Abstract. 2 x = 1. It is shown by contradiction that the uniqueness

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1 Copyright by A. A. Frempog Fermat's Last Theorem Proved o a Sigle Page "5% of the people thik; 0% of the people thik that they thik; ad the other 85% would rather die tha thik."----thomas Ediso "The simplest solutio is usually the best solutio"---albert Eistei Abstract Hoorable Pierre de Fermat was truthful. He could have squeezed the proof of his last theorem ito a page margi. Fermat's last theorem has bee proved o a sigle page. Two similar versios of the proof are preseted, usig a sigle page for each versio. The proof is based o the Pythagorea idetity si θ + θ =, or si x+ x =. It is show by cotradictio that the uiqueess of this idetity excludes all other -values, > from satisfyig the equatio c = a + b. Oe will first show that if =, c = a + b holds, followed by showig that if > ( a iteger), c = a + b does ot hold. For the first versio, let a, b, ad c be three relatively prime positive itegers which are the legths of the sides of the right triagle, ABC, where c is the legth of the hypoteuse, ad a ad b are the legths of the other two sides. Also, let the acute agle at vertex A be deoted by θ. For the secod versio of the proof, ratio terms were used to begi the tructio of the proof, without referece to a triagle. The secod versio cofirmed the proof i the first versio. It is also exemplified that if some of the legths are ot positive itegers but positive radicals, the derived ecessary coditio for c = a + b to hold is applicable. Each proof versio is very simple, ad eve high school studets ca lear it. Perhaps, the proof i this paper is the proof that Fermat wished there were eough margi for it i his paper. With respect to prizes, if the prize for a 50-page proof were $75,000, the the prize for a sigle page proof (iderig the advatages) usig iverse proportio, would be $07,50,000.

2 Proof: Versio Pythagorea Idetity Postulate: There exists oly a sigle fudametal trigoometric idetity such that siθ = ( a positive iteger). Give: c = a + b ( a iteger; ab,,ad care relatively prime positive itegers) Required: To prove that c = a + b does ot hold if >. Pla: Oe will first show that if =, c = a + b holds, followed by showig that if > ( a iteger), c = a + b does ot hold. Proof: Let ab,,ad cbe three relatively prime positive itegers which are the legths of the sides of the right triagle i the figure below, where c is the legth of the hypoteuse, ad a ad b are the legths of the other two sides. Also, let θ deote the acute agle at vertex A. The a = csiθ () b = cθ () c = a + b () B c = ( csi θ) + ( c θ) c c = c si θ + c θ (4) a c = c(siθ + θ ) (5). Left-had side (LHS) of equatio ( 5 ) equals θ right-had side (RHS) of (5) oly if A b C siθ = That is, a ecessary coditio for (5) to be true is that siθ = If =, c = c(siθ + θ ) is true sice si θ + θ =.Therefore equatios (5) ad () are true. Sice there exists oly a sigle Pythagorea idetity (a postulate) such that siθ =, ad si θ + θ =, with =, there are o other positive itegers,, such that θ + siθ =. Therefore, equatios (5) ad () will be true oly if =, ad there are o other positive itegers, > which will make equatios (5) ad () true. Therefore, c = a + b holds oly if =, ad does ot hold if >. The proof is complete. Coclusio Fermat's last theorem has bee proved i this paper. Note above that the mai criterio is i equatio (5) above, which requires that siθ =, if c = c(siθ + θ ) ad c = a + b are to hold. Perhaps, the proof i this paper is the proof that Fermat wished there were eough margi for it i his paper. About the Pythagorea Idetity Postulate Sice si θ + θ = is true, for ay other value such that siθ =, siθ = siθ + θ, would imply that if, siθ = siθ + θ would be a false statemet. For example, if =, siθ + θ = siθ + θ would imply that = (equatig the expoets), which is false; ad by cotradictio, siθ + θ siθ + θ ad equetly, si θ + θ. Therefore, siθ = oly if =. Thus, there exists oly a sigle Pythagorea idetity such that siθ =, ad it is si θ + θ =, with =.

