Integral points on the rational curve


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1 Integrl points on the rtionl curve y x bx c x ;, b, c integers. Konstntine Zeltor Mthemtics University of Wisconsin  Mrinette 750 W. Byshore Street Mrinette, WI Also: Konstntine Zeltor P.O. Box 480 Pittsburgh, PA 503 Emil Addresses: ) )
2 . Introduction In this work, we investigte the subject of integrl points on rtionl curves of the form y x bx c x ; where, b, c re fixed or given integers. An integrl point is point (x,y) with both coordintes x nd y being integers. Not tht if (i, i ) is n integrl point with i ; on the grph of the bove rtionl function. Then (i, i ) lso stisfies the lgebric eqution, of degree ; 0. And conversely, if (i, i ) is n integer pir with i ; stisfying this x xy bx y c lgebric eqution. It then lies on the grph of the rtionl function y x bx c x. () Since () 4(0)=>0 (coefficient of the x term is ; coefficient of the xy term is ; nd coefficient of the term y is 0); the bove degree lgebric eqution describes hyperbol on the xy plne. In L.E. Diskson s book History of the Theory of Numbers, Vol. II (see reference []); there is welth of historic informtion on Diophntine equtions of degree ; in two or three vribles. In reference [], the reder will find n rticle published in 009; nd deling with some specil cses of integrl points on hyperbols. In Section, we prove key proposition, Proposition. Proposition lys out the precise (i.e. necessry nd sufficient conditions, for qudrtic trinomil with integer coefficients to hve two integer roots or zeros. This proposition plys key role in estblishing Theorem in Section 5; the min theorem of this pper. According to Theorem, if the integer b c is not zero (which is to sy tht the integer is not zero of the qudrtic trinomil x +bx+c). Then the grph of the rtionl function in () contins exctly 4N distinct integrl points; provided tht b+c is not equl to n integer or perfect squre; or minus n integer squre. If on the other hnd b+c=k ; or k ; where k is positive integer. Then the grph of y=f(x) contins exctly 4N distinct integrl points. In both cses bove, N stnds for the number of positive integer divisors of b c ; divisors not exceeding b c. In Section 6, we pply Theorem, in the cse were b c p p pp,,, or. Where p is prime; nd p p re distinct primes.
3 In Section 3, we present n nlysis of the grph of y=f(x) in the specil cse b 4c=0. In this cse the rtionl function f(x) reduces to ( x d) f( x) x ; where d is n integer such tht b=d nd c=d. We nlyze (in Section 3) this cse from the Clculus point of view. As it becomes evident, there re relly only two types of grph of y=f(x); in the cse b See Figures,, 3, 4, nd 5 4c 0.. A Proposition nd its Proof The following proposition, Proposition, is the key result used in estblishing Theorem in Section 5. This proposition nd its proof cn lso be found in reference [3] (rticle uthored by this uthor). But, for resons of completeness nd convenience for the reders of this pper; we include it here. We will mke use of wellknown lemm in number theory, (see [4]). Lemm (Euclid s Lemm) Let n, n, n 3 be nonzero integers such tht n is divisor of the product n, n 3. Then if n is reltively prime to n ; n must divide n 3. Proposition Let, b, c be integers; with the integer being nonzero. g x x bxc. Consider the qudrtic trinomil (i) (ii) (iii) Then, The trinomil The trinomil n integer squre. The trinomil gx ( ) hs either two rtionl zeros (or roots) or (otherwise) two irrtionl zeros. gx ( ) b 4c hs two rtionl zeros (or roots) if, nd only if, the discriminnt is gx ( ) hs two integer zeros if, nd only if, the following conditions re stisfied: b 4 c k, for some integer k. And, the integer is divisor of both integers c nd b Equivlently, is divisor of the gretest common divisor of b nd c 3
