ANOTHER PROOF FOR FERMAT S LAST THEOREM Mugur B. RĂUŢ Correspodig author: Mugur B. RĂUŢ, E-mail: m_b_raut@yahoo.om Abstrat I this paper we propose aother proof for Fermat s Last Theorem (FLT). We foud a simpler approah through Pythagorea Theorem, so our demostratio would be lose to the times FLT was formulated. O the other had it seems the Pythagoras Theorem was the ispiratio for FLT. It resulted oe of the most diffiult mathematial problem of all times, as it was osidered. Pythagorea triples existee seems to support the laims of the previous phrase. Key words: Pythagorea triples; Fermat s Last Theorem; Pythagoras Theorem; iteger umbers. 1. INTRODUCTION Fermat s Last Theorem postulates that there are ot three itegers a, b, whih satisfy the Diophatie equatio: a b, a, b,, Z, (1) for ay atural value of greater tha two. Apparetly simple, this theorem statemet has give muh trouble to may geeratios of mathematiias. Although it was formulated i 1637, after several partial demostratios, the suessful proof omes later, i 1995, after some papers published by the British mathematiia Wiles, [1] ad []. A brief history of the usuessful efforts to solve FLT is foud i ref. [3]. I our opiio, the demostratio diffiultess of this theorem osists i its outless possibilities of approah. I this otext we preset, i the followig, aother demostratio of this theorem, i terms of Pythagorea Theorem.. PYTHAGORAS THEOREM VS. FERMAT S LAST THEOREM I order to make a easy demostratio, whih otherwise it would be redued to a edless series of attempts by givig values to itegers a, b ad, hee impossible to prove i fat, we must redue our problem to aother problem i whih a, b ad must gai sigifiae. The best way is to redue equatio (1) to a geometry problem. Cosequetly, a oveiet way is to redue the problem to a relatio betwee sizes of a retagular triagle. Therefore let us osider a, b ad the sizes of a retagular triagle. It is kow that the relatio betwee them is, aordig to the Pythagorea Theorem: a b ()
Aother proof for Fermat s Last Theorem where a, b, R ilude both egative ad positive values. The sizes a ad b a take positive or egative values if we oveiet hoose the referee system i whih we built the retagular triagle. Regardig the hypoteuse, give that it is a sum of squares: it is obvious that it a also take egative values too. Let us ow defie the trigoometri futios ( a b si a / ad: os b/ with ( 0,) beause a ad b must take both positive ad egative values. For a better uderstadig why a ad b must take positive ad egative values, we a imagie a irle like trigoometri irle i whih we outlied i eah quadrat a symmetri triagle. I quadrat I, a ad b are positive ad α is a aute agle. I quadrat II a is positive ad b is egative, the agle α is already a obtuse agle. You might thik ow that sharp agles sum is ot equal to π/. The fulfillig oditio for a retagular triagle is broke. Begiig with quadrat II agle α would be already obtuse ad this is the reaso why triagles aot be oeived i these quadrats. But this is ot exatly so. I quadrat II, for example, the agle symmetrial with α will have the value π-α. If the opposed agle will have the value α-π /, the their sum will be exatly the oditio that must be fulfilled by a retagular triagle. Thus our irle will ot be a simple referee for the measure of a ad b, positive ad egative. Hee we obtai four idetial retagular triagles, with idetial sizes ad agles. The oly differee betwee them is that some of them are positive, other egative, somethig ovetioal, atually illustratig their perfet symmetry from the eter of oordiate axes. With these established otios we a thik ow a way to trasform equatio () i a relatio lose to equatio (1). Oe way is to multiply equatio () by ad usig trigoometri futios defied above i order to obtai a equatio i whih the power of the ukows a, b, ad to be multiplied by oe order. Remakig the multipliatio times we have therefore: It would have bee easier to write a ) 1 si b os (3) a b si os (4) but we hoose the form (3) beause the futios formed by sie ad osie of high power i (4) aot be defied if the trigoometri futios have egative values. We have therefore (3) i whih we a already operate the restritio a, b, Z. We distiguish two mai ases. The first ase orrespods to the geeral situatio i whih: p (5) Here p must be see ot eessarily a iteger but rather as a umber that make (5) a iteger. Cosiderig (5), equatio (3) beomes: p p p p p a si b os (6) Obviously, for p=1 it results equatio () from whih we started. a, b, ad are itegers this time, ad we a say that we have already demostrate the viability of the geeral relatio () for iteger umbers. But we are iterested i higher powers tha two of, respetively higher tha oe of p ad therefore i a equatio (6) of the form (1). Note that we a do this, ad this is to write equatio (6) of the form: with d asi ad e bos, d, e Z, oly if the oditio p p p p p d e (7)
