A solution of certain Diophantine problems

Transcription

1 A slutin f certain Diphantine prblems Authr L. Euler* E7 Nvi Cmmentarii academiae scientiarum Petrplitanae 0, 1776, pp Opera Omnia: Series 1, Vlume 3, pp Reprinted in Cmmentat. arithm. 1, 189, pp. -9 [E7a] Prblem 1. Find tw pairs f squares xx, yy and tt, uu s that (xx + yy)(ttxx + uuyy) as well as (xx + yy)(uuxx + ttyy) becmes a square number. Analysis. 1. First, it is well knwn, whatever number pairs will have been fund fr x, y as well as fr t, u, likewise their multiples, such as αx, αy and ζt, ζu als satisfy the questin; and s it is cnvenient t restrict the prblem, s that bth x and y as well as t and u are numbers [relatively] prime t each ther.. Let us begin with the first expressin (xx + yy)(ttxx + uuyy), which having been set equal t the square (xx + yy) xxyy (pp + qq) results in ttxx + uuyy = xxyy( xx + yy)(( pp qq) + ( pq) ) frm which it is cncluded and thus it will be 3. Nw fr the ther frmula, since it is it will result that tx = xy( x( pp qq) + pqy); uy = xy( y( pp qq) pqx) t = xy( pp qq) + pqyy; u = xy( pp qq) pqxx. ty = xyy( pp qq) + pqy ; ux = xxy( pp qq) pqx 3 3 ttyy + uuxx = xxy ( pp qq) + pqxy ( pp qq) + ppqqy + x yy( pp qq) pqx y( pp qq) + ppqqx which expressin, because it clearly is divisible by xx + yy, is changed int * Translated by Cynthia J. Huffman, Department f Mathematics, Pittsburg State University, Pittsburg, KS. 017 (With much thanks and appreciatin t the annymus reviewer.)

2 ( xx yy)( xxyy( pp qq) pqxy( xx yy)( pp qq) ppqq( x xxyy y ) ).. Nw since this expressin multiplied by xx + yy shuld prduce a square number, we will have the fllwing expressin t be reduced t a square: ppqqx pq pp qq x y + p 6p q + q x y + pq pp qq xy + ppqqy 3 3 which clearly is a square, if x=y hwever, frm here n, we reasnably exclude this easiest case; since the full questin returns in this case t prving that ( tt + uu) is square. 5. But setting the frmer frmula equal t this square after the matching terms have been deleted it is ( pqxx ( pp qq) xy + pqyy) p 6ppqq+ q x y + pq( pp qq) xy = p + 6ppqq+ q x y pq( pp qq) xy 3 3 hence 8 pq( pp qq) y = 1ppqqx frm whence this slutin f the prblem is deduced: x= pp qq ; y = 3 pq; and frm here furthermre t pq p ppqq q u= pq pp qq. = 6 ( + + ) and 6. Ntice therefre extending the first slutin infinitely, because it is allwed t take by chice the numbers p and q; f curse after reducing the numbers t and u t lwest terms, and because it is in the same way whether they may be psitive r negative, we will have * 3 x = pp qq t = p + ppqq + q = xx + yy 1 y = 3 pq; u = ( pp qq) = xx ; 3 ; frm here is fund, xx + yy = p + ppqq + q 9 ttxx + uuyy = xxyy ( xx + yy)( pp + qq) = xx ( xx + yy) xx + yy 16 1 uuxx + ttyy = ppqq( xx + yy)( p + 7 ppqq + q ) = ( xx + yy) xx + yy. 7. In rder that we find ther slutins, let us put the square rt f the abve frm(ula) ** : * Nte: The riginal article has pp in the expressin fr t instead f p. ** Nte: riginal article has first term as ppxx.