3 a = csiθ b = cθ c = a + b c = ( csi θ ) + ( c θ) c = csiθ + c θ c =. c (si θ + θ ). Equatio ( 5 ) is true oly if si θ + θ = For (5) to be true si θ + θ =. If =, c = c (si θ + θ ) is true sice si θ + θ = ad therefore, equatios (5) ad () hold. There exists a sigle idetity such that si θ + θ =, ad si θ + θ = with =, there are o other positive itegers such that si θ + θ = Therefore, equatios (5) ad () will be true oly if =, ad there are o other itegers, > makig eqs (5) ad () true. c= a + b holds oly if =, ad does ot hold if >. QED Adote Proof: Versio i the Margi Fermat was truthful. He could have squeezed the proof ito the page margi. If Fermat were reicarated, he would be pleased.

4 Proof: Versio (Usig ratios) Cofirmatio of Versio Proof Pythagorea Idetity Postulate: There exists oly a sigle fudametal trigoometric idetity such that si x = ( a positive iteger). Give: c = a + b ( a iteger; ab,,ad care relatively prime positive itegers) Required: To prove that c = a + b does ot hold if > Pla: Oe will first show that if =, c = a + b holds, followed by showig that if > ( a iteger), c = a + b does ot hold. Oe will begi by applyig ratio terms. c = a + b () (Give) Example o ratio terms a + b = c If 4+ 8=, ad () (rewritig) the ratio terms are a = rc () ( r is a ratio term) ad, the b = sc (4) ( s is a ratio term) ( r+ s =) 4 =, rc + sc = c (5) (substitute for a ad b from () ad (4) 8= c( r+ s) = c ; ad the (6) sum of the ratio terms is Now, by the substitutio axiom, sice r+ s =, r+ s ca be replaced + = by ay quatity =. Oe ca therefore replace r+ s by si x+ x, Other equivalet idetities sice si x+ x =. The equatio (6) becomes Note: "magic "umber,. c(si x+ = c sec (7) x ta x = csc If =, (7) becomes c(si x+ = c x cot x = (8) x+ si x = c = c(si x+ (8) (rewritig) x x = Equatio (8 ) is true sice si x+ x =. Cosequetly, equatios (8) ad () hold. Therefore, if =, c = a + b Elimiatio of the. ratio terms r ad s Geeralizig equatio (7), oe obtais c(si x+ = c (9) The author was impressed i which the ecessary coditio for (9) to hold is si ad gratified by the x =. substitutio axiom which Sice there exists oly a sigle fudametal Pythagorea idetity permitted the ( a postulate) such that si x =, ad si x+ x =, with itroductio of the much eeded ecessary coditio =, there are o other positive itegers,, such that si si x = i x =. Therefore, equatios (9) ad () will be true oly if Versios of the =, ad there are o other positive itegers, > which will make proof. equatios (9) ad () true. Therefore, c = a + b holds oly if =. ad does ot hold if >. The proof is complete. Coclusio Fermat's last theorem has bee proved i this paper. Note above that the mai criterio is that si x =, if c = c(si x+ ad c = a + b are to hold. Perhaps, the proof i this paper is the proof that Fermat wished there were eough margi for it i his paper. About the Pythagorea Idetity Postulate Sice si x+ x =, for ay other value such that si x =, si x = si x+ x, would imply that if, si x = si x+ x would be a false statemet. For example, if =, six+ x = si x+ x would imply that = (equatig the expoets), which is false; ad by cotradictio, six+ x si x+ x ad si θ + θ. Thus, si x = oly if =. 4

