A SOLUTION TO THE ARITHMETIC PROBLEM OF FINDING A NUMBER WHICH, DIVIDED BY GIVEN NUMBERS, LEAVES BEHIND GIVEN REMAINDERS

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1 A SOLUTION TO THE ARITHMETIC PROBLEM OF FINDING A NUMBER WHICH DIVIDED BY GIVEN NUMBERS LEAVES BEHIND GIVEN REMAINDERS LEONHARD EULER 0 Prolms o this sort an oun in ommon ooks o arithmti rywhr ut thir solutions rquir mor stuy an lrnss than thy appar. Although in many algorithms ha n propos y whih w an otain a solution through ingnuity nrthlss thy ar ithr not gnraliz to ry as an aus th irumstans o th qustion ar too inrquntly ix thy ar o no gratr us; or thy ar n just als. Th onstrution o magi squars has n han own long ago in th arithmti traition; howr n th grat ingnuity o La Hir an Sauur is insuiint or ining a solution. Likwis y a similar mtho th prolm rsol or narly all ass so that w in a numr whih ii y an las as a rmainr ut is iisil y 7 without rmainr: truly nowhr has a suital mtho or soling prolms o this typ n shown; a solution may oun in a grat numr o ass an through tsting alus. 2 I th iisors ar small so ar as in this xampl th solution is asily oun y trial-an-rror; prolms o this sort woul mor iiult howr i th propos iisors wr ry larg. An so th mtho or soling prolms o this typ n now is not gnral xtning qually to oth larg iisors an to small ons; not in ain o I ha onin in my ollt work a mtho y whih without tsting alus suh prolms an rsol or n th largst iisors. 3 Hnorth I will prsnt my work on th prolm in th orr in whih I ha onsir it an I gin with th simplst as in whih a singl iisor is gin an w sk a numr whih whn ii y sai iisor las a gin rmainr. To wit a numr z is sought whih whn ii y a numr a las p as a rmainr. Th solution to this qustion is ry simpl or it will z = ma + p with m noting any intgr; it is lar that this solution is unirsal an all solutions ar inlu in this xprssion. Moror rom this w an isrn that i w know on solution thn ininitly many solutions an oun whn w a multipls o a to it or i it is possil sutrat multipls o a. But th smallst solution will p that is 0a + p whih gis th solution a + p an thn 2a + p 3a + p 4a + p an all th numrs onstituting th arithmti progrssion with onstant irn a ollow. 4 Haing propos this w pro to th as in whih multipl ix iisors with orrsponing rmainrs ar propos. For howr many propos iisors thr will th qustion an always ru to th as in whih two iisors ar gin. How I will show in th ollowing pags. Consir th numr z whih whn ii y a las p an y las q; an lt a gratr than. Thn aus z was rquir to la rmainr p whn ii y a it must o th orm ma + p. From th othr onition in whih z ii y las rmainr q w ha z = n + q. For this rason sin ma + p = n + q 0 Originally pulish in Commntarii aamia sintiarum Ptropolitana 7 (740) pp.4-. Translat y Crystal Lai Bnjamin Russll an Kathy Yu unr th suprision o Stphn Knny. Rrring to P. La Hir s (40-78) Ds quarrz ou tals magiqus pulish in 93 an J. Sauur s (53-7) Constrution gnral s quarrs magiqus pulish in 70.