3 pqxx pp qq xy pqyy + Ayy after the square f which has been set equal t the frmer [expressin], the equatin will prduce AA Apq yy A pp qq xy + Apq ppqq xx = 0, in this case, if A=pq the preceding slutin appears; while putting A=pq it is + 3pqy + pp qq x = 0, which crrespnds equally with the frmer. 8. Let us set A= ppand this equatin will appear pp + pq yy + pp qq xy pq + qq xx = 0 which divided by x+y gives pp + pq y pq + qq x = 0 frm which results = ( + ) and y q( q p) x p p q ( )( 3 ) = + then certainly u= pqq p+ q qq+ pq+ pp. 9. Ntice therefre anther slutin happens, different frm the preceding and infinitely extending, the numbers t and u having been reduced t lwest terms = ( + ); t = p( q+ p)( pp+ pq+ 3qq) x p p q = ( + ); u= q( p+ q)( qq+ pq+ 3pp) y q q p and hence it is discvered: xx + yy = p + p q + 8ppqq + pq + q 3 3 ( ) ( ) ttxx + uuyy = p + q q + p pp + qq xx + yy ( ) uuxx + ttyy = ppqq pp + pq + qq xx + yy 10. Having put A pp divided by y = prduces x gives ( pp pq) y ( pq qq) x = 0and therefre = ( ); t = p( p q)( pp pq+ 3qq) x p p q = ( ) ; u= q( p q)( qq pq+ 3qq) y q p q pp pq yy pp qq xy + pq qq xx = 0 which

4 xx + yy = p p q + 8ppqq pq + q 3 3 ( ) ( ) ttxx + uuyy = p q p q pp + qq xx + yy ( ) uuxx + ttyy = ppqq pp pq + qq xx + yy This slutin hwever is nt different than the preceding ne; nr d the cnditins A= qqand A= qq prvide different slutins. 11. Methds exist, f which ther [slutins] can be fund by the help frm ne knwn slutin; hwever they bring abut excessively cmplex calculatins. Therefre, it is permitted t btain ( 3 ) ( 3 ) x = q pp qq pp qq y = pp + qq p pp + qq ± q pp + qq certainly paragraph supplies apprpriate values fr t and u. Slutin I. x pp qq In this slutin, the rati f the numbers x and y is = whence they are fund frm y 3 pq t 3xx + yy the perpendiculars f a right triangle; then, certainly the rati = ; frm which the u xx simpler slutins are: 1. x = 5; y = ; t = 91; u = 5. x = 7; y = 5; t = 7; u = 9 3. x = 5; y = 9; t = 399; u = 5. x = 3; y = 10; t = 7; u = 9 5. x = 11; y = 1; t = 117; u = x = 15; y = 7; t = 871; u = 5 7. x = 16; y = 5; t = 17; u = 6 8. x = 16; y = 9; t = 73; u = 6 9. x = 7; y = 18; t = 13; u = x = 13; y = 0; t = 107; u = 169 Slutin II.

5 Here the rati f numbers x and y is ( + )( ) ( + )( ) t p q p pp pq qq = u q p q qq pq pp ( + q) ( + ) x p p = y q q p ppy + pqy = qqx + pqx, it is discvered that hwever [the rati] f numbers t and u [is] if we bserve when the numbers x and y are given, fr p q ; x y+ xx xy+ yy = y frm which the character f the numbers x and y is established in this, s that xx xy + yy is square; f whatsever kind f numbers are fund with ease; [let] it be xx xy + yy = zz ; and it will be hence it results p x y+ z x p z y = = ; r = ; q y y x+ z q x z p+ q x y z z+ y p q+ p z y z+ x y = = and = q+ p z y+ x z+ x q p+ q z x z+ y x s that ( z y)( z+ x y) = ( z+ x y)( z x y)( z x)( z+ y x) = ( z+ y x)( z x y) frm which Next is ( ) ( ) p q+ p z+ x y =. q p+ q z+ y x ( x y) + ( x z) ( x y) + ( z y) pp + pq + 3qq z + y x = = ; qq + pq + 3pp z + x y frm this finally it is elicited that ( z+ x y)( z x+ y) ( + )( + ) t = u z y x z x y. Thus after the numbers x and y have been fund f the type that make xx xy + yy = z ratinal: hence it is btained that ( ) ( ) t = z+ x y z x+ y = xx+ yy+ x y z u= z+ y x z y+ x = xx+ yy x y z

6 indeed likewise in the same manner ( ) ttxx + uuyy = xx + yy xx xy + yy + x + y z uuxx + ttyy = xx + yy xx xy + yy x + y z 1 ttxx + uuyy = 9 xx + yy x + y + z z x y 1 uuxx + ttyy = 9 xx + yy x + y z z + x + y ( ) ( ) Nw hwever we therefre easily discver apprpriate values fr x and y, let us cnsider x as having been given and [if] we set z = y vit will be xx vv vv xx xx xy = yv + vv and y = = ; v+ x v x Having assumed whatever value f x itself, the integer cases fr y must be rted ut: hwever it must be nted, that x cannt be assumed ddly even [i.e. nt cngruent t md ], because y wuld als be even:. x y z t u

