LATTICE POINTS ON THE PLANE AND THE DIOPHANTINE SYSTEM. By Konstantine Hermes Zelator. Page 1 of 27

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1 LATTICE POINTS ON THE PLANE AN THE IOPHANTINE SYSTEM B Konstntine Heres Zeltor Pge of 7

2 Lttie Points on the plne n the liner iophntine sste Introution In lost ever introutor nuer theor ook there is setion evote to the tretent of the liner iophntine eqution in two unknowns or vriles n n with integer oeffiients n.the wor iophntine refers to equtions to e solve in the set of integers usull enote Z. Also the se wor iophntine hs its histori erivtion in the ne of iophntos Greek thetiin (estite to hve live in the perio 5A-5A who live in Alenri Egpt. iophntos tkle rel life proles whose solution relie on the solution of ertin equtions to e solve in integers.the ooks Eleentr Nuer Theor with Applitions (see [] Kenneth Rosen n Clssil Alger (see[] Willi J.Gilert n Sott A.Vnstone re eellent ourses for the stuent who is unfilir with the ove liner iophntine eqution; the re replete with interesting eerises not onl on this topi ut throughout n the ephsie the oputtionl spets of nuer theor in ler n onise nner. As with ever other tetook wiih ontins presenttion of the ove eqution the oplete solution set is presente (pretri solutions in ters of n integer-vlue preter long with the relevnt onitions n fts. Fro geoetri perspetive when one tries to solve the iophntine eqution one hs the ojetive of eterining ll the lttie points if n tht lie on the stright line esrie the eqution.in given oorinte plne with oorinte es n lttie point ( is sipl point in whih oth oorintes n re integers. Lter on in this pper we will ke use of the pretri solution to the ove liner iophntine eqution. Now let us turn our ttention to three-iensionl spe equippe with oorinte sste of three utull perpeniulr or orthogonl es n ; n igine for oent lttie points ( in spe; tht woul e point with the rel nuers tull eing integers. The prole of eterining the set of ll lttie points tht lie on given plne esrie the eqution (with eing integers is tntount to fining ll integer solutions to the ove eqution. This eqution often rises in prtil proles s well. Onl rrel oes n Pge of 7

3 introutor nuer theor ook evote setion to this eqution. Even ore rrel oes it ontin setion on the sste of two (siultneous liner iophntine equtions Geoetrill speking we woul seek to fin the set of ll the lttie points whih lie on the intersetion line of two plnes. Of ourse there e not line of intersetion (prllel plnes or if there is one it not ontin n lttie points. This is then the i of this rtile: to present (in step--step nner the pretri solutions to the ove three-vrile iophntine eqution s well s the sste of the two three-vrile iophntine equtions; long with interesting oputtionl eples. The orgnition of this pper in sequentil setions is s follows:. Terinolog. Review: Solving the iophntine eqution 3. The Four For uls 4. Eples 5. Eplntions n Proofs 6. Chrt of Speil Cses. Terinolog A nottion of the for ( kl n e interprete three ws: s point in the oorinte plne (with es n ; seonl s the gretest oon ivisor of two integers k n l.an thirl s solution to the iophntine eqution (here we eep t the open intervl nottion. S n hve three enings: s point in spe (equippe with n oorinte sste s the gretest oon ivisor of three integers k l r or (thir ening s solution to the liner iophntine eqution or the iophntine sste iilrl nottion of the for ( k l r ( It will lws e ler fro the tet whether for eple k l r refers to the gretest oon ivisor of the integers k l r ; or to point in spe or solution to the ove tree vrile iophntine eqution or sste. Note tht the ifferene etween the enings of the triple ( k l r s point or s solution to the ove eqution (or sste is onl senti; for ll prtil purposes we regr the frist n the thir enings s inistinguishle. Thus the onl istintion tht nees to e e ler is tht etween Pge 3 of 7

