A alteratve way to wrte Ferat s equato ad otes o a eleetary roof of FLT C. Sloae Vctora Uversty, NZ (chrssloae70@gal.co) (Feb 08) Abstract We dscovered a way to wrte the equato x +y -z =0 frst studed by Ferat, owers of 3 other varables defed as; the su t = x+y-z, the roduct (xyz) ad aother ter r = x +yz-xt-t. Oce x +y -z s wrtte owers of t, r ad (xyz) we foud that 3 cases of a re factor q of x +yz dvded t. We realzed that fro ths alteratve for of Ferat s equato f all cases of q dvded t that ths would lead to a cotradcto ad solve Ferat s Last Theore. Itrgued by ths, we the dscovered that the fourth case, q=3s+ also dvdes t whe usg a lea that uquely defes a asect of Ferat s equato resultg the followg theore: If x + y - z = 0 ad suose x, y,z are arwse co - re the ay re factor q of (x + yz) wll dvde t,where t = x+ y - z. Coyrght Chrs Sloae 08
ON FERMAT S LAST THEOREM Itroducto There have bee thousads of attets to solve Ferat s Last Theore (FLT) usg Eleetary Nuber Theory (ENT) over the cetures. Naturally whe cosderg a roble that ca be easly stated ad uderstood oe would assue a relatvely easy roof ENT would exst. However, oe were foud wth the equato as t s wrtte, wth the exceto of Adrew Wle s roof usg oder uber theory techques. Whe 993 Adrew Beal cojectured that there were oly coo factor solutos to the geeral case of FLT aely x a +y b -z c =0 whe a,b,c> oe the assued that there were coo factor solutos to FLT but obvously these solutos would cacel out ad be o-exstet the secal case. We therefore wodered what would be a good way of showg coo factors or ore secfcally what ter s re coosto would gve coo factors f they shared a re? We foud 3 good caddates x +yz, y +xz, ad z -xy because f they shared a re factor (q) wth owers of x, y, z or xy, xz, yz or xyz the we get coo factor solutos q. We ca t see how to use ths wth Ferat s equato as t s wrtte but whe we were tryg to factor x +yz to the =3,5,7 equato we tally foud a searato of the ters (x+y-z) ad (xyz). We the wodered whether ths was ossble for all. What we wated to do was see f we ca ut ths equato ters of( x+y-z) ad (xyz), or ore secfcally owers of (x+y-z) ad owers of (xyz) ad deed we could f we troduce a ew ter we call the syetrc r= x +yz-xt-t whch haes to have a x +yz cooet. For exale we have Ferat s equato for =7, 7 7 7 x y z 0 ad the ew reresetato we have for =7, 7 5 4 3 3 9t 56t r 35( xyz )t 35r t 35( xyz )t r 7tr 7( xyz ) t 7r ( xyz ) 0 Oe ca see ths s wrtte owers of the 3 ters t, r,(xyz) ad these ters coletely relace the owers of x, y ad z to becoe the arguets or varables the roble. We the studed ths ew equato ad realsed that f we showed all the re factors of x +yz or y +xz, or z -xy dvded t we could solve Ferat s Last theore because ths leads to a cotradcto x +yz t but x yz t FLT as the case ot. We frst show that t 0od3 ad recogzed that f we take a re factor (q) of x +yz we ca easly show that for oe case of q ad two sub-cases of q aely, q 3s+ q=s+, s M3 q=3s+, s M whe s re () we get t 0odq or we get coo factor solutos for these cases. The 4th case q=3s k + s ore dffcult but we develo ethods to deal wth t. We use a lea (lea 5) that defes a artcular roerty of Ferat s equato aely; x+y=c, z-y=a, z-x=b. The, alog wth the ossble solutos whe q=3s+, we show that these q s ust also dvde t. We further geeralze to all k usg a exoetato ethod that cobed wth lea 5 shows all q(k) dvde t. We therefore ed u wth t 0odq for all ossble cases of q. Whe we look at our ew reresetato of Ferat s equato, we ca show that wth the decoosto of, q( q ) q( q ) q( q ) q( q ) x yz q 3 q q 3...q where q s re ad q(q ) the hghest ower dvdg x +yz, that these hgher ower ters ust also dvde t but we have that x yz t whch obvously elates teger solutos. Reark: The rese behd solvg ths roble s qute sle - all we are showg s all the re factors of a artcular ter dvde t or we get coo factor solutos, resultg the theore; If x + y - z = 0 ad suose x, y,z are arwse co - re the ay re factor q of (x + yz) wll dvde t where t=x+y-z Coyrght Chrs Sloae 08