4 (iv) If, then gx ( ) hs two integer zeros if nd only if If, then gx ( ) hs two integer zeros if nd only if b b 4cis n integer squre. 4c is n integer squre. Proof b (i) If r nd r re the zeros of gx ( ). Then, r r (ii) ; (nd lso rr c ). b Since is rtionl number; it is cler from the lst eqution tht if one of r, r is rtionl; so is the other. First, suppose tht b 4c is the squre of the squre of n integer; b 4c D, where D is nonnegtive integer. The two zeros or roots of the trinomil gx ( ); re the rel numbers b D nd b D ; tht is, the numbers b D nd b D ; which re both rtionl since b, D, nd re ll integers. Now the converse. Let r nd r be the rtionl zeros of gx ( ); nd T the discriminnt. We hve, b T b T r, r ; nd T b 4c () We write the rtionl number r in lowest terms: u r, where u nd v re reltively v prime integers nd v nonzero From () nd (3) we obtin, u bv v T (4) (3) If T 0, then T b nd we re done: T is n integer squre. If T is nonzero, then since v is lso nonzero; nd so by (4), so is the integer u bv. According to (4), the nonzero integer squre v divides the nonzero integer squre ( u bv). This implies tht positive integer v must divide the positive integer u bv (This lst inference cn typiclly be found s n 4
5 exercise in elementry number theory books. It cn lso be found in reference []). Therefore the integer is divisor of the integer u bv. And so, u bv t v, for some integer t (5) or equivlently, u v ( t b) If t=b, then by (5) we get u 0which implies u 0 in view of being nonzero. But then r 0 by (3). Since gx ( ) hs zero s one of its roots. It follows tht the constnt term c must zero; c=0, which implies T b by (). Agin T is n integer squre. If t b in (5). Then since ll three integers,v, nd t b re nonzero; so must be the integer u by (5). So ll four integers, u, v, t b re nonzero. By Lemm (Euclid s Lemm) since v is divisor of the product u ; nd v is reltively prime to u. It follows tht v must be divisor of the integer. So tht, vw, for some nonzero integer w (6) Combining (6) nd (4) yields, v w b v T ; T w b, which proves tht T is integer squre. The proof is complete. (iii) Suppose tht the trinomil gx ( ) hs two integer roots nd zeros. Then by prt (ii) it follows tht, b 4 c k, (7) for some nonnegtive integer k And the two zeros of gx ( ) re the numbers 5
6 b k b k r nd r where r nd r re integers (8) So, from (8) we hve, r b k; ( r b) k (9) From (9) nd (7) we get, b 4br 4 r b 4 c; nd since 0, br r c (0) From (8), we lso hve, ( r r ) b; ( r r ) b, Which shows tht is divisor of b. Thus, b, for some integer () By () nd (0) we obtin, r c r ; r ( r ) c; which proves Tht is lso divisor of c. We hve shown tht the integer is common divisor of b nd c. Now the converse. Suppose tht is common divisor of both b nd c; nd tht k. Then, b b nd c c, for some integers b nd c. b 4c k, for some 6
7 And so, b 4 c k ; ( b 4 c ) k () According to () ; k. is divisor of k which implies tht is divisor of k. And so k k, for some Using b b, c c, k k ; nd the formuls in (8). We see tht the two zeros r nd r of gx ( ) re the numbers r b k b k nd r (3) Also from, (), nd k k ; we hve b 4c k (4) Eqution (4) shows tht b nd k hve the sme prity; they re either both odd or both even. Hence by (3), it follows tht both rtionl numbers r nd r ; re ctully integers. (iv) This prt is n immedite consequence of prt (iii). We omit the detils. 3. The Cse b 4c 0: A Clculus Bsed Anlysis Then function f( x) x bx c x = (, ) U(, ). hs domin, Df x x R nd x  Now, suppose tht the integers b nd c stisfy, b b 4c0; 4c The integer b must be even; b d, for some integer d. Which yields c d. We hve, 7
8 x dx d f( x) x ( x d) f( x) x. ; Cse : b d In this cse the function f reduces to, f(x)=x+ The grph of the y f ( x) is the grph of the stright line with eqution y x but with the point ( o, ) removed. ( x ) Figure y f ( x) ; with 0 x Figure 0 y f ( x) x x Figure 3 8
9 y ( x ) f ( x) x, with < o ( x d) f( x) x Cse : b d (i) Asymptotes The grph of y f ( x) hs no horizontl symptote. It hs one verticl symptote, the verticl line x. Also, lim f( x), while lim. x x The grph of y f ( x) lso hs n oblique symptote; the slnt line with eqution y x d. If we perform long division or synthetic division of (x+d) ( x d) ( x ) ( x d ) ( d ) ; ( x d) ( d ) f ( x) x d ; for x. x x with x+; we obtin So tht lim f ( x) ( x d ) 0 lim f ( x) ( x d ). x x (ii) Intercepts The point (d, 0) is the x intercept on the grph of y=f(x); while the point 0, d is the y intercept. (iii) First Derivtice, Open Intervls of Increse/Decrese, nd Points of Reltive Extremum. We use the quotient rule to clculte the derivtive f (x) of f(x): 9