2 2 LEONHARD EULER 0 intgrs must trmin an sustitut in pla o m an n suh that ma + p = n + q an haing oun suh intgrs ma + p or n + q will th solution z. 5 Baus w ha ma + p = n + q n = ma+p q an haing st = p q thn n = ma+. Hn m shoul in suh that ma + is iisil y. Baus a > lt a = α + ; thn n = mα + m+ ; an lt m + amit iision y. Sin α an ar known quantitis oun y iision o a y thn α is th quotint an is th rmainr. Thn lt A = m+ so m = A ; osr that A must iisil y. I is iisil y th algorithm an omplt. [Without loss o gnrality] lt A = 0 so that m = an z = a + p. En i w want to aoi ngati numrs th xprssion amits ininit positi solutions or z. I howr is not iisil y whih woul ha ma A ii y an all th quotint β an th rmainr. By this A ought to an intgr lt us all it B whih will mak A = B+ B = 0 an A = β an m = an intgr 2 st = β + ; 3 that is = Aβ + A = m an A. Now shoul iisil y mak B+. I howr is not iisil y st = γ + ; an A = Bγ +. An st B+ = C so that B = C. Now i is iisil y st C = 0 an it will that B = an A = γ βγ an m = ; shoul C not yt an intgr st = δ + an thn B = Cδ + ; an mak C = D so that C = D+ until it is sn whthr is iisil y or not an in lattr as th algorithm ought to rpat as ao. 7 Baus a > an > an > an so on th squn a... is rasing so it ought to onrg to som alu whih is a iisor o. Th squn... is th ontinuing rmainr o th stanar algorithm y whih th gratst ommon iisor o a an is ustomarily oun whih I pla hr: n = ma+ a α a = α + m = A β = β + A = BC+ γ = γ + B = C δ = δ + C = D+ ɛ = ɛ + g D = E g g ζ = ζg + h E = F g+ h h g η g = ηh + i F = Gh i i h θ h = θi + k G = Hi+ k k 8 This algorithm y whih it is ustomary to in th gratst ommon iisor o th numrs a an shoul ontinu until it rahs th rmainr whih ought to ii. By th ollowing mtho whih w ha alray oun w instigat m. I is iisil y thn A = 0 an m =. I is iisil y thn B = 0 an A = an m = = β aus = β +. Thus th alus o m ar mor asily oun or th alu o A is. () m = A (2) m = B+β (3) m = C (+βγ) 2 By whih is mant it is asy to pik A suh that w nsur A is an intgr. 3 In th original txt th β w ha hr is a ariant ta. Howr it rrs to th sam ojt as whn Eulr uss β itsl thror or th sak o larity w ha us β in ths situations.

3 A SOLUTION TO THE ARITHMETIC PROBLEM 3 (4) m = D+(δ+βγδ+β) (5) m = E (δɛ+βγδɛ+βɛ+βγ+) g () m = F +(δɛζ+βγδɛζ+βɛζ+βγζ+ζ+δ+βγδ+β) h... Osr that th sign o altrnats thus: Thn th oiints o ollow this pattrn: β γ β βγ δ + βγδ + ɛ δ + β βγδɛ + δɛ ζ + βɛ + βγ +... whos progrssion is th sum o th pring trm multipli y th pring suprsript an th trm pring that on. 9 I is iisil y thn m = 0; i is iisil y thn m = aus A = 0; i is iisil y thn lt B = 0; an thn m = β. Whn th ollowing law is gin: I an intgr thn m = 0 m = m = + β m = (βγ + ) m = + (βγδ + δ + β) g m = g (βγδɛ + δɛ + βɛ + βγ + ) h m = + h (βγδɛζ + δɛζ + βɛζ + βγζ + βγδ + ζ + δ + β) t. Now i ths alus o m ar sustitut in th quation z = ma + p th ollowing is otain: I an intgr thn z = q + = q + z = q α z = q + (αβ + ) z = q (αβγ + α + γ) z = q + (αβγδ + αβ + αδ + γδ + ) g z = q g (αβγδɛ + αβγ + αβɛ + αδɛ + γδɛ + α + γ + ɛ t. 0 To in th numr z whih ii y a las p an y las q with p q = propos w ha th ollowing rul: th algorithm is uilt to in th gratst ommon iisor o a an an rom this iisor w an prou a rmainr whih is th iisor o itsl an whatr quantity w otain y th iiing y th rmainr all it Q is whr th algorithm ns. Thn th quantitis α β γ t. rsulting rom this iision an writtn as a squn ma o ths numrs: th nw sris α αβ + αβγ + α + γ t. whih is orm rom th prious squn o numrs an in this ashion it shoul ontinu. Unr this nw squn altrnating signs ar writn: + + t. an th last trm along with its sign is multipli y Q an also y th smallr iisor ; an to this prout is a th rmainr q orrsponing to th iisor. In this mannr th solution will onstrut. W an in on solution z in this way rom whih many othr solutions ar oun. For i z ii y a las p an y las q th numrs a + z 2a + z an ma + z will ha th sam quality. Multipls o th prout a an ontinously a or sutrat i a an ar rlatily prim. But i a an ar omposit numrs thn it suis to tak thir last ommon iisor; aing an sutrating multipls (rom z) yils mor solutions suh that i th smallst ommon iisor is M all solutions w want ar o th orm mm + z.