3 ON FERMAT S LAST THEOREM Hstorcal Note: Although, FLT s a acet roble t s oly relatvely recetly (30 years) that we have foud the geeralzed verso of the roble alost certaly has oly coo factor solutos. If acet atheatcas had kow ths, they would have realsed that coo factor solutos to the secal case would ot exst ad would be a good way of solvg FLT. It s dffcult to fd coo factor ethods workg wth three deedet varables. However chagg the for of Ferat s equato to cororate secfc ters lke x +yz creates a evroet fredly to coo factor aroaches. Wth ths alteratve verso of Ferat s equato also ukow to atheatcas utl ow, the the roble ay ot be outsde the reals of eleetary uber theory after all.. Deftos We defe the deedet varable t as, Aother way of wrtg t s to let, x y C, z y A, z x B. t= x+y-z (.0) t = A B+ C x = A+ t y = B + t z = C - t (.0) (.03) (.04) (.05) We defe the syetrc r geeral as, We ca also wrte ths as, r(v) = x + yz - xt + vt = y + xz - yt + vt = z - xy + zt + vt r(v) = xz + yz - xy + vt (.06) (.07) I ths work we wll oly be usg v= -, The syetrc arts are defed as, r(-) or r = xz + yz - xy - t (.08) We ca use ay of I ths work we wll use, r(x / t) = x + yz, r(y / t)= y + xz, r(-z / t) = z - xy, r(0) = xz + yz - xy (.09) r( x / t ) x yz, r ( y / t ) y xz, r( z / t ) z - xy, to cota our re factors q r(x / t)or r' = x + yz (.0) We use catalzato whe referg to these deftos x, y, z. e x x, y y, z z Hece, T = x + y - z (.) R = x z + y z - x y -T (.) R' = x + y z (.3) Coyrght Chrs Sloae 08
4 ON FERMAT S LAST THEOREM Reark: M stads for ultle of at soe laces ths work. We derve the ew for of Ferat s equato usg cobatoral arguets. Ths roof s qute log ad as oe kows cobatoral roofs take a log te to work through (over 4 ages ths stace). Hece ot to dstract the reader wth ths cubersoe roof we wll state the results Theore. ad Corollary, for brevty. Oe ca use a calculator or couter to check the valdty of these equatos for ay ut. If oe requres the rgorous roof lease see extract Proosto We ca wrte x y ( z ) ters of (xyz), r ad t Startg wth, x y z ( A t ) ( B t ) (C t ) We tally factor A +BC-At-t ths dervato ad covert back to x +yz-xt-t. I geeral we ed u gettg for ad l; odd eve # eve odd ( 3 ) ( 5 ) ( ) ( 5 ) ( ) ( 3#) ( 5#) ( )......... 3 ((... )t xyzr!! 0! ( )!!! #!!( )! ( 5 ) ( 7 ) ( 7 ) ( 7 ) ( 7 ) ( 5#) ( 7 #) ( 7 )......... 9 (... )t ( xyz ) 3 r! 3! 0! ( )! 3!! #! 3!( )! ( ) ( 4 ) ( ( 3 ) ( 4 ) ( ( 6 ) ( ( 3 ) ( )( )...( ) ( )( )...( ) (...!! 0! ( )!!! ( #) ( ( 4#) ( ( 3 ) ( )( )...( ) 3 ))t ( xyz ) r (.4) ( #)!!! odd odd * eve, eve ( ) ( 4 ) ( ) ( 4 ) ( ) ( *) ( 3*) ( )......... ((... )t r! 0! 0! ( )! 0!! *! 0!( )! ( 4 ) ( 6 ) ( 4 ) ( 6 ) ( 4 ) ( 3*) ( 5*) ( 4 )......... 6 (... )t ( xyz ) r...!! 0! ( )!!! *!!( )! ( ) ( 4 ) ( ( 3 ) ( 4 ) ( ( 6 ) ( ( 3 ) ( )( )...( ) ( )( )...( ) (...!! 0! ( )!!! ( *) ( ( 3*) ( ( 3 ) ( )( )...( ) ))t ( xyz ) ( * )!!! Ths gves, r ( s ) 3 3 / ( )/ ( )! t x y ( z ) ( )t ( xyz ) r s 0 0 3 ( s )!! s!( )! od 3 ters for (.6) (.5) Coyrght Chrs Sloae 08