10 x d f( x) ; x ( x d) ( x ) ( x d) ( x d) ( x ) ( x d) f '( x) ; ( x ) ( x ) ( x d) x d f '( x). ( x ) (5) We see from (5) tht the function f, hs two criticl numbers in its domin, ( ). These re the numbers d nd (d)=d. Note tht these two numbers re distinct since nd d re distinct. Furthermore, when <d, then d<<d; nd so the function f is incresing on the open intervls (, d) nd ( d, ); decresing on the open intervls (d,) nd (,d). The point (d,0) on the grph of y=f(x); is point of reltive mximum. While the point (d, 4(d)) is point of reltive mximum. Note tht On the other hnd, if d<. Then (d ) f ( d ) 4( d ). d d d. The function f increses on the open intervls, d nd d, ; nd it decreses on the open intervls d, nd, d. The point d,4( d ) on the grph of y f ( x) ; is poiont of reltive mximum; while the point d,0is point of reltive minimum when d< nd vicevers when >d. (iv) Second derivtive, open intervls of concvity, nd inflection points As we will see below, the grph of y f ( x) hs no inflection points. We compute the second derivtive. First we write the numertor in (5); in expnded form. f '( x) x x d d ( ). ( x ) Applying the quotient rule gives, ( x ) ( x ) ( x ) [ x x d( d)] f "( x) ; 4 ( x ) ( x ) [ x x d( d)] f "( x) ; 3 ( x ) 0
11 [ x x x x d( d)] f "( x) ; 3 ( x ) ( d) f "( x) ; nd since 3 ( x ) ( d) 0, On ccount of d. It cler tht f "( x) o on the open intervl (, ) ; while f "( x) o on the open intervl (, ). Thus, the grph of y f ( x) is concve downwrds over the open intervl (, ) ; nd it concve upwrds over the open intervl (, ). There re no inflection points on the grph of y f ( x). (v) Two grphs Figure 4 d. And so, d+<d<<d y ( x d) f ( x) x
12 Figure 5 d. And so, d d d
13 4. Integrl points in the cse b 4c o As we hve seen in the previous section; when The function y f ( x) reduces to, b 4c, then b d nd c d ; d n integer. y ( x d) f ( x) ; with x x (6) Cse : =d. We hve y x ; nd so obviously there re infinitely mny integrl points on the grph of y f ( x). Set of integrl points, S ( x, y) x t nd y t ; t, nd t Cse : The integers nd d re distinct Eqution (6) is equivlent to, y ( x ) ( x d) (7) with x Clerly, the pir (d,0) is solution to (7); note tht d, since d. Now, if xdin (7). Then ( x d) o. So either y nd x+ re both positive integers; or they re both negtive integers. So we split the process into two cses. Cse : y ( x ) ( x d) ; x, yre integers such tht y, x+ (7) Let be the gretest common divisor of y nd x+. Then, y z, x z ; for some reltively prime positive integers (7b) z nd z From (7) nd (7b) we obtin z z x d (7c) ( ) 3
14 According to (7c), the positive integer must divide x d divides the positive integer ( x d) ; which implies tht x d z3, for some positive integer z3 (7d) Combining (7c) with (7d) yields, z z z (7e) 3 Since the positive integers z nd z re reltively prime nd their product, ccording to (7e), is equl to perfect squre or integer squre; ech of them must be n integer squre. Recll tht more generlly; if the product of two reltively prime positive integers is equl to n nth integer power; ech of those positive integers must equl n nth integer power. This result cn esily be found in number theory books. For exmples, see reference [4]. Thus by (7e) we must hve, z v, z v, z3 vv ; for reltively prime positive integers v nd v (7f) Going bck to (7d) nd (7b) to obtin y v, x v, x d v v (7g) First, suppose tht <d. And so >d. Since x>. We hve x>d, nd so by (7g) it follows tht x d x d vv. Which further implies by (7g) gin; tht v v v v d; v ( v v ) d ; d v ( v v) ; with v v nd d . By setting m v nd vv n; v m n. We obtin y m n ( ), x m ; with m nd n being reltively prime integers. Clerly the conditions y nd xin (7) re stisfied. Clerly gcd( mn, ) ; since gcd( v, v). Next, suppose tht d ; nd so d. There re two 4