4 4 LEONHARD EULER 0 2 Baus this algorithm is st illustrat y xampls w sk a numr whih whn ii y 03 has a rmainr o 87 an whn ii y 57 has a rmainr o 25. Lt a = 03; = 57; p = 87 an q = 25 an = 2; hn I isplay this algorithm: = 3 = Q Now 9 3 = 279; an on solution is ; sin this is ngati a to it that is an thn = 735 whih is th smallst suh numr; thus ry solution is o th orm m Hratr w sk a numr whih whn ii y 4 las a rmainr o 0 an whn ii y 29 las a rmainr o 28. In this xampl I mploy a shortut aus in othr similar omputations it will ha grat utility: whn iiing y 29 th rmainr woul 28 - this an also onsir rmainr o y th sam iision i it is known that th quotint is gratr y. I us as th rmainr o iision y 29; thn a = 4 = 29 p = 0 an q = ; hn =. So I pla th algorithm low: = = Q Thn 7 = 87; an a solution is Sutrating y yils =. Ery solution an writtn in th orm m Mor gnrally th algorithm ontinus so that atr th numr Q is multipli y th last numr o th squn this prout is thn ii y th largr iisor a. Th rmainr o this iision multipli y th iisor an thn inras y th rmainr q will gi th solution. Furthr th solution oun y this mtho is in at th smallst numr that satisis th quation. Thrror y this iision opration w an gt a positi rmainr n i th iin is ngati. In th irst xampl rom stion 2 w ha 279 whih ii y 03 las +30. Thn th smallst solution is = Now that this is possil xampls whih amit no solutions at all may prsnt suh as i a numr is sought whih whn ii y 24 has a rmainr o 3 y 5 a rmainr o 9; in y th lattr onition suh a numr ought to iisil y 3 ut y th ormr it shoul not. Th ollowing

5 A SOLUTION TO THE ARITHMETIC PROBLEM 5 prinipl las to th sam onlusion that no suh rmainr whih is lt hin whn that is 4 iis is arri at xluing 0 as th ollowing algorithm shoul show In w annot apply our algorithm to xampls o this kin i th iisors a an ar rlatily omposit; ut shoul thy oprim thn solutions ar always al to oun. Furthrmor shoul th iisors a an omposit numrs an not iisil y th gratst ommon iisor o a an thn th prolm always las to asurity. This is th ritrion y whih it an i whthr th prolm amits a solution or th algorithm is gun. Now that this unirsal mtho y whih ry ariation o th prolm is al to asily sol has n xplain anothr rul an orm rom it whih is asily arri out ut is simplr. Moror this ollows i haing isor th alus o z ao in pla o α β γ t. thir alus rom th quations a = α + = β + t. ar sustitut. For i th algorithm to in th gratst ommon iisor o a an is gun thn rom it th rmainrs t. ar oun; all this numr z = ( q + a a + + ) ontinuing this sris until is iisil y anothr ator o th nominator. For xampl i w sk a numr whih whn ii y has a rmainr o an ii y 9 has a rmainr o 7 thn a = = 9 p = q = 7 an =. Hn: So z = 7 9 ( ) 7 2 = = 3 = 47. Thror all numrs m that is m satisy th ontraints; an th minimum possil alu is 97. As ( ao a gnral ormula ) or z an xprss in this mannr in whih th sris o rations z = p a + + is ontinu until th alu o z oms an intgr. 