5 ON FERMAT S LAST THEOREM Therefore we ca wrte, Theore. t deedet equato v = -, ( > 0) ( - s - - ) 3-3- / ( - )/ ( )! x y ( z ) (- ) (- ) ( )t ( xyz ) r (.7) - 3-0 s 0 0 ( - s )!! s!( )! od Where r s the v= - syetrc x +yz-xt-t 3 Makg / ( )/ 3 ( s )! 0 s 0 0 ( s )!! s!( )! x y ( z ) ( ) ( ) ( )t ( xyz ) r (.8) od Corollary ( 3) 4 ( 4 )( 5 ) 6 3 3 z x ( z x ) ( z x ) zx ( z x ) z x ( z x ) z x! 3! 4 ( )( )... / / / /... ( z x )z x! ( =eve) 3 ( )( )... ( )/ ( )/... ( z x )z x ( )! ( =odd) (. 9 ) Corollary!!!! x y ( x y ) x y xy x y x y... ( )!!( )! ( )!!!( )!!!! / / x y x y... x y ( eve )!( )! ( )!!!!! ( )/ ( )/... x y ( odd ) (. 0 ) ( ) ( )!! Coyrght Chrs Sloae 08
6 ON FERMAT S LAST THEOREM Frst Exales v = -... 6 6 6 0 9 8 7 6 3 6 5 4 4 4 3 3 x y z ( t 6t r 6xyzt 5t r 4xyzt r 0t r 3( xyz ) t 36xyzt r 5r t 0( xyz ) t r 4xyzt r 3 3 5 4 3 6 3 4 6 0( xyz ) t 6t r 9t ( xyz ) r 6txyzr t( xyz ) r r 6( xyz ) r 3( xyz ) )( xyz ) 5 5 5 0 8 7 6 5 4 3 4 3 4 3 x y z ( t 5t r 5xyzt 0t r 5xyzt r 0t r 0( xyz ) t 5xyzt r 5r t 5( xyz ) t r 5xyztr 3 5 5 5( xyz ) t r 5( xyz ) r )( xyz ) 4 4 4 8 6 5 4 3 3 4 4 x y z ( t 4t r 4xyzt 6r t 8xyzrt 4r t ( xyz ) t 4xyztr r 4( xyz ) r )( xyz ) 3 3 3 6 4 3 3 3 x y z ( t 3t r 3xyzt 3t r 3xyztr r 3( xyz ) )( xyz ) 4 x y z ( t t r xyzt r )( xyz ) x y z ( t r )( xyz ) 0 0 0 x y z 3 x y z t x y z 3t r 3 3 3 3 x y z 4t 3tr 3xyz 4 4 4 4 x y z 7t 8t r 4xyzt r 5 5 5 5 3 x y z t 5t r 0xyzt 5r t 5xyzr 6 6 6 6 4 3 3 x y z 8t 30t r 8xyzt 5r t xyztr r 3( xyz ) 7 7 7 7 5 4 3 3 x y z 9t 56t r 35xyzt 35r t 35xyzt r 7tr 7( xyz ) t 7r xyz 8 8 8 8 6 5 4 3 3 4 x y z 47t 04t r 64xyzt 80r t 80xyzt r 4t r 0( xyz ) t 4xyztr r 8( xyz ) r 9 9 9 9 7 6 5 4 3 3 3 4 3 3 x y z 76t 89t r 7xyzt 7r t 80xyzt r 66t r 45( xyz ) t 8xyzt r 9tr 7( xyz ) rt 9xyzr 3( xyz ) 0 0 0 0 8 7 6 5 4 3 4 3 4 3 x y z 3t 340t r 0xyzt 355r t 380xyzt r 70t r 00( xyz ) t 0xyzt r 35t r 90( xyz ) rt 40xyztr 3 5 0( xyz ) t r 5r ( xyz ) 9 8 7 6 5 3 5 4 3 4 3 x y z 99t 605t r 374xyzt 75r t 78xyzt r 407t r 09( xyz ) t 56xyzt r 0t r 4( xyz ) rt 3 3 5 4 3 54xyzt r 33( xyz ) t r t 66r ( xyz ) t xyzr ( xyz ) r... Couter Verfcato. Oe ay care to verfy these results by couter where t = x+y-z ad r =x +yz-xt-t =y +xz-yt-t =z -xy+zt-t There are ay corollares but otable corollares requred for FLT are as follows: Corollary 3 Whe s a ultle of 3 the the equato eds wth ±3(xyz) /3 3 Fro (. 7 ) wth 0, we have 0 the 3 hece, -(+) ( )! ( )! 3 ( xyz ) 3( xyz ) -3 0!!( )!! Coyrght Chrs Sloae 08
7 ON FERMAT S LAST THEOREM Corollary 4 For =M3- or =M3+ ad l=0 the the coeffcets of (xyz) r or (xyz) r resectvely s ± 3 Fro (. 7 ) wth 0, we have the 3 hece, -(+) ( )! ( )! ( xyz ) r ( xyz ) r ( xyz ) r ( xyz ) r -3 0!!( )!! 3 the 3 4 -(+) ( )! ( )! ( xyz ) r ( xyz ) r ( xyz ) r ( xyz ) r -3 0!!( )!!! Corollary 5 For =M3+ ad l= the the coeffcet of (xyz) t s ± ad for =M3- ad l= the the coeffcet of (xyz) t s ±(+3)/ ( )/3 ( ) ( )! (3 ) The t sequece for x y ( z) t( xyz) r (.) 3 0 eve( odd)!!( )! odd eve (3) for 0 we get, ( ) ( )!! t( xyz) t( xyz) 3!!!( )! The t sequece for x y ( z) ( )/3 ( ) ( ) ( )! ( )! (3) ( ( ) ( )) t( xyz) r (.) 3 3 0 eve( eve)!!( )! 0!!( )! odd odd (3) for 0 we get, ( ) ( ) ( )! ( )! ( ( ) (! 3 )) t ( xyz) ( ) t ( xyz) ( ) t ( xyz) 3 3!!!!( )! 0!!( )! Corollary 6 The total su of the exoets each ter add to ( > 0) ad ( < 0) f we clude the x,y,z degree (xyz) = 3, r = ad t = as a weghtg factor. 3 Equato (.7) the total su s 3 3 We therefore have for = M3, loe (xyz) /3 ters (Corollary 3) For =M3-we have ( xyz ) 3 t ad ( xyz ) 3 r ters > 0 ad vce versa < 0. For =M3+ we 4 4 have ( xyz ) 3 t ad ( xyz ) 3 r ters > 0 (Corollares 4,5) Coyrght Chrs Sloae 08