Mugur RĂUŢ is fulfilled. By replaig p 1/ i equatio (6) we get immediately the equatio: asi bos whih turs i equatio () through trigoometri futios sie ad osie defiitio relatios. So, our equatio (6) a be writte as equatio (1), but oly for p=1/. For higher powers of p, p 1/, we always have: p p whih leads to the olusio that betwee three umbers a, b, Z, for ay p>1, there is oly a relatio of the form (6). The form (6) does ot exlude therefore a relatio like: p k m d e with k m p, but with d ad e real umbers. The quatities d ad e aot be itegers beause they are a produt of a iteger ad a real umber. Partiular ases of extreme values of trigoometri futios are exluded beause the quatities iluded i equatio (6) are the sizes of a retagular triagle for whih the sum of the sharp agles is always equal to π /. A speial ase is the oe i whih: For p we have ( 1/ ) p 1 1 si os, ad equatio (6) beomes p p p 1 1 p p a b Note that for p=1 the result is equatio (). But we are iterested i higher values of p. Therefore, to have a equatio of the form (7) it have to p 1 p Hee p=-1 for whih d e equatio that a be writte as 1 1 1 d e or ' ' ' d e with ', d', e' Z. The seod mai ase is orrespodig to: p 1 ad brigs othig speial i the spetrum of results. We obtai the equatio: p1 p1 1 p p1 1 p a si b os (8) for whih we do the same kid of reasoig. Equatio (8) beomes the type: p1 p1 p1 d e (9) with d asi ad e bos, d, e Z, uder oditio: p 1 1 p. This oditio easily leads to the oly possible result, equatio (). We are iterested i all situatios but espeially i p 1 1 p But this leads to the olusio that betwee the three iteger umbers a, b, Z ad p> 1 ould ot be ay relatio but that illustrated by the form (8). The partiular ase i whih the sie ad osie values are equal is desribed by the equatio: p1 p1 p1 p1 a p1 b (10)
4 Aother proof for Fermat s Last Theorem The oditio: leads to equatio: whih a be writte as: with ', d', e' Z. p 1 p 1 d e ' d ' e ' 3. PARTICULAR CASE A simplest ase, whose solutio is obvious, is whe trigoometri futios are positive, ( 0, / ) I equatio (3), writte i the form (4): a b si os we a do the restritio a, b, N. For = we a see that the resultig that equatio () it results very easily. For > we must see what happes with trigoometri futios withi paretheses. Notig that: 1 ad: also: a arise oly as: 0 si 1, 0 os 1, 0 si 1, 0 os 1 si d e,,os N, Q d, e R where d a si ad e b os. From equatio (4) it might exist a situatio. Cosider: 3, si 0, 1 ad a xy...z0, where x, y, z are ompoet iphers of the atural umber a. It is possible, therefore, the situatio i whih d N. I this situatio we also kow that e R ad N. As you a see, this situatio aot exist beause a atural umber aot be the sum of a iteger umber ad a real oe. So this situatio is impossible. The same olusio applies also for =3k, for k atural umber ad siα=0,0 01, with a=x0 0. As for the situatio whe siα=osα, It is treated the same way ad lead to the same results as i the previous ases, = p ad = p +1. By summig, ultimately, all the olusios of all ases i a sigle oe there are o three itegers a, b ad that satisfy equatio (1), for >. The solutios of this equatio are oly for =.
Mugur RĂUŢ 4. DISCUSSIONS Whereas equatios (6) ad (8) result from equatio () by multiplyig with, you would thik that we've prove othig so far, these equatios are equivalet to equatio (). We assume that equatio () is valid from the very begiig ad we got fially that oly () may be valid. It is what you might all a trivial ase. But thigs seem to stad exatly opposite.. We assume, i the hypothesis, that equatio () is ideed valid, but a, b, are real umbers. The result is that the equatio () is ideed valid, but i ase we are iterested, whe a, b, are itegers. Other ases, for higher powers of, are impossible. It is a orret reasoig beause i the FLT statemet equatio () is assumed valid, for ay iteger triplet, other ases beig impossible. I our reasoig we start with what is kow, but the umbers a, b ad are assumed to be ot itegers. This is the result we reahed. We iitially assumed that the equatio () is valid, whatever be three real umbers a, b ad. 3 3 3 By usig this reasoig, if we assume that the Pythagorea Theorem is of the form a b, we have reahed the same olusio: there aot be three itegers a, b, whih satisfy the equatio 3 3 3 0 0 0 a b. The same olusio is if the Pythagorea Theorem is a b. No matter what form has therefore the Pythagorea Theorem, the result is of the form of Pythagorea Theorem. This seems to be strage at first sight, but i our opiio, is due to the similarities of the two theorems form. Up to a poit you might get the impressio it is oe ad the same theorem. You might thik that FLT is just a existee theorem for equatio (), ad there is o a lear lie of separatio betwee them. Ad this is, perhaps, beause Fermat s Theorem was ispired by the Pythagoras Theorem, ad tries to apply it oly to some umerial values without ay meaig. He sueeded to reate, hee, the most diffiult mathematial problem, aordig to some, of all times. By the same blur delimitatio betwee the two theorems we use i our approah. We have show that there a be o triples of itegers satisfyig the equatio (1), for higher powers of. The solutios are oly for =. 5. Colusios Based o the similarities that exist, to ertai extet, from Pythagorea Theorem ad Fermat s Last Theorem, we attempted, i this paper, to demostrate the later through the former. It has bee show that give equatio (1), it has o solutio a, b, itegers, but oly for =. Case orrespodig to Pythagorea Theorem. 6. REFERENCES 1. WILES A. (1995), Modular ellipti urves ad Fermat s Last Theorem, Aals of Mathematis 141 (3), 443-551;. TAYLOR R, WILES A. (1995), Rig theoreti properties of ertai Heke algebras, Aals of Mathematis 141 (3), 553-57; 3. RIBENBOIM P. (1979), 13 letures o Fermat s Last Theorem, Spriger Verlag, New York.