15 possibilities to consider in this cse. One possibility is d x. The other possibility is x d ; which requires tht d ( ) d d. For if d=, then there is no integer in the open intervl (, d). Also, clerly x d, s implied by the conditions in (7). So, if d x. Then x d 0; x d x d. So (7g) implies x v d v v ; d v ( v v ) ; with v v v. By setting m v nd vv n ; v m n. We obtin y m n ( ), x m ; under the conditions m n ; nd with gcd( mn, ). Note tht the conditions y nd xin (7) re stisfied. Lstly, if x d; with d. We hve xd 0 ; x d (since x+d is n integer). And then (7g) gives, And so, x v d v v d v ( v v) We get, v n, v m n; with m nd n being reltively positive integers with mn. We obtin y m n ( ), x n. We cn now stte Result, which summrizes the results in Cse (the cse with y ) Result Consider the rtionl function y ( x d) f ( x) x, where nd d re distinct integers. The set of ll integrl points (x,y) on the grph of y f ( x) ; with y ; is finite set which cn be described s follows: (i) If d. Then the set of ll integrl points on the grph of y f ( x); with y, nd thus with xs well; cn be prmetriclly described in the following mnner: y m n ( ), 5
16 d x m ; where m nd n re reltively prime positive integers such tht mn ; nd is positive integer divisor of d. (ii) If d. Then set of ll integrl points on the grph of y f ( x) ; with y cn be prmetriclly described in the following mnner. Those integrl points with y nd xd re given by, y m n ( ), x m ; where m nd n re reltively prime integers such tht d mn; nd with mn ; positive divisor of d. And those integrl points (such points exist only when d ) with y nd x d cn be described s follows: y m n x n ( ), Where is positive divisor of d; nd m, n re d reltively prime positive integers such tht, mn ; nd with mn. Remrk Note tht when d, nd mn, d. We obtin the integrl point ( x, y) ( d, 4( d )); which is one of the two points of extremum tht the curve y ( x d) f ( x) ; s we sw in Section 3. x Now, we go bck to (7) nd we consider the second cse. Cse b: y ( x ) ( x d) ; xy, re integers such tht y, x (7h) If gcd( y, x ). Then, y z, x z; z, z re reltively prime positive integers. (7i) From 7h nd 7i one obtins, 6
17 zz z3, z 3 x d ; z 3 positive integer. (7j) From this point on the rest of the nlysis/proof/procedure for this cre (Cseb) is very similr to tht of Cse. We omit the detils nd stte of Result. Result Consider the rtionl function y ( x d) f ( x), x where nd d re distinct integers. The set of ll integrl points ( xy, ) on the grph of y f ( x); with y ; is finite set which cn be described s follows: () i If d, then the set of ll integrl points on the grph of y f ( x); with y, nd thus with x s well; cn be prmetriclly in the following mnner: ( ), ; where is positive divisor of d; y m n x p m d reltively prime positive integers such tht mn. nd m, n re ( ii ) If d, then the set of ll integrl points on the grph of y f ( x); with y ; cn be prmetriclly described in the following mnner. Those integrl points with y nd x d re given by nd n re reltively prime positive integers such tht mn. y m n x m ( ), ; where m And those integrl points (such points exist only when d ) with y nd d x cn be described s follows: y m n x n ( ), ; where m nd n re reltively prime positive integers such tht n m. We conclude this section by stting the obvious result below. 7
18 Result 3 Let ( x d) f( x) ; x where nd d re integers. () i If d, the only integrl point on the grph of y f ( x); with y coordinte zero; is the point ( d,0). ( ii ) If d, the set S of integrl points on the grph of y f ( x) is the set, S ( x, y) x t, y t; t, t n integer. 5. The Generl Cse nd the Min Theorem x bx c We now go bck to y If is zero of the trinomil x. Then, the other zero is the integer b+. So in this cse we hve x bx c b c ; i.e. if 0. ( x ( b))( x ) y x ( b) x b. All integrl points in this cse on the bove curve re x the points of the form (x,y)=(t, t+b); t cn be ny integer other thn. Next, ssume tht is not zero. An integrl point (i, i ) will be on the curve precisely when, i i ( b i ) c i 0, (8) nd i According to (8), i is n integer zero of the qudrtic trinomil (with integer coefficients, b i, c i ) t ( b i ) c i ; nd so the other zero must lso be integrl. Thus by Proposition (iii) we must hve (nd only then): b c ( b i) 4 ( c i) K, for some nonegtive integer K. ( b i) K ( b i) K The zeros of (8) re nd i one of these two zeros. (8) From (8), we hve, i ( b ) i (4 c K ) b 0 (8b) 8