8 ON FERMAT S LAST THEOREM Corollary 7 The frst ter coeffcet s gve by the Lucas sequece over ad for = (re) s cogruet to od, all the other ters are cogruet to 0od. The frst ter coeffcet s geerated fro the Lucas fucto ad hece L s cogruet to od f s re [] We have whe ad 0 fro th. / s0 ( s)! t s!(( s)! Ths s a forula for the Lucas sequece hece ad for, L s cogruet to od f s re [] We ca see fro. that f the the deoator factorals are always less tha sof the there s a ter the uerator hece all ters whe are cogruret to 0 od Corollary 8 We ca aly the t, r, (xyz) reresetato to ay three varable equato of the for Ax a +By b - Cz c =D f we ake T equal the equato questo T=D ad X=Ax a,y=by b, Z=Cz c T deedet equato, ( s ) 3 3 ( )! / ( ) / X Y ( Z ) ( ) ( ) ( )T ( XYZ ) R (.3) 0 s 0 0 3 ( s )!! s!( )! od Where X,Y,Z reresets the ters the equato ad R X YZ XT For exale FLT we have that T= x y z 0 sot s 0 ad R x y z x T T x y z Hece, T Y XZ YT T Z XY ZT T ( s ) ( )! 3 / ( ) / 3 ( x ) ( y ) ( z ) ( ) ( ) ( )T ( x y z ) R 0 s 0 0 3 ( s )!! s!( )! od Hece f T 0 the we ecessarly have 0 ad we get the T deedet equato ( ) 3 / 3 ( )! ( ) ( x ) ( R ) 3 0(, eve ) 0!!( )! (, odd ) ( x ) ( y ) ( z ) y z Where R x y z (.4) (.5) Lea If x yz q ad x y z the x y z q y x z q z x y 0 od q 0 od 0 0 od, 0 od, x y z ca be factored to x yz sce s odd hece x y z 0 od q Wth x y z 0 the x y z x ( z y ) y z x z y ( z x hece f 0 od so ust 0 od x y z q y x z q Slarly, x y z x ( z y ) y z z ( x y ) x y z x y hece f 0 od so ust 0 od x y z q z x y q ) y x z Coyrght Chrs Sloae 08
9 ON FERMAT S LAST THEOREM Lea If x, y, z 0 od q ad x y - z 0, ad where q 0 s a re factor of ay of the syetrc arts R x y z 0 od q or R y x z 0 od q or R z x y 0 od q we ca wrte; g x od q 3 3 g y od q 3 3 g z od q Where g s defe d as the ultlcatve rtve root set geerator q has a rtve root g ad we use the rtve root as the geerator of the ultlcatve set of tegers od ulo q or g geerates all resdues od q, for 0 q Lets choose a g actg o x such that the resdue s od q hece, 3 3 3 g x od q Wth x y z y x z z x y 0 od q fro lea we have g x g y z 0 od q x g y z 0 od q ad wth g y z 0 od q hece, x g y 0 od q g y 0od q ad wth, g z y 3 x g z 0 od q 3 3 g z 0od q Therefore, a 0 od q g x od q (.6) 3 3 g y od q (.7) 3 3 g z od q (.8) 3 3 3 3 3 3 we have x y odq, z y od q, z x od q (.9) Lea 3 If g x od q the, g y od q ad g z od q therefore g y, g z od q 3 3 3 3 Whe g x od q, g y od q, g z od q the, ( g y )( g y g y ) 0 od q, ( g z )( g z g z ) 0 od q. If g y od q or g z od q, the g z od q or y od q resectvely fro g x g y g z 0 The fro g x g y z 0 od q we get 3 0 od q whch t s ot hece g y od q, g z od q, P Lea 4 We have quadratc cogrueces y ad z wth uque solutos for y, z 3 3 We ca wrte y z 0 od q, ( y z )( y y z z ) 0 od q If y z od q the g y od q, g z od q fro g x +g y -g z =0, 4g y 4g x z 0 od q, ad g x 0 od q, whch s a cotradcto 3 0 od q hece y z od q so y y z z 0 od q. So we have quadratc cogrueces y ad z wth uque solutos for y, z Coyrght Chrs Sloae 08
