaty

.                                                             

 

Case#2.  What, if    a + b = c

 

Let’s put for clarity |a|<|b|<|c|, where |a|=mod a and all integers are positive  (>0).

Then     a2 + b2 <  c2 = (a + b)2 = a2 + b2 + 2ab                                                                  (7)

and by applying to the expression (9) the same operations as to the case (1) above we are obtaining for a + b = c         

                                                                an + bn <  cn .                                                            (8)

 

In power of similarity condition (1) to condition (9) all possible combinations for the negative integers considered in Case #1 will be merely all the same. That’s why I don’t repeat it here: it was promised the enjoyable proof – not the boring one. J

 

Case#3. What, if  a +b > c

(some triangle can be constructed)

 

Well, the good news is that this case is  the last one: no more relationships between  a, b, c are available (except a sign of course). One more good news is the number of different type of triangles is finite and not too much.  J

Let’s start from the simplest: a=b=c , the equiangular triangle. Obviously a+ a>a and

2(an)>an when an >0; or 2(an)<an when an <0  for every n ≥ 2.

The next obvious thing is a<b=c ,  i.e. one of the possible isosceles triangles. Obviously a+ b>c and an+ bn > bn  when an >0  and bn >0; or an+ bn < bn , if either an<0 or both  an, bn <0 for every n ≥ 2.

 

 


                                                 a’

 

                                                                          b’                          Drawing. 2

                                         a                         b    

 

 

 

 


                                                              c

As we see from Drawing. 2 the three type of triangles is representing the rest of possible relationships between integer a,b,c .

 

Right-angled triangle.

 

The key point here is the right-angled triangle of course, for which

          a2 + b2 =  c2.                                                                   (9)

 

Again as usually let’s put for clarity |a|<|b|<|c|, where |a|=mod a .

Let’s multiply the expression (10) by c :

a2c + b2c=  c2c.                                                                (10)

However such as a3< a2c and b3< b2c we can conclude that

 

a3 + b3 <  c3                                                               (11)

 

and by applying further the approach of Mathematic Induction we will come to conclusion (3) again: an + bn <  cn .

Certainly because of the inequity (11) taking place the consideration for negative integers will be all the same as in Case#1 and #2 : it’s still no chance for equity (one can check it out).

 

Obtuse-angled triangle. Oh, this is a simple one after the right-angled triangle was done right now: just take a look on Drawing. 2. Does anyone want a detailed consideration ? I’m simply writing down the obvious results:

 

a2 + b2 < c2 ,  a3 + b3 <  c3 and an + bn <  cn                                        (12)

 

plus the inequities (5), (7) and (8), if some from integers a,b,c happen to be negative and referring the readers to the corresponding place in  Case#1.

 

Scalene triangle. This case is the most complicated. From Drawing. 2 the first what we see is

 

a2 + b2 > c2.                                                         (13)

 

The approach of multiplying by  c  and then replacing the members of inequity on smaller

values won’t be applied here. So, even for n=3  we can  not make any general conclusion.

The direct calculations for some well known numbers demonstrate that direction of inequity sign will depend on absolute values of a,b,c and  n .

For example, for n=3 , a=4, b=5, c=7 ( something close to Pythagoras’ triangle 32 +42=52 )

43 +  53 =280 < 73 =343,  while 42 + 62 = 52 > 72 = 49 as it should be according to the general inequity (12). But as soon as a,b,c each becomes larger than 10 then inequity will have the same direction as for square power by starting already from cube:

 

103+113=2331>133 =2197.

 

 Let’s take one more look at another Pythagoras’ triangle 62 +82=102.

For closest scalene triangle 72+82=113>102, while 73+83=855<103 and even

7100 + 8100=2.037*1090  < 10100=100*1090 and the difference is just increasing with increasing n.

From other side: 103+113 =2331>123=1728  but <133=2197.

So, generally saying for scalene triangle the direction of inequity can take both sides:

 <  and  >.

Nevertheless the geometric images is not only help us to embrace and systematically consider all possible cases for Fermat’s theorem but now it gives us an opportunity to develop a general strategy to write down some expressions allowing us to deduce some useful conclusions concerning on verification of Fermat’s Last Theorem.