7 I will now onsir som spial ass in whih a an ha a gin rlation; irst lt = a or a = + an lt th rmainrs o th numr in qustion ii y a an p an q as thy wr or. Thn = ; an y th prious rul z = p a = p ap + aq. I aq + p > ap gis th minimum solution thn this is th xprssion (or z): an i aq + p < ap thn th smallst solution will a 2 a + p ap + aq. All solutions ar xprss y th gnral ormula ma 2 ma + p ap + aq an analogously m 2 m + q p + q. For whatr alu o m th rmainr o this quantity ii y 2 + will th smallst solution. 8 In this way y ining this xprssion in trms o th gin rmainrs an thn what rmains atr iision y an + th unknown numr itsl may oun as (Mihal) Stil writs in his Commntary on

6 LEONHARD EULER 0 th Cossi Art (Algra) o Ruol 4. His rul is as ollows: i thr is a rmainr p o th unknown alu ii y + an a rmainr q o it ii y multiply q y + an p y 2 an thn ii th sum o ths prouts q + p( + ) y 2 + : what rmains atr iision h laims is th solution. This rul howr ollows rom our gnral ormula i m is st qual to p or thn w ha 2 p +( +)q whih ii y 2 + las th solution. 9 In th man tim th smallst solution may oun in th ollowing way: multiply th rmainr q lt rom iision y y + an a to this prout o an + or 2 + ; thn sutrat th prout rom th rmainr q multipli y ; i this alu is < 2 + it will th solution. I it is > 2 + thn sutrat 2 + an that will th solution. I or instan w sk a numr whih ii y 00 las 75 an y 0 las 37 thn a 000 to th prout o 75 an 0 or 7575; thn w ha 775 an w sutrat rom it th prout o 37 an 00 (or 3700) whih las us with 3975 rom whih i 000 is sutrat prous 3875 whih is th smallst solution. 20 I w sk th numr whih ii y las q an y n + las p thn = an th solution z = p a = p ap + aq = (n + )q np sin a = n +. All solutions ar inlu in this xprssion: mn 2 + m + (n + )q np. I th xprssion is iisil y n + th rmainr is th minimum solution. 2 Th as in whih th rmainrs p an q ar qual mrits urthr mntion. In this as lt = 0 an w ha a solution: z = p. Thn i M is th last ommon iisor o a an all solutions ar o th orm mm + p. This ormula works or any iisors a i th sam M nots th last ommon iisor o all th iisors. Thn all solutions ar in this way onstrut suh that thy la p whn ii y M. 22 Hn a wll-worn prolm is al to sol in whih w sk a numr whih whn ii y has a rmainr o an y 7 a rmainr o 0. All numrs whih whn ii y ha a rmainr o possss th proprty that whn ii y 0 whih is th last ommon multipl o an thy la a rmainr o. Thror th prolm rus to isoring th numr whih whn ii y 0 has a rmainr o in it is iisil y 7; thn a = 0 = 7 p = q = 0 an =. Whn th algorithm is omplt: = = Q Erg. z = m an i m = thn z = Th prolm in whih w sk a numr whih whn ii y th numrs has a rmainr o rsptily ut is al to ii y 7 sms to mor iiult aus th propos rmainrs ar not qual. But that qustion is quialnt to this on: to in a numr whih whn ii y has a rmainr o an y 7 a rmainr o 0. Th ormula 0m satisis thos onitions: aus w sk a numr whih whn ii y 0 has a rmainr o ut y 7 las 0 st a = 0 = 7 p = = 0 an =. Thn y th algorithm ao Q = whih turns 7 into +7 an this multipli y gis 9 th numr w sk. 4 Rrring to M. Stil s (487-57) Di Coss Christos Ruols pulish in 553.