0 ON FERMAT S LAST THEOREM Ferat s Last Theore x +y -z = 0 has o o zero teger (ad hece ratoal) solutos whe >. Make re ( ). We assue that x, y, z have o coo dvsors for f they dd we could factor the out ad fd a ew soluto to the equato. If oe of x, y, z M 3 the the other varables ust be od 3 to satsfy x +y -z 0.e. (M3) ( M 3 ) ( M 3 ) 0 x y - z t 0 od 3 (5.0) If x, y, z M 3 the oly ( M 3 ) ( M 3 ) ( M 3 ) 0 s allowed, hece t M 3 M 3 M 3 M 3 t 0 od 3 (5.0) x, y, z 0 ad t x y z so f x y z the z x y d ad x y ( x y d ) 0 a equalty, hece t 0 Wth z x, y ad r' x yz r' 0 ad r' s odd as oe of x, y, z ust be eve ad t s eve. Furtherore, x yz t.e ( z y t ) yz t. We ca also show ths for r( y / t ), r( z / t ). ( 5. 03) Usg r' x yz, lets ake q a re decoosto factor of r' whch s odd 3 Proosto 5 For all the cases of q we show that t 0 od q or x, y, z share coo factor q or we get a cotradcto odulo q. If q 3 the t 0 od q as above, otherwse We eed to defe cases (lus sub -cases) whe q 3 : ) q 3s (5.05) b) q s, s M 3 (5.06) c) q 3s, s M (5.07) ) q 3s (5.08) Case. Wrte l uq v ad ake u v. Ths s a exteto of Bezoult's lea where q, are co-re or f q the GCD s ( v M ). Hece, l v q v v q q (5.09) Choose v such that v( q ) q l where l M 3 ad fro our ( T deedet) reresetato (Corollary 8) wth T 0 l 5 l 7 l 7 l 9 l l 3 ( l 3) ( )( ) ( l 9) ( )( )( )( ) l5 l l l ( x ) ( y ) ( z ) 0 l( xyz ) (( R ) ( xyz ) ( R ) 4 ( xyz ) ( R ) 3! 5! ( l ( )) ( l ( 4 )) ( l ( 3 )) ( )( )...( ) l3 ( )... ( xyz ) ( R ) ) (5.0)! LHS t od q.e. ( x y z ) Mq t Mq t od q f x, y, z Mq (fro Ferat's lttle theore) RHS 0 od q. ( R x y z 0x 0 Mr ) t 0 od q (5.) Reark : If oe of x, y, z cota q the so do the other varables ad we have a coo factor soluto whch ust factor out. Coyrght Chrs Sloae 08
ON FERMAT S LAST THEOREM Case b) Wrte l uq v ad ake u v 3, l v 3 q v v q 3 q (5.) l vs 3q where s s eve 3 l s odd M3 hece fro (C.9), T 0. l 5 l 7 l 7 l 9 l l 3 ( l3) ( )( ) ( l9) ( )( )( )( ) l5 l l l ( x ) ( y ) ( z ) 0 l( xyz) (( R) ( xyz) ( R) 4 ( xyz) ( R) 3! 5! ( l ( )) ( l ( 4)) ( l (3 )) ( )( )...( ) l3 ( )... ( xyz) ( R) ) (5.3)! 3 3 3 LHS x y z odq f x, y, z Mq RHS 0 od q Hece we get coo factor solutos ths case. Case c) Wrte l uq v ad ake u v, l v3s q where s s eve q M3 l s odd M 3. -3( xyz) 0 od q (5.4) l v q v v q q (5.5) l 5 l 7 l 7 l 9 l l 3 ( l 3 ) ( )( ) ( l 9 ) ( )( )( )( ) l5 l l l 4 ( x ) ( y ) ( z ) 0 l( xyz ) (( R ) ( xyz ) ( R ) ( xyz ) ( R ) 3! 5! ( l ( )) ( l ( 4 )) ( l ( 3 )) ( )( )...( ) l3 ( )... ( xyz ) ( R ) ) (5.6)! LHS t od q f x, y, z Mq RHS 0 od q t 0 od q (5.7) Case ) Wth q 3s, we ca factor r' fro R x ( yz ) by Lea For case we eed to uquely defe x +y -z =0 as oosed to x +y -z 0odq. Ths s doe va ths lea 5 Lea 5, If x +y -z =0 the Case ) We ca wrte x+y=c, z-y=a, z-x=b f x +y -z =0 f does ot dvde x,y,z Case ) If oe of x,y,z=m the we ca wrte z-y= - a,z-x= - b,x+y= - c resectvely Case ) Wth factor out x y C therefore C ust dvde z Make z cw where c s ay coo dvsor of ( x y) ad z Fro Corollory (see extract ) we have, - -3 ( - 3) -5 x y z ( x y){( x y) - ( x y) xy ( x y) x y...! ( -)/ ( -)/... ( x y )} z Hece, ( x y ) c otherwse xy would share all coo factors wth z (excludg ) whch s ot ossble the secal case. Therefore, f does ot dvde C the x y C c ad z cw ad z s dvsable by all of x y Slarly, z - y A ust dvde x ad fro Corollory (see extract ) z - y A a ad x au ad z - x B b ad y bv for a, b, c 0 (5.) Coyrght Chrs Sloae 08
ON FERMAT S LAST THEOREM Furtherore, f oe of A, B, C cota the we have, x au A t a t ad a / t (5.3) y bv B t b t ad b / t (5.4) z cw C - t c t ad c / t (5.5) Furtherore, ( x y) z M hece C z M but C z t / t Moreover we ow ca wrte, A B C -t Hece, a b c t abc (5.6) Case ) If dvdes, say C, ad hece z the we have C c However, the other ters A, B wll ot cota otherwse we have a coo factor. Because we have a coeffcet the last ter of corollary the shared coo factor betwee x y ad z does ot eed to be to the ower but oe less -. 