In attempt to build up some systematic way for calculations we may use our experience obtained during the current consideration. Indeed, let’s find out for any given c  a pair of a and b, for which  an + bn = cn . As we found out by using Drawing. 2 the possible solution, if it exists at all can be only “between” right and scalene triangle built on given c.

The easiest and certain way is to start out of the equiangular triangle having sides equal to

c-1, c-1 and  c (at the beginning of the Case#3 consideration it was shown that it makes no sense to get anything useful in isosceles triangles with b=c.J)

δ = (c-1)n + (c-1)n - cn = 0.                                                   (14’)

 

Of course it will be ¹0 but we are looking for δ=0 and therefore we’ll make our next step:

δ = (c-1)n + (c-2)n - cn = 0 then δ = (c-1)n + (c-3)n - cn = 0 and so on until inequity will change its direction or until (c-t)=1.

Then we will put δ = (c-2)n + (c-2)n - cn = 0 and check out all possibilities to get  δ=0.

Well, all that is god for computer but not for mathematics.

For mathematics we should make a suggestion:

 

(c-s)n + (c-t)n = cn,                                                                                              (14)

 

i.e. there are some s  and  t , for which the equation (14) is correct.

Basically the parentheses can be expanded  by means of Newton’s binomial formula however without loosing commonness and for better visibility I’ll make it for n=3 and 4, and for s=1, t=2 only:

 

(c33c2 +3c1) + (c3 - 6 c2+12c –8) - c3 = 0                                (15)

                             (c4-4c3+6c2-4c+1) + (c4-8c3+24c2-32c+16) - c4  = 0.                               (16)

 

Thus the consideration of Fermat’s Last Theorem is reduced to the problem of the polynomials comparison. As it is stated in the polynomials theory: “The two polynomials are equal then and only then when the pairs of their corresponding members are equal.” It is introducing by definition what means it is free of proof necessity.

As we see on equations (15), (16) after the corresponding contraction of c3 in (15) and  c4 in (16) staying alone with the appropriate ones in either of parentheses there is no chance for those two parentheses to be equal. even at s=t because of specificity (or originality) of general equation (14)  and in force of definition for polynomials equity. Well, this was a good chance to put a big fat point in  Fermat’s Last Theorem history. But as justifiably remarked to me one of our mathematic fellow ( I am not naming him since I don’t have his approve on it): the polynomials could not be equal, while sometime their values do – like the Pythagoras equation (now it’s my turn to make remarks: no more examples are available doesn’t it ? J). That’s why below I’ll try to use that point for further development.

        If we apply the theory of limits to the expressions (15), (16) we will see that at given n>2 when c→∞ then lim δ→∞ too (exprs. (14’, 15, 16). It means: the bigger c, the less chances that the curves in parentheses would intersect at all. When c→0  then lim δ→Ω

         Similar conclusions can be made for given c and growing  power n.

Thus once an inequity δ (14’) turns its direction its most the like that it never get back.

         For better understanding the nature of intersecting of curves, which is not continuous let’s consider again the case

(c-s)2 + (c-t)2 = c2

(c-s)2 = c2 - c2 +2ctt2 = 2ctt2.                                                        (17)

This is possible intersection of square “parabola” and a straight “line”.

Example: s=2, t=1  and details are on Drawing 3. below :

(c-1)2 + (c-2)2 = c2

 

 

 

 

                            

 

                                   (c-s)2                                                            t/2<s

                                                                                                                Drawing 3.

                                                                                      t/2>s

                                                                                                             c-s)2=y=2ctt2

 

                                                      s                                                       (c=0; y= – t2 )

                                            t/2                                             c                (y=0; c= t/2)                                                             

                            -t2                                   t/2            c0                                                                     

                                               

 

                                   -t2         

 

 

                                                                    (c-s)2k+1

         

                             y

                                                             

                                                                       c2k+1-(c-t)2k

                                                                                                                         Drawing 4.