4 ( k l rs point (or solution n s gretest oon ivisor of k l r we will e writing ( k l r or soe other letter. With regr to the nottion Z whih pper on osion throughout this pper it sipl ens elongs to the set Z or is eer of the set Z (in plin lnguge is n integer. Finll the onjuntive e use etween stteents whih sipl ens n. An one ore lrifition. If we stte tht the integer k ivies the integer l we will en tht k etl ivies l ; tht is k t l for soe t Z. The se ening we will ttriute to the epression k is ivisor of l. The nottion n lso e eploe k l ens k ivies l ".. Review: Solving the iophntine eqution First ssue tht oth n re nonero integers.the speil se (n the geoetri interprettion of (i.e. ( ( or ( is isusse in the speil ses setion.(setion 6 The grph esrie the eqution is slnt (stright line whose slope is equl to. The eqution hs solutions in the set of integers Z if n onl if the gretest oon ivisor ( is lso ivisor of : (. If oes not ivie then the set of solutions is the ept set (no solutions We hve ; the lst iophntine eqution is equivlent to the originl one; tht is the hve the se solution set. The two reue integers n re reltivel prie or oprie nother w of sing tht their gretest oon ivisor is equl to : iophntine eqution then ll the solutions re given. If ( is prtiulr solution to the ove liner Solution set S: ( (For further etils proof et see either of the two referenes [] or [] Where is n integer preter; in other wors n tke n integer vlue; we n sipl write Z.Thus for eh speifi integer vlue of new prtiulr solution is proue. Pge 4 of 7

5 So for ( (. Thus if we know one speifi or prtiulr solution ll the rest n e foun. But how oes one eterine prtiulr solution (? Soeties it is es to fin one inspetion or short tril n error proess espeill if the oeffiients n re sll in solutel vlue. But there is fil proof proeure tht works in ll ses lthough it severl steps in orer to rrive t prtiulr solution (.Essentill this proess is wht is known s the Eulien Algorith. We n eplin how it works without eoing unul theoretil. For sipliit let us put ;(.We hve the eqution. If or we re one; for eple s (i.e. or - n suppose.we solve for to otin. So the solution set S is given S: ( ( Z. A siilr piture eerges in the ses or. Now suppose tht > n >. Beuse of the onition ( it follows tht n ust e istint positive integers: either > or <. For the ske of isussion suppose tht < : we solve the eqution for (if it were > we woul solve for : We perfor the ivisions n ; we hve q r Q R : : Where qq re the quotients of the ivisions n r R the two reiners. As we know for the ivision theore r < n R <. We fin tht q ( q r ( Q R r R Q Oviousl for to e n integer the rtio r R r R. r R ust e n integer; s Pge 5 of 7

6 Note tht in the lst eqution R R <. Thus setting we n repet the proess solving for the vrile (whih is one tht hs the sller in solute vlue oeffiient. The Eulien Algorith ensures us tht the proess ust terinte fter finite nuer of steps otining in the lst stge n eqution of the for kn n μ whih hs the integer solution ( μ.b k sustitution we n tre prtiulr solution ( to eqution. Let us illustrte this with n eple. Consier the liner iophntine eqution 4 38 We hve 4 38; ( ( 4 ; n 59. The originl eqution is equivlent to Oviousl 5 5 < 7 7 so we solve for : The ivisions 59:5 n 7:5 iel 59(3(56 n 7((59. Therefore [(( 3 5 6] [(( 5 9] ; Net we set n liner iophntine eqution with unknowns 5 n ; we otin 5 9 6; 9 9 < 5 5so we solve for : ( Agin we set liner iophntine eqution in the unknowns n : 9 we hve ; note tht 3 3 < 9 6 n so solve for : Pge 6 of 7

7 ( ( 3 We oul ontinue for few ore steps till we get n eqution of the for k μ (whih hs the integer solution (μ ; ut it is not rell neessr for the n n lst iophntine eqution hs n ovious prtiulr solution: n. Tring k we fin n 3 3 ( 3 Thus ( ( 3 is prtiulr solution to the iophntine eqution n hene the originl one. All the solutions re given ; Solution set S((--735 Z In the Eples Setion in ll the eples (eept for one the prtiulr solutions re esil foun inspetion. However we wnte to give n illustrtion to the reer of how this proess works in less thn ovious ses. 3. The Four Foruls In foruls n 3 elow ssue tht the oeffiients re nonero (for the ses in whih see the Speil Cses Setion. Forul If t lest on of the positive integers is equl to (i.e. ± or ± or ± ll the solutions to the iophntine eqution n e otine erel solving for the orresponing vrile. For eple if solution set S : ( ( n n where n n re integer- vlue preters. While in the se - S : n n with n Z Solution set ( ( The net forul pertins to the se in whih > > > n t lest one of the gretest oon ivisors ((( is equl to. First oserve tht the iophntine eqution will hve solutions if n onl if the nuer ( is ivisor of. Otherwise if oes not ivie the eqution hs no integer solutions. If the given iophntine eqution is equivlent to Pge 7 of 7