5 5 5 5 So lets say 5 ad x y ad z w So fro corollary 5 4 5 w {( ) ( ) xy x y } w {( 5 ) 4 ( 5 ) xy x y } hece w M ad tur xy M gvg coo factor solutos - Therefore, x y c 3 3 3 3 where c s the other shared factors as above. If g x od q, g y od q, g z odq ad q 3s the fro lea 3, 4 ether: or s s s g z y odq ad g x z odq ad g y x od q (5.7) s s s g z y od q ad g x z od q ad g y x od q (5.8) Lea 6 s(4) s(7) We ca also wrte g (3l) s There ust exst oe 3l such that g z y odq Wrte, s g z y od q g z y od q g z y od q z y odq for l 0,,...( ) etc. for each of these cogrueces. s s( ) g z y od q s(3( ) ) g z y Factorg we get, od q s s( ) s( ) ( g z y)( g z g z y... y ) 4s 4 s( ) 4 s( ) ( g z y)( g z g z y... y ) s(3 ) s(3 )( ) s(3 )( ) ( g z y)( g z g z y... y ) We have rows wth each oe havg a uque soluto otherwse we get coo factor solutos f they share the sae soluto. Sce there are at ost uque solutos the secod brackets there ust be oe soluto the frst braket o ay artcular row. Coyrght Chrs Sloae 08
3 ON FERMAT S LAST THEOREM Lkewse for the other relatos so we ca wrte a table as follows; Table g s s s s g z y od q g x z od q g y x od q 4s 4s 4s g z y od q g x z od q g y x od q 7s 7s 7s g z y od q g x z od q g y x od q s s g z y od q g x z od q g 3 y x od q s or g s (3 ) s (3 ) s (3 ) s g z y od q g x z od q g y x od q g s s s z y od q g x z od q g y x od q 5s 5s 5s g z y od q g x z od q g y x od q 8s 8s 8s g z y od q g x z od q g y x od q s s g 4 z y od q g 5 x z od q g 6 y x od q s (3 ) s (3 ) s (3 ) s g z y od q g x z od q g y x od q Lea 7: Whe oe of x +yz 0odq, y +xz 0odq, z -xy 0odq the two of the solutos to lea 6 ust fall o the sae row. There are a uber of ways to show ths. We have x yz 0 od q hece we have, g x y od q for soe g f x, y, z Mq g x g yz 0 od q y( x g z) 0 od q so we have g y xod q ad g x z od q (so 3 the to table) For y xz 0 od q, 3 For z xy 0 od q, For x +yz ad our two lea 7 solutos beg o the sae row, they ust be o the s or s row. because by lea5 take together we have g ls (x+y) (x-z)odq g ls c -b odq But we ust have a g c -bodq f a,b,c Mq hece rasg both sdes to eas l=m There s oly oe exoet o each of the tables f s M ad ths occurs /3 artto ots; s ad s. For Case of lea 5 deedg o whch x,y,z s dvsble by we choose a decoosto ter such that we have two of a,b,c gvg the /3, /3 artto ots; If x=m the choose x +yz If y=m the choose y +xz Coyrght Chrs Sloae 08
4 ON FERMAT S LAST THEOREM If z=m the choose z -xy These /3, /3 ots ea t 0odq as follows. We work out what a s ters of b ad c for exale. Ths s deedet o what the frst colu soluto s ad s oly deedet o the s or s soluto gve by lea 5 ad 7 Wrte s s g z g x a od q s ss s ( ) ( ) od g g z g z x a q ad wrte s s g z g y c od q s ss s ( ) ( ) od g g z g z y c q s s g b g a a c od q (5.9) s s (Hece t s deedet of ) but g c b od q 4s ( g ) a ( g ) c od q (5.30) s s s 3s Now ( g )( g ) g od q sce g od q Next we have fro (5.9) 4s 4s s s s s g b g a g a b od q b g a od q or g b a od q Hece we ca wrte; s g ( t) t od q s s g a c od q (5.3) 4s s s ( g ) b g ( g ) a od q (5.3) s s s g a g b g c c a b od q ( g )t 0 od q ad sce ( g ) 0 od q t 0 od q (5.33) Therefore, t ust also be dvsble by q=3s+, s M. However, ths s ot ecessarly true whe q=3s + for >. To geeralze to all we do the followg. Coyrght Chrs Sloae 08