                                                                     

                               t2k+1                                                                                                          

                                               s                                                                           for  n=2k+1 

                                                          c                     y=(c-s)2k+1 = +mc2k - …+t2k+1

                                                                                                          

 

 

 

 

                                                                                                                        

                                                                      

                                                                                                                             

                                                                                                

                                                                                                                          

                                     y                                                (c-s)2k

                               

                                                                                                                             

                                                                                           c2k-1+…- t2k

 


                                            

 

                                                     s                   

                                                                                               c                           

 

                                                                                                                            Drawing 5. 

                                                               

                                       -t2k                                                                                                              for  n=2k

 

 (c-s)2k=  y = +mc2k-1 - … -t2k

 

As we see at the Drawing 3. for the same functions (17) dependently from parameters s and t , which

physically just changing the long of smaller triangle sides built on basis of given c (Drawing 2) the necessary conditions for intersection may be created (when t/2<s) or may be not (t/2>s).

          I said “necessary conditions” because the visible impression like on Drawing 3 does not guarantee yet that the curves indeed intersect because they are not continuous in differ of real numbers. This constitutes the magic of integer numbers and that’s why it is possible to establish the necessary conditions only, not the sufficient ones, which would be the proof of  Fermat’s Last Theorem.

          Further I will examine a possibility (14’): whether δ = (c-1)n + (c-1)n - cn = 0.  For example, if we assume for n=3 that there are all three solutions for integers and expand

               (c – x)(c – y)(c – z) = c3 – c2(x + y + z) + c(xy + xz + yz) - xyz                             (18)

              

and then compare coefficients with the Newton’s ones from expression (15) we will receive the hole system of equations or conditions for (x+y+z), for (zy+xz+yz) up to x*y*z=s3+t3,

which means one of the well know condition: possible solution(s) could be (but not should be!) a divisor of sum of the two cubes. Again we can consider only every particular case for given s  and  t and find out again that the sum may be a simple number like 22+32=17, which does not have divisors except itself and 1, while the divisor(s) of another sum can exist but may not be a solution(s) for equations like (15) or (16): otherwise saying, if we mark a divisor as c0 the polynomials (15) or (16) may not be necessarily transformed into product of (c - c0) on polynomials of (n-1)-power as it follows from discussion of expr.(18) because the two more conditions must be fulfilled for that. Again the divisor of the sun of two cubes was only the necessary condition, while the sufficient one(s) can not be formulated for about 350 years.

 

          Now it’s getting easier to consider all possible “necessary conditions” expressed in (15,16) and the like: it always will be possibility of  “intersection” or “not possibility” similar to Drawing 3 with a little difference: instead of “strait line”, for the odd powers it will be some parabola (square- or cubic-like one) shown on  Drawings 4 , 5 . One more difference: it will be impossible to calculate the coordinates of “crossing” the axes…

          Well, how one expects to prove “invisible”, “not existing”? I don’t know and hence  propose the “Obushenko-Fermat Machine™”, while I showed a pair of principle things: i) the calculations can be arranged to be automatic and without any gaps or arithmetic mistakes; ii) the calculations is not unlimited in differ of what was thought so far.

 

DISCUSSION

 

 

What was seen above reminds me more and more an attempt to catch very small baby size fish by using the net for playing soccer or volleyball. The results can be improved little bit by substituting the net with one for playing a ping-pong: this is what Pythagoras equation does for integers. And the larger c and  n the less chances on success obviously: the holes are getting larger and larger…

From that point I think the Fermat’s Last Theorem is a problem for probability & statistics guys rather than for the numbers theory ones . But I’m not sure whether the number “0” sounds familiar to the formers. J

In “the PROOF” above it has been shown that: “If we apply the theory of limits to the expressions (15), (16) we will see that at given n>2 when c→∞ then lim δ→∞ too (exprs. (14’, 15, 16). It means: the bigger c, the less chances that the curves in parentheses would intersect at all. When c→0  then lim δ→Ω.

         Similar conclusions can be made for given c and growing  power n.