5 Discussio About Versio of the proof (Usig ratios) From equatio (6) i Versio proof, oe could replace r+ s by each of the equivalet idetities as show below. Note that r+ s =; c( r+ s) = c (6) Alteratives for equatio (6) i Versio proof c(sec x ta = c c(csc x cot = c c(x+ si = c c( x = c Other equivalet idetities Note the umber, i various positios sec x ta x = csc x cot x = x+ si x = x x = Justificatio for usig a right triagle i Versio of proof Versio proof bega with referece to a right triagle ad ot a oblique triagle. If it were a oblique triagle, the equatio would be c = a + b ab C (A) For Fermat's Last Theorem, three terms are ivolved: c = a + b (B) Comparig equatios (A) ad (B), ab C = 0 i (B), which implies that m C = 90. That is, C is a right agle ( 90 = 0). Uiqueess of si x+ x = ad si x+ x = If =, si x+ x = becomes si x+ x =, which is true. If =, si x becomes six+ x. Now, if it is assumed that si x+ x =. the, six+ x = si x+ x (C) (sice si x+ x = ), ad equatio (C) implies that = (equatig expoets), which is false; ad by cotradictio, six+ x si x+ x, ad si x+ x. Similarly if = 4, oe will obtai the false statemet, 4 =, ad si 4 x+ 4 x. Similarly, if = 567,,,... oe would obtai respectively, the false statemets 5 =, 6 =, 7 =,..., ad by cotradictio, each of these -valies will ot make si x equal to si x+ x ad equetly, will ot make si x = true. Therefore, If, there are o positive itegers which will satisfy si x = Note: sim x+ m x = si x oly if m =. Algebraic example If a + b = a + b, the = or If =, the a + b = a + b Questio: If =, is a + b = a + b? Aswer: No. Note: If A= B ad C A, the C B. 5

6 Overall Coclusio Fermat's last theorem has bee proved i this paper. I the first versio of the proof, oe bega with referece to a right triagle; but i the secod versio of the proof, the proof tructio bega with ratio terms without referece to a triagle. The ratio terms were later o "miraculously" elimiated from the equatios. The ecessary coditio for the relevat equatios ivolved to be true is that si x = (or siθ =). Thus, if c = c(si x+ ad c = a + b are to hold, si x = or siθ = must be satisfied. First, the author determied, why the equatio, c = a + b is true if =. It was determied that the ecessary coditio is si x = or siθ =, ad this coditio is satisfied oly if =, to produce si x+ x =. If = 45,,,..., this ecessary si x = or siθ = is ever satisfied. From the proof, the oly coditio for c = a + b to hold is the ecessary coditio derived i this paper. Therefore, c = a + b holds oly if =, ad does ot hold if >. Oe should ote above that versio proof cofirmed the proof i versio of the proof. Perhaps, the proof i this paper is the proof that Fermat wished there were eough margi for it i his paper. About the umbers, a, b, c, beig positive itegers B The equatio c = a + b will be true if siθ = If =, siθ = becomes si θ + θ = 5. Let oe apply the ecessary coditio to the dimesios of θ the triagles i the figures to the right. A 4 C The legths of the sides of triagle ABC are, 4, 5 (pos.itegers) E si θ = ; si 9 θ = ; θ = 4 5 ; 5 5 θ = 6 5 ; si θ + θ = + = The legths of the sides of triagle DEF are,, (oe radical) 4 4 si θ = ; si θ = ; θ = ; 4 θ = 4 ; si θ + θ = + = If =, for triagle ABC, si If =, for triagle DEF, si 64 9 θ + θ = + = θ + θ = + 8 = If =, each of the sets of the dimesios of the two triagles satisfies the ecessary coditio, siθ =; but if = ( > ), the ecessary coditio is ot satisfied, that is, siθ The legths of the sides of triagle ABC are all positive itegers, but ot all the legths of triagle DEF are positive iteger (oe legth is a radical). Therefore, the ecessary coditio is applicable eve if some of the umbers ivolved are positive radicals. Questio for a mathematics fial exam for the 06 Fall semester. Bous Questio: Prove Fermat's Last Theorem. Adote D θ F 6