5 ON FERMAT S LAST THEOREM Startg wth q 3s we ca wrte fro the lea 6 s s s g z y od q g x z od q g y x od q 4s 4s 4s g z y od q g x z od q g y x od q g z yod q g Lea 8 If q 3s ad x y z 0 the we also have x y z 0 Table 7s 7s 7s x z od q g y x od q s s 3s g z y od q g x z od q g y x od q (3 ) s (3 ) s (3 ) s g z y od q g x z od q g y x od q s Assue the g soluto ad,, are solutos resectvely. 3 Oe ca see f q 3s we have the exoets of Fro lea 5 we have x au, y bv, z cw g beg ultles of s There s o loss of geeralty f we wrte g x z dq where d 0 od a or d 0 od u, sce x au ad f z M ( a, u) they would share coo factors a, u. s Moreover, there ust exst a such that g od q for soe, 0 od q where s the resdue od q whereby rsg t to we have zod q Exale, f z od5 the f 5 s Also wthout loss of geeralty we ca wrte, g hq ( h 0 od a or h 0 od q) s We ca add g h to both sdes for h 0,,..., 0 a or 0 u thereby gvg us, Hece, where ' h g ( h ) ( h g h ) q (5.34) s s s s g h g h q z f q ( ') ( ( ) ) ' (5.35) We ca get all resdues od a or od u o the RHS by adjustg h such that f ' d od a or f ' d od u because z, g, q 0 od a or 0 od u (Note: f a g or u g the choose aother rtve root as q s ow large 3s ad has ay rtve roots) Therefore, f we ake the resdue d od a or d od u we have s s g x g Moreover, sce x au the Ma or Mu Exale : s Let 5, 5, 7, od5, 5 od, (0,,, 3, 4, 5, 6) 0,,, 3, 4, 5, 6 od ' Maq, or Muq (5.36) q a z d q f h g a ( 0 q) (5 od 7) q, ( q) ( od 7) q, ( q) (4 od 7) q, ( 3 q) (od 7) q, a a ( 4 q) (6 od 7) q, ( 5 q) (3 od 7) q, ( 6 q) (0 od 7) q Hece f d d ' a 5 the we choose ( 0 q) 5 od a, 3 Furtherore, we ca get all resdues the coefcets of the a, a, a... a ters od a (that beg d ', d ''...etc.) 3 or u, u, u... u od u by addg ultles of a, u resectvely ad sce d ' (0,... a ) od a we ca ake ' '... ' s s M a or a a gvg g x g ' ' a... ' a hece we ca ake x M a. Lkewse x Mu Exale:as above f we add 7 q, (0 7 ) to both sdes of [5.9] for ( 0 q) 5 od a, we get ((3 od 7)7 5 od 7) q, ((od 7)7 5 od 7) q, ((6 od 7)7 5 od 7) q, ((4 od 7)7 5 od 7) q, (( od 7)7 5 od 7) q, ((0 od 7)7 5 od 7 ) q Coyrght Chrs Sloae 08
6 ON FERMAT S LAST THEOREM So we ca add ultles of aq, a q... a q ad uq, u q... u q to both sdes of (5.34) to get the exoetato of x ( au) ( x ') where x ' s obvously less tha x or we could vew t as x s a ower. Reark, we ca do ths because we have g our table whch ca be elated (5.36). If we dd ot have the exoet the we could ot elate t leavg g ters eag a, u would ot ecessarly dvde x. Lkewse, we ca do ths for the other colus of table to get y ( bv) Hece, we ca wrte ( y '), z ( cw) ( z ') x ' y ' z ' 0 (5.37) Oe ca see fro Lea 5 x ' y ' c ', z ' y ' a ', z ' x ' b ' ad a ', b ', c ' do ot ecessarly equal abc,, Now we have fro Lea that x y z y x z z x y ' x ' y z ' y ' x ' z ' z ' x ' y ' 3 3 3 Therefore, we get x ' y ' z ' od q We have fro table ad lea 6 s s g x ' z ' od q, g y ' x ' od q However, ow M for we have z ' x ' b ', x ' y ' c ' whch s the sae as s so f we have ' ' od the ad ths s the / 3, / 3 soluto ots. Therefore ' ' ' ' 0 od ad t x y z 0 od q. g x z q M t x y z q 3 3 3 3 We ca reeat ths for q 3s to gve x '' y '' z '' 0 etc.... q 3s......... x''' y''' z ''' 0 ad we get saller trles x ''' x '' x ' x etc. as creases. The above arguets hold for all therefore all q ' s dvde t Theore. If x y z 0 ad suose x, y, z are arwse co re the ay re factor q of ( x yz ) wll dv de t where t x y z Corollary 39 Theore. s vald for ay re factor q of ( y xz ) or ( z xy ) Ths follows fro the syetry of the roble ad ethods above Closg Arguet If we have coo factors, the secal case x + y - z 0 loses o geeralty assug that the greatest coo dvsor of x; y ad z s. Hece t ust cota all the re decoostos q of ( x yz ). We ca ow use 3 equalty arguets for exoet x y z cogruret to odulo 3, cogruret to odulo 3, ad exoet q. We eed our equato. res ad Corollares 9,5,8. We wll wrte hghest ower of q( q ) q( q ) q( q 3 ) q( q ) r x yz q q q 3...q q dvdg x yz. where q s re ad q( q ) s defed to be the Coyrght Chrs Sloae 08