8 ; n Beuse of the ove reution in Foruls n 3 we strt with n eqution whih is lre reue; ( Forul Assue ( > n t lest one of ( ( ( eing equl to. If ( ll the solutions to the iophntine eqution re given ( n ( n with n eing integervlue preters n ( eing solution to the liner iophntine eqution L ikewise if ( ll the solution n e epresse the two-preter foruls ( n ( n with ( eing solution to the eqution. S iilrl if ( ll the solutions n e given ( n with eing solution to ( n ( The net forul Forul 3 els with the ses in whih not onl eh of the three oeffiient is greter thn in solute vlue; ut lso eh of the gretest oon is greter thn (so we n not use Forul. ivisors ( ( ( There is ft fro nuer theor tht is use in Forul 3 nel (( ( for n nonero integers. We leve this s n eerise for the r eer to show. In our se ( (( Pge 8 of 7

9 Forul 3 Assue > n ( ( ( > n let ( All the solutions to the iophntine eqution re given the foruls n n ( t ( t Where n re integer-vlue preters ( is solution to the liner iophntine eqution ; iophntine eqution t. n ( t is solution to the liner To finish this setion we stt Forul 4 whih pertins to the iophntine sste entione in the introution. Forul 4 ssue tht t ost one of the si integers onsier the liner iophntine sste A is ero n here n re lso integers. i. If ε ε ε where ε or - then the ove eqution hs solutions if n onl if ε ; tht is if ε the solution set is the ept set; while for ε the solution set to the ove sste is equl to the solution set of the liner iophntine eqution whih n e solve with the i of Foruls or 3; provie tht ( is ivisor of (otherwise there re no solutions; if one of or hppens to e ero refer to the Speil Cses setion. W Pge 9 of 7

10 ii. Assue tht t lest one of the onitions ε ε ε hols true s well s ( (. Let 3 e the eterinnts respetivel of the tries : 3 n. n 3 n not oth e ero (see Eplntions n Proofs setion n so we n efine ( 3 3 n 3 Uner the onition (see notes n elow. The ove iophntine sste will hve solutions (i.e. nonept solution set onl if oth onitions stte elow re stisfie: Conition : The integer ivies (or is ivisor of the integer 3 Conition : is ivisor of the integer for n solution ( to the liner iophntine eqution 3 (see note 3. If oth onitions re et the solution set to the ove sste n e esrie λ λ λ 3 3 where λ is n integer-vlue preter n ( is solution to the liner iophntine eqution 3 (whih will hve solutions onition. Note : Note tht the nuer-frtion is inee n integer: n es 3 oputtion shows tht Oviousl oth n 3 re integers sine is the gretest oon ivisor of n 3.Siilrl the nuers n re 3 oth integers the ver efinition of the integer. Pge of 7

11 Note : We ssue sine if it were ; hpothesis we know tht t ost one of the oeffiient is ero n so neessit we woul hve.we n sipl rene the oeffiients: eoe n onversel. Note 3: With regr to Conition : It will either hol for ever solution none; this will eoe evient in the Eplntions n Proof setion. ( or for When one is fe with speifi liner iophntine eqution (or sste of two three or ore vriles one solve it without ppling n ssteti generl etho or foruls; ut if one follows ifferent pth for solving the ver se sste one en up with ifferent set of pretri foruls esriing the se solution set. Assuing tht no error ourre in the proess oth sets of pretri foruls woul e orret; the woul e equivlent: one oul in finite nuer of lgeri steps erive one fro the other. We ke this rerk in light of the possiilit tht soe reers wnt to eperient n iprovise with speifi liner iophntine equtions or sstes n isover this ft in their own eperiene. 4. Eples Below we present seven eples. Eple i. Fin the solutions to the iophntine eqution 3 4 ii. eterine the lttie points tht lie on the plne esrie the ove eqution n whih lso lie in the interior of or on the ounr of (sphere the ll whose enter is the origin ( n whose rius is R. Solution i We re in the siplest of ses nel those elt with Froul. We solve for to otin 3 4 ; solution set S : 3 4n where n re integer-vlue preters. ( ( n ii For point ( to lie insie or on the given ll it is neessr n suffiient tht its istne fro the enter not eee R; sine the enter is the point ( we ust hve 4 It follows tht 4 ( ( Pge of 7