The direct calculations by means of  the “Obushenko – Fermat Machine” ™ brought up the following results:

for n=3: 7^3-6^3-5^3 = 2,  9^3-8^3 -6^3=1,  12^3-10^3-9^3= -1

for n=4: 6^4-5^4-5^4=46, (by the way from Drawing 4. it’s getting clear why for the even “n” the δ is larger, than for the odd ones: the potential intersect could occur at considerably larger “c”, than for odd  n” – look at Drawing 5.)

for n=5: 17^2 - 16^5 -13^5= -12  and

for c>20  and/or  for n>5:   the differences become larger than 1000 and further rapidly grow in full correspondence to the just said.

 

CONCLUSIONS

 

       A new approach for considering the Fermat’s Last Theorem had been disclosed above and shown to be useful. The Theorem had been considered in a comprehensive manner.

       It was found the ranges of a, b, c , for which the  Theorem has been proved strictly and unambiguously

(a + b)<c     and         (a + b)=c

      

It was circled the range of a and b for every given c where the solution(s) for Fermat’s equation could be, if they would exist (I’m sure they are NOT). That “circle” is given by the following inequity

 c2< a2+b2<2c2

 

|c|<|a+b|<2|c|

 
                                                      (18)

,

      

where |a|=mod a .

During the process of consideration a series of boundaries and limitations were established, which were useful because of availability to turn them into computers’ algorithm; its functioning was partially described in Case#3 of the consideration. Based on the current consideration that algorithm called the “Obushenko – Fermat Machine” ™ made able to perform the fully automated verification of the Theorem and ultimately TO PROVE IT TO BE RIGHT:

 

for the integer numbers at n>2  an + bn ¹ cn .

 

       Publishers of books on Mathematics should consider an option to provide larger margins on the pages.  

 

 

Acknowledgments

 

      First of all I would like to thank you to my dear school teacher, who spent the time of her own on additional  facultative (just optional) lessons, which were far beyond the school program and toll free like all the education (well, University – too J) ; I didn’t become a mathematician - however I hope my work on Fermat’s Last Theorem would be somewhat worth-while feedback. Her name is (I expect not “was”…) Faïna  Markovna Andreeva and she would appreciate at least my sincere try. It was almost 40 years ago in city of Kiev, school #107 lowest ranked in a district. Now that school is not functioning anymore and the empty building stays in a wait to be dismounted. However there were a few really good teachers graduated from the best University in a country named by Taras Shevchenko, who didn’t have the “ties” and who was headed by the degraded diplomat as the school principal.

      I am thankful to my mama Nadine (a Hope in English) who made my everyday life in my childhood comfortable and trouble free, who did not survive to those fair days and did not see her name listed together with mine in Who’s Who in a World.

      Thank you to Mr. Pierre de Fermat’s nephew for understanding so good the importance of the math papers of his uncle for the world.

      Also thank you to the mentioned in introduction Dr. Phys.-Math. Sci. Natalie Okolitenko for keeping Ukrainians on pulse of what world is thinking about.

       Special “thanks” to my countless interviewers stolen my time with help of annoying “recruiters”. To the interviewers who always found my merits below (sometime far below) of their expectations and sometime taken a “liberty” to tell it into my face and always baffled me with all the same “smart” question: ”Where  do you see yourself for 5 years since now?”. Do you know what guys ? I still don’t know… However now you may (I hope J) see where I am, don’t you?  And would you give me to be compensated as good as you (almost, of courseJ)) are I’d be care free too from such small thing as the Fermat’s Last Theorem does. Do you remember me, guys? Do you? Well, I’ll tell you a good (for me) news: I survived.

      And thank you, Academy for preserving me busy and first of all - to the head of my Department for granting me a position out there.

      Amen.

References

 

1.      Dictionary of Science. Edited by Peter Lafferty & Julian Rowe. Brockhampton Press, London, 1997. ISBN  1 – 86019 – 501 – 6.

2.   Fermat’s Last Theorem. Article by: J J O'Connor & E F Robertson, February 1996.

      WebPage. http://www-history.mcs.st-andrews.ac.uk/HistTopics/Fermat's_last_theorem.html