7 ON FERMAT S LAST THEOREM Lea 9 : For exoet 3 cogruret to od ulo 3 ad q ex oet the t M r. If q( q ), q( q ), q( q )... q( q ) are all equal to the t M r but x yz t fro [5.03]. 3 Hece, we have a equalty ad cotradcto. Therefore oe or ore of q( q ), q( q ), q( q )... q( q ) Lets frstly assue q( q ) 3 x y z 3 Mq ( xyz ) t ( x yz xt t ) ( xyz ) ( xyz ) t Mq because the last ter ( x yz xt t ) ( xyz ) Mq t Mq ( xyz ) q x; y z t M r x yz t q( q ) t Mq t r 3 6 tr Mq 6 4 ( xyz ) t Mq t x yz x yz t q( q ) 4 t Mq 8 s becoe Mq hece, 8 ad 8 q( q ) q( q ) t Mq. q x yz q( q ) t q( q ) Fro corollares 9,5,8 we have 0 (5.38) Oe ca see hece because gves us coo factors ad the but hece we have a equalty ad cotradcto as before. Lets ake 3 we stll get so the the hgher ters or x y z Mq ( xyz ) t ( x yz xt t ) ( xyz ) we wrte, 0 (5.39) Hece, ad but hece we have a equalty ad cotradcto as before. Next ake 4 we stll get ad our hgher ter x y z Mq ( xyz ) t ( x yz xt t ) ( xyz ) t Mq ad we get our cotradcto aga. Therefore, for ay we get As creases owers creases owers so the ust the hgher ters cotag hgher t, r cobatos whch tur creases t ad we cotue to get the cotradcto as q( q ). Oe ca see ths s true for all q( q ), q( q ), q( q )... q( q ) so we ca coclude t M r whch s a cotradcto r t. 3 3 Lea 0: For exoet 3 cogruret to od ulo 3 ad q ex oet the t M r. If q( q ), q( q ), q( q )... q( q ) are all equal to the t M r but x yz t fro [5.03]. Hece we have a equalty ad cotradcto. Therefore oe or ore of q( q ), q( q ), q( q )... q( q ) Lets frstly assue q( q ) 3 3 3 Fro corollares 9,5,8 we have x y z Mq ( xyz ) t ( xyz ) ( x yz xt t ) 0 (5.40) q( q ) t Mq t r 3 t r 6 Mq hece, xt Mq the last ter r but x gves us coo factor q x; y ad z so t Mq the t M r but x yz t hece we have a equalty ad cotradcto as before. Lets ake 3 we stll get ad hgher ters ad we wrte, 6 3 x y z Mq ( xyz ) t ( xyz ) ( x yz xt t ) 0 (5.4) 3 3 hece ad we get a cotradcto as before 3 4 4 q( q ) t Mq xt Mq t Mq q( q q( q ), we get t Mq ) q xt Mq t Mq Next ake 4 we stll get ad hece ad ad we get a cotradcto aga Therefore, for ay 3 ad we get a cotradcto as ( q ). Oe ca see ths s true for all q( q ), q( q ), q( q )... q( q ) so we ca coclude t M r whch s a cotradcto r t. Fro Corollary 0, that the coeffcets of all ters are cogruret to 0 od excet the frst we eed to ake sure the cotradcto stll works for q or r cotas a ower of Coyrght Chrs Sloae 08
8 ON FERMAT S LAST THEOREM Lea For exoet 3 ad q the lea, are uchaged; t M r Fro corollary 0 the coeffcets of all ters are cogruret to 0 od excet the frst whch s cogruret to od so f t M the we dvde out leavg the relavet ed ters coeffcets equal to so we have the sae for of the equato but wth the frst ter t deshed by so that ter s M however ths ter s rrelevat the above arguets so our cotradcto holds ths case too. For 3 we get drectly 3xyz 0 od q f q 3 hece coo factor solutos aga. If q 3, ad sce t M3 the 3xyz M3 therefore, oe of x, y, z M3 ad the so ust the other varables hece share a coo factor 3. (5.4) Reark t 0 eve f x or y s egatve for we would just rearraged the equato for odd exoets.e f y s a egatve teger x y z becoes y z x ad x becoes the hgher ter but that case we would just wrte x z Wth 4 solved by Ferat we ca coclude there are o dscrete solutos to Ferat's equato for. Coyrght Chrs Sloae 08