12 These onitions re neessr ut not suffiient for point ( to lie in or on the given ll.the lous tht the lst three inequlities esrie is the set of ll points in spe whih lie insie or on the retngulr o whose ounr surfe is ue entere t ( n whose eges hve length The given ll is insrie in tht o. Appling prt (i we rrive t the neessr onitions ( 3 4n n. A quik serh n e one sstetill piking vlue of (-- n heking to see whether there re vlues of n stisfing the first n the thir inequlities. Seven points re proues: (-- (- (- ( (-- (-- n (-. However onl three lie in or the given ll (tull ll three lie insie the ll: (- ( (-- Eple In how n ws n person p the ount of 8 ents using onl ies nikels or qurters? Solution If is the nuer of ies the nuer of nikels n the nuer of qurters we ust hve We re seeking the nuer of nonnegtive solutions to the lst iophntine eqution; we ust hve whih le to the onstrints This is Forul se solving the eqution iels 6 5n n where n n re integer-vlue preters. Appling the onstrins iplies 8 5n 6 n 3 A serh proues etl twent solutions; the triples (6((6(3(4(9(4((7((3 (35(3(48(43(56(5(64(7 n (8 Therefore there re etl twent ws of ping the given ount. Eple 3 i Fin the integer solutions to the eqution ii Fin those lttie points tht lie on the plne esrie the eqution of prt (i n in the interior spe oune or on the ue whih is esrie the inequlities Pge of 7

13 Solution i. Sine (3 (37 (7 we n ppl Forul with ore thn one hoie; we ppl it with ( (3 ; in our se 3 7 n 3. All the solutions n e esrie ( n ( 3 7 n where n Z re the two preters n is n integer solution to the eqution 3. ( B inspetion ( ( is solution. Hene Solution set S : ( ( 3 7 3n3 7 n. ii. We ppl the onstrins to the solutions we foun: ( n n n 6 7 n ( A serh shows tht onl n 3 iel vlues of n tht stisf the first n seon inequlities. Speifill for we otin n -3 n n -4; while for the vlues of n re onl n -6; n for 3 those vlues of n re n - n n. Going k to the pretri foruls for n we otin etl five points. ( (3 (3 (4 ( 3 (-3 The thir point in the list lies in the interior spe oune the ue wheres the other four lie on the ue itself. Eple 4 Fin the solutions to the iophntine eqution Solution We hve thus (. Sine > 3 > 5 > n ( > the relevnt forul to use is Forul 3. ( ( First we nee to fin n integer solution ( where ( 3. to the liner eqution B inspetion the eqution 5 hs solution ( ( integer solution ( t ( Forul 3 we otin 3. Net we fin n is solution. Appling the pretri foruls in Solution set S:( (-6-3 5n - n 3 with n Z Chek: 6 (-6 3 5n -5 (- n ( 3 4 Pge 3 of 7

14 Eple Fin the solutions to the liner iophntine sste Solution We hve Note tht ( ( n ll the oeffiients re nonero. Also oviousl the requireent tht ± or ± or 3 ± 3 is stisfie so we n ppl Forul 4 prt (ii. We opute the four eterinnts to fin Also 3 3 ( 3 ( n ((3 ( We hek the two onitions: Conition : 3 oviousl ivies -35. So this onition is lso stisfie. Conition : First we fin solution ( to the eqution ; ( (53 is solution. We opute stisfie. whih is ivisile 3; so the onition is Thus the sste hs solutions. To fin the we ust fin solution ( eqution 3 ( ; to the liner ; n ovious solution is ( (4. Appling this together with the other infortion we lre hve to the foruls in Forul 4. we fin Solution set S : ( (5 λ 3 λ 4 48λ; λ Z Chek: 6 ( 5 λ 4( 3 λ 3( 4 48λ 3 An 3 ( 5 λ 6 ( 3 λ ( 4 48λ 5 Pge 4 of 7