7 Prblem. Find tw pairs f squares xx, yy and tt, uu, s that ( ttxx + uuyy )( uuxx + ttyy) is a square number. Slutin. This prblem btains by lt the same slutin, as the preceding, and the same numbers taken fur at a time t be fund fr x, y, t, u satisfy. Thence therefre the simplest slutin is x = 3; y = 5; t = 11; u= 5 [Nte: bth the riginal E7 and the reprint E7a have a typ f x =.] frm which results and therefre ttxx + uuyy = ; uuxx + ttyy = 3 65 ( ttxx uuyy )( uuxx ttyy) =. Mrever this slutin is particular, nt nly because f that reasn, fr which it was such, but likewise this prblem is seen t admit infinite slutins, which d nt agree with the preceding. Indeed it is pssible t ccur, that this frmula ( ttxx + uuyy )( uuxx + ttyy) is square, even if neither f the preceding ( xx + yy)( ttxx + uuyy), and ( xx + yy)( uuxx + ttyy) shall be square, f which it suffices t give a single example: x = 973; y = 63; t = 973; u= 181 fr indeed uuxx + ttyy = is duble a square and duble a square. ttxx + uuyy = is

8 T this pint, behld laying ut anther slutin mre bradly, x = 3n + 6 mmnn m ; t = mx x = 3m + 6 mmnn n ; u = ny The ratinale f the discvery f which is easily understd, fr having put t = mx and u = ny, it is *, ttxx + uuyy = mmx + nny and uuxx + ttyy = xxyy mm + nn and thus this is the frmula which shall be reduced t a square ( mm + nn)( mmx + nny ) which by having made x= v+ z and y = v z leads thrugh t that slutin; hence mrever the preceding slutins are nt btained. Prblem 3. Find tw pairs f squares xx, yy and tt, uu as much this number ttxx + uuyy as this ne ttyy + uuxx is a square. Slutin. By the recently delivered slutin t the previus prblem, fr m and n having been assumed numbers f the kind, s that mm + nn is square. Thus if it happens m = and n = 3, it results in x = 851; t = 30 y = 1551, u = 653. [Nte: bth E7 and E7a have u = 3653.] But frm the first great prblem very elegant slutins are achieved, t which even mre than the tw prescribed cnditins and this third is fulfilled s that xx + yy is als square. While a secnd slutin f the first prblem prvides ne case, where xx + yy certainly becmes square frm whence results x = 8; y = 15; t = 99; u= 190 xx + yy = 17 ttxx + uuyy = ttyy + uuxx = If in additin this cnditin has been added, that xx xy + yy as well wuld be square the same slutin wuld cmplete the wrk. Hwever a slutin f this kind is drawn ut by means f the * Nte: the reprint E7a has a missing expnent n the first equatin: mmx + nny.

9 3 3 desired numbers x and y s that this expressin x x y+ xxyy xy + y shuld be square t which frmerly as befre it ught t be fund the numbers t and u. This prblem f Diphantus t be discussed has presented a Gemetry prblem prpsed by Schtenius [Frans van Schten], where the triangle base, perpendicular, and a rati f sides having been given, the sides themselves are sught. This prblem admits a duble slutin, either f which in rder t prduce the sides expressed ratinally, the task is induced by that prblem f Diphantus. If in fact the base f the triangle is set = a, the perpendicular = b, and the sides in rati m:n; the sides having been called mz and nz; it is first necessary fr a and b t be expressed as a= ( mm nn)( xx+ yy) and b= mnxy, Then certainly fr z this duble expressin is btained: ( ) ( ) z = xx+ yy m n xx+ m+ n yy z = xx+ yy m+ n xx+ m n yy and that the tw m+ n= b; m n= uare in fact ratinal results frm ur prblem f Diphantus. Frm which therefre it will be a mst simple case t have btained frm which these facts are prduced: x = 3; y = 5; t = 5 and u= 11, the rati f the sides m: n = 8 :17 the base f the triangle a = 33; and the perpendicular = 8, frm which are discvered, the sides themselves: r if integers are t be btained will btain the rati f the sides 8:17 keeping either mz = and nz = r mz = and nz = 5 5 the base a = 95; and the perpendicular b= 0 either mz = 700 and nz = 5 r mz = 109 and nz = 663.