7 A SOLUTION TO THE ARITHMETIC PROBLEM 7 24 From ths two xampls it appars or qustions o this sort in whih any numr o iisors ar propos or whih two rmainrs ar lt th gin rstritions may rlas: or th qustion an ru immiatly to th qustion or two iisors: an i all th rmainrs ar qual th solution may pursu as though only on iisor was propos. An i th rmainrs ar inqual thn nrthlss rpating ths algorithms in whih w tak on two iisors at a tim th solution may otain. For it must irst satisy two o th iisors thn a thir thn th ourth until all ar satisi. This is most onnintly xplain with xampls. 25 Lt us sarh or th numr whih ii y 7 las y 9 las 7 y las 8 an y 7 las. From ths our onitions lt us tak on th irst two an w will instigat all numrs satisying thm. W ha a = 9 = 7 p = 7 q = an = so that w arry out th algorithm as ollows: Q =. an it will that z = = 34 All alus satisying ths onitions ar inlu in th quation 3m + 34 suh that ii y 3 th rmainr is Thror th prolm has n ru to this so that w sk th numr whih ii y 3 las 34 y las 8 an y 7 las. O ths thr onitions w tak on th irst two an ha a = 3 = p = 34 q = 8 an = 2 rom whih w gt th ollowing algorithm: Q = 2 2 = thror z = m y whih th smallst solution is oun with m = 4; it will z = = 349. Thus all solutions an writtn in th orm 93m that is numrs haing th proprty that ii y 93 thy la 349 as a rmainr. 27 Finally th prolm is ru thus that a numr shoul trmin whih whn ii y 93 has a rmainr o 349 an whn ii y 7 has a rmainr o. St a = 93 = 7 p = 349 q = an = 348 an with this inormation prorm th ollowing algorithm: Q = =

8 8 LEONHARD EULER z = 93 7 m So th smallst solution shoul appar st m = 5 thn z = = 735 whih is th smallst numr satisying th our gin onitions. All solutions ar gin y th ormula 78m From this xampl mor than nough is larn in that all qustions o this kin ought to sol. 28 W now ha suiint knowlg to sol th amous hronology prolm o ating th irth o Christ gin th numr o yls o th sun an moon pass as wll as th Roman inition o that yar 5. Lt th numr o solar yls th rmainr whih ariss rom iiing th numr o yars y 9 plus 28; moror lt th numr o yls o th moon th rmainr whih ariss rom iiing th numr o yars y plus 9; also lt th inition th rmainr whih ariss rom iiing th numr o yars y 5 is a to 3 an th ollowing solution appars. Lt thr p solar yls q lunar yls an r inition [sin th passing o th nt]; p is multipl y 4845 q y 4200 an r y 9; ths thr prouts ar summ togthr with 327 an thn ii y 7980; th rmainr will th yar w sk. I w want th yar in trms o th Julian prio thn th algorithm is arri out in th sam way xpt th numr 327 shoul lt out; th mtho is gin hr. 29 Th algorithm or many iisors rquirs mor work i w simply ru th prolm y rasing th numr o iisors on y on as w i in th prior xampl; ut an asir an muh shortr way ollows rom this algorithm an th prolm an ru to th as o two iisors (rom howr many iisors thr wr). W sk a numr whih ii y a (whih I posit ar rlatily prim) las rsptily th rmainrs p q r s t. Thn th numr Ap + Bq + Cr + Ds + Et + ma satisis th prolm in whih A is th numr whih ii y th ator has rmainr 0 y a las ; B is th numr whih ii y a has rmainr 0 y las ; C is th numr whih ii y a has rmainr 0 y has rmainr ; D is th numr whih ii y a has rmainr 0 y has rmainr ; an E is th numr whih ii y a has rmainr 0 y has rmainr. Thus ths numrs an oun y th gin algorithm or two iisors. 5 An inition is a gin yar in a 5-yar yl us to at oumnts in Mial Europ.g. th 7th Inition woul th 7th yar in th yl. Howr th 5-yar yls thmsls ar usually not numr hn it is not immiatly apparnt uring whih spii yar a oumnt was writtn