15 Eple 6 Solve the iophntine sste Solution Oserve tht this sste flls in the tegor of the Speil Cses setion sine two of the si oeffiients re ero nel n. We n solve this sste first fining the generl solution to the seon eqution n sustituting (for into the first eqution. B inspetion is solution to the seon eqution. The generl solution n e given 7λ 5λ; λ Z Sustituting into the first eqution gives ( 5λ λ To fin solution ( λ to the lst eqution we solve for : we hve 55 λ 3λ 4λ λ Let t 3t λ 4t 3 t An ovious solution to the lst eqution is t λ 3λ 3 Thus 4λ We onlue tht ( λ ( is solution to the liner iophntine eqution 3 55λ. Hene ll integer solutions to the lst eqution re given -(-55 n λ 3. We sustitute for λ in the previous equtions for n to fin. Solution set S: ( ( Chek: 3(55 ( n -5 (-5-9 7( Eple 7 Let p e positive integer. i esrie the set of ll triples ( with the two properties. The su of the three integers is equl to p n. The integers (in tht orer re suessive ters of n ritheti progression. Pge 5 of 7

16 ii Wht onition ust p stisf in orer tht e positive integers? iii Fin the sllest vlue of p for whih re positive integers. iv Wht onitions ust p stisf in orer tht e the sie lengths of tringle? v For p 85 fin those triples ( with >. Whih of those triples orrespon to tringle with eing the sie lengths? Solution i if the nuers re onseutive ters of n ritheti progression the ile ter ust e the verge of the other two. Thus we otin the iophntine sste p p ( ( p ( p ( 7 p 4 3 3p 3[ ( ] 3p In the lst sste we solve the seon eqution for to otin 3 ; lerl 4 ( ( 9 p 3p is solution n so ll the solutions to the seon liner iophntine eqution re given 9 p 3 n 3p 4 where is n integer-vlue preter. Sustituting k in the first eqution (of the lst sste ove for we fin tht Solution set S ( ( 9 p 33 p 4 p 7 : Z ii Setting ( > > > ( 9 p 3 > 3p 4 > p 7 > 9 p 3p p < < < p 9 p sin e p > n < < < < is the neessr n suffiient onition for n to e positive integers. Wht this rell ss is tht in orer for n to e positive integers it is neessr n suffiient tht the open intervl p 9 l p p (here we use stnr prelulus/ 7 3 lulus nottion ontin t lest one integer. For eh suh vlue of the preter positive integer triple ( will e generte. Pge 6 of 7

17 iii iv An es lultion show tht p 3 is the sllest vlue of the (positive integer p for whih the intervl l p ontins n integer. For p l p l3 ; is the onl integer tht flls in l 3. For p 3 n 7 3 the foruls prt (i proue ( (. If ( orrespons to tringle the three tringle inequlities ust e stisfie: ( > > >. Note tht these three inequlities lone ipl > > > ( for eple the first two inequlities eer wise to otin >. This ens tht we shoul otin n intervl (tht epens on p whih ust ontin t lest one integer n whih is suintervl of the intervl otine in prt (ii. Inee using the ove three tringles inequlities n pretri foruls for ( in prt (i we now hve ( p 7 < p 7 8 p 3 > 9 p 3 p 4 > 3p 4 3p 7 p 5p > > > p 3p sin e p > n < < < We see tht the neessr n suffiient onition for ( to orrespon to 7 p 3p tringle is tht the open intervl J p ontin n integer No te s epete tht sine < < < the intervl J p lies entirel within (the intervl in prt (ii. l p v For p 85 l85 n J 85. Also sine we see tht the internl l ontins three integers; the integers ; wheres the intervl 85 J 85 ontins two integers nel Sustituting for p 85 n in the foruls in prt (i we fin tht ( (3737 (4734 (35 Pge 7 of 7

18 The first two triples orrespon to tringle ut the thir one oes not sine 3 34 is in ft less thn 5; this is s epete virtue of the ft tht the integer 58 oes not lie within the intervl J 85. Also note tht in prt (iii where we foun tht for p 3 there is one positive integer triple ( nel ( (; we see tht p 3 is lso the sllest vlue of p for whih triple ( orrespons to tringle; whih for p 3 is the equilterl tringle of sie lengthe. 5. Eplntions n Proofs. Of Forul Not uh to show here one sipl solves for the vrile whose oeffiient is or -.. Of forul We onl nee eplin the erivtion of Forul in the se (. Sine ( the liner iophntine eqution hs solutions let ( e prtiulr solution. We hve. We put where n e n integer; so tht [( ] [ ( ] (( whih shows tht the pir ( is solution to the liner iophntine eqution. Consequentl ll the integer solutions of the lst eqution re given ( n ( n n Z. But the initil iophntine eqution is oviousl equivlent to ( Z. Hene ll the solutions to the initil eqution re given ( n ( n where n re integervlue preters. We re one. 3. Of Forul 3 The iophntine eqution is equivlent to. Sine is the gretest oon ivisor of n we ust hve n where the integers n re reltivel prie; tht is. We see tht Clerl the lst eqution will hve solutions if n onl if is ivisor of -. Evientl the originl eqution is equivlent to the sste of iophntine equtions Pge 8 of 7

19 t t t t ( ( the vriles eing n t. Oserve tht eh eqution in the lst sste hs solutions: the first hs solutions sine. The seon eqution hs solutions s well virtue of ( (( ( whih is true hpothesis. If ( is n integer solution then oviousl ( t t t Z is solution to eqution (. Furtherore let ( t e prtiulr solution to (; then ll the integer solutions to eqution ( re given t t Z But for given integer vlue of t ( t t is prtiulr solution to eqution (; therefore ll the integer solutions to eqution ( while ( lso ho ls true n e esrie t n t n; where n Z n t t. We onlue tht the solution set to the originl eqution n e esrie ; ( t n ( t n Where n re integer-vlue preters ( iophntine eqution 4.. Of Forul 4 is prtiulr solution to the liner t prtiulr solution to t. n ( i This prt is ovious it is self-eplntor. ii First note tht the integers n 3 n not e oth ero: for if tht were the se we woul hve ; then none of oul e ero sine if one the were ero nother woul lso hve to e ero (for eple if then whih is not llowe hpothesis sine t ost one of the si oeffiients n e ero. Thus ll si oeffiients woul e nonero n so we woul hve Pge 9 of 7

20 ; ut lso ( where n re reltivel prie ( ( ( integers (i.e. n is the gretest oon ivisor of n ; thus whih iplies (see rerk ( μ μ λ λ for soe integers μ n λ ; we see tht μ λ ( n in view of the ssuptio n tht gretest oon ivisor of n is equl to we onlue or -; Likewise the hpothesis ( iplies or -. Oviousl there re four ointions of vlues of n ut in ll ses ε where ε or -. Consequentl ( ε ε ε ε Contrr to the ssuption of prt (ii. It is now ler tht t lest one of 3 ust e nonero n therefore their gretest oon ivisor 3 eists n is of ourse positive integer. We hve sine st n ( ties eq.plus 3 (3 (4 We see tht in orer for eqution (4 to hve integer solutions it is neessr n suffiient t ht 3 ( 3 e ivisor of in other wors onition ust hol true. Pge of 7

21 If is n integer solution to eqution (4 then ll integer solutions to (4 re given ( Z (5 Sustituting for n in eqution (4 iels ( n sine the quntit in the rkets in equl to 3 (see note we rrive t ( 3 (6 It is ler tht the originl iophntine sste is equivlent to the pir of equtions (5 n (6 in the vriles n. For eqution (6 to hve integer solutions it is neessr n suffiient tht 3 e ivisor of the integer ( ; in other wors onition ust hol true. The integer solutions to (6 given 3 λ λ where λ is n integer-vlue preter n prtiulr solution to (6. Sustituting for n in (5 we fin ( 3 3 λ λ 3 λ Pge of 7

22 Rerk. Note the ssuption ( (. Oviousl if we re given sste of two liner iophntine equtions in three vriles the first thing to o woul e hek whether the gretest oon evisor of the oeffiients of the unknowns in eh eqution ivies the orresponing onstnt ( or on the other sie of the eqution. If either eqution fils the test sste hs no integer solutions. If on the other hn oth eqution pss the test we reue eh eqution the ( iviing oth sies with so tht we hve ( for i i i i i Rerk. In the eplntions of forul 4 we e use of the following ft fro α γ nuer theor: if four non ero integers α β γ stisf the onitions n β α γ ( γ then ( α k γ β k for soe nonero integer k. Inee iplies β α βγ ; sine γ is reltivel prie to n γ ivies the proutα it ust ivie α ;this ft fro nuer theor is tpill overe in the first or 3 weeks of n introutor ourse in eleentr nuer theor. We hve α kγ for soe non ero integer k ; n froα βγ ( k γ βγ (sineγ β k. 6. Chrt of Speil Cses A. The liner iophntine eqution with. i If n then the eqution hs no solution if is not ivisor of. If on the other hn is evisor of then the solution set onsists of ll pirs of the for where Z(ll lttie points on the horiontl line. ii If n there re no solutions if is not ivisor of ; otherwise (if ivies solution set S : ( ( Z (ll lttie points on the vertil line iii If there re no solutions unless in whih se the solution set onsist of ll lttie points on the plne. Nel S : ( ( n n Z B. The liner iophntine eqution with. Let [ ] e the tri of the oeffiients. In eh group elow we onl present the solution for the first (s viewe fro left to right tri (of the oeffiients in the group. The others re trete siilrl. Pge of 7

23 i Group : Etl one of the oeffiients is ero. Possile tries re[ ]. [ ] [ ] [ : The eqution esries plne whih is prllel to or ontins the -is. If oes not ivie there re no solutions. If on the other hn ] ( ( ivies the solution set is given n n n S ; ( : Z n solution to (. ii Group : Two of the oeffiients re ero. The thir non ero [ ] [ ][ ] [ ]: If is not ivisor of there re no solutions. If oes ivie solution set n n S ; ( : Z. The eqution esries plne prllel or oinient to the - plne. iii Group 3: [ ] There re no solutions unless in whih se λ λ ( ( : n n S Z (ll lttie points in spe. C. The liner iophntine sste with t lest two of the oeffiients eing ero. Mtri of oeffiients is. Etl two of the si oeffiients re ero i Group: Two eros in olun. : Eh eqution esries plne tht is prllel to or ontins the -is. If or equivlentl ; then the sste hs no solutions unless in whih se the proleis reue to the single eqution with (see prt B se (i. Pge 3 of 7

24 If there re no solutions unless the integer ( is oon ivisor of the integers ( n ( ;if tht is the se solution set S ( : Z. ii Group : Two eros in row. : The first eqution esries plne prllel to of oinient with plne. If is not ivisor of the sste hs no integer solutions. If is ivisor of the sste will hve solutions onl if ( is ivisor of the integer ; in tht se ll the solutions re given ( : S where is solution to (. iii Group 3: Two eros not oth on the se row or olun. : If ( is not ivisor of the sste will hve no solutions.if is ivisor of the sste will hve solutions onl if is ivisor of the integer where n e n prtiulr solution to ; if tht is the se the solution set to the sste n e esrie ( λ : S n Pge 4 of 7

25 λ where is solution to ( n λ is n integer-vlue preter.. Etl three of the si oeffiients re ero i The two of the three eros re on the se olun n two ong the lie on the se row. Possile tries: : The first eqution esries plne prllel to or oinient with -plne; the seon eqution esries plne prllel to or ontining the - is. The sste will hve no solutions onl if not ivisor of. If on the other hn the sste will hve solutions onl if is ivisor of the integer. In tht se Solution set ( : S ii Two ong the three eros lies on the se row ut no two lie on the se olun. Possile tries: : The first eqution esries plne prllel to or oinient with the plne; the seon eqution esries plne prllel to or ontining the -is. If is not ivisor of there re no solutions. If on the other hn the sste will hve solutions onl if (. If tht is the se the solution set n e esrie Pge 5 of 7

26 S ( : Z; where is solution to (. iii Three eros on the row. : If there re no solutions while if the sste is reue to with whih hs lre een overe Foruls or Etl four of the si oeffiients re ero i Group : : There re solutions onl if n ; if so there re solutions onl if ; if tht is the se Solution set ( : n S where n Z (lttie points on the plne. ii Group : : The first eqution esrie plne prllel to or oinient with the plne while the seon eqution esries plne prllel to or oinient with -plne. There re solutions onl if n in whih se solution set ( : S ; Z iii Group 3: Pge 6 of 7

27 : If there re no solutions. If the se reues to prt B(i. iv Etl five of the si oeffiients re ero : If there re no solutions. If the se reues to prt B(ii. 5. All oeffiients re ero : If either or is non ero there re no solutions. If solution set S : ( ( n k ; n k in other wors eh lttie point in spe is solution. Referenes [] Kenneth Rosen Eleentr Nuer Theor with Applitions n eition Prentie Hll. [] Willi J. Gilert & Sott A. Vnstone Clssil Alger Wterloo Universit Press. Pge 7 of 7