19 Eqution (8b) shows tht since the integer i is one of its two zeros; so must be the other zero. Applying Proposition (iii) gin, for some nonnegtive integer M. The zeros of (8c) re the integers b M nd b M ; i is one of the two integers. 4( b ) 4 b (4 c K ) 4 M, (8c) Tking the eqution in (8c) further we get, b 4b 4 b 4 c K M ; 4( b c) M K ( M K)( M K) (8d) Recll tht b+c is nonzero. It is cler from (8d) tht the nonnegtive integers M nd K must hve the sme prity: either they re both odd; or they re both even. Moreover, since M 0 nd K 0; we hve M K 0. Therefore (8d) implies 4 b c ( M K) M K. (8e) Since M, K hve the sme prity, M+K nd MK re both even integers. Also b c is positive integer, since b+c is nonzero. Thus we must hve, M K d nd M K d where nd re positive divisors of such tht dd b c ; nd with d d d d b c (8f) Note tht d cnnot exceed d, since M K M K, on ccount of the fct tht both M nd K re nonnegtive. The equl sign holds only in the cse K=0 nd M ; or in the cse M 0 nd K (clerly t most one of K, M cn be zero, by (8e) nd the fct tht b c ). Also, the cses K 0, M ; or K, M 0 cn only occur in the cse in which b c is perfect or integer 9
20 squre. When this hppens, d cn equl d : d d b c ; nd with either K 0 nd M ; or lterntively K nd M 0. Otherwise, when b c is not perfect squre. Both M nd K re positive integers; nd so, M K M K ; nd d d. Looking t (8f) we see tht in ll cses, b c d d d d d And so,. d b c Thus ll the possible such pirs of divisors of d nd d ; re obtined by choosing positive divisor d b c not exceeding b c. Then of ; d b c d. Furthermore from (8f) we must hve, M d d nd K d d; when 0 K M While, M d d nd K d d; when 0 M K (8g) Combining (8g), (8c), nd (8d) we see tht, when ctully hve M d+ d, K d d; nd with either i b M b ( d d). or lterntively, i b M b ( d d) And so from (8h) nd (8), we lso get, or i or i or i either i ; b ( b ) ( d d ) ( d d ) b ( b ) ( d d) ( d d) b ( b ) ( d d) ( d d) b ( b ) ( d d) ( d d) (8i) (8h) b c is positive; we must 0
21 The bove four possibilities in (8i) relly reduce to the following four possibilities s strightforwrd clcultion shows: i d or i d (8j) or i d or i d Combining (8j) with (8h), we see tht when b c is positive integer; then for ech pir of divisors d nd d in (8f). Four integrl points re generted. These re, ( i, i ) ( d, b ( d d)), ( d, b ( d d)), (8k) ( d, b ( d d)), ( d, b ( d d)). By inspection, these four integrl points re distinct; with exception of the cse where b c is perfect squre; with the choice points then. d d b c The four points reduce to two distinct. Wht hppens when b c is negtive integer? By (8d) we must hve 0 M K. And so by (8g) we hve M d d nd K d d. Combining (8g), (8c), nd (8d) one gets either i b ( d d ) or i b ( d d ). After tht one combines this with (8) to obtin the four integrl points corresponding to the choice of divisors d nd d stisfying (8f) (see prt (iii) of Theorem stted below). We omit the detils. Also, if we choose nother pir of divisors nd stisfying (8f); with the pir (, ) being distinct from the pir ( d, d ). Then it is cler tht the four points generted by the pir (, ); will be distinct from the four points generted by the pir ( d, d ). This follows from (8k) (nd the corresponding formuls in the cse b c 0) nd (8f). We omit the detils.
22 We now stte Theorem. Theorem x bx c Consider the function f( x), with domin ll rels except ;, b, c being integers. x Then, (i) If b c 0. There re infinitely mny integrl points on the grph of y=f(x). These re the points of the form, ( x, y) ( t, t b ), t n integer, t. (ii) If 0. When there re exctly 4N integrl points on the grph of y f ( x). Except b c in the cse where points. b c is perfect squre, in which cse there re exctly 4N such Where N (in either cse) is the number of positive divisors of the integer not exceed b c b c ; divisors which do. All 4N integrl points cn be prmetriclly described by the formuls, ( x, y) ( d, b ( d d )), ( d, b ( d d )), ( d, b ( d d )), ( d, b ( d d )). Where d, d re positive divisors of b c ; such tht d d nd dd b c (iii) If b c 0, there re exctly 4N integrl points on the grph of y f ( x), unless b c is equl to minus perfect squre, in which cse there re exctly 4N such points. Where N (in either cse) is the number of positive divisors of the integer b c ; divisors which do not exceed formuls, b c. All 4N integrl points cn be described by the ( x, y) ( d, b ( d d )), ( d, b ( d d )),
23 ( d, b ( d d )), ( d, b ( d d )) Where d nd d re positive integers such tht d d nd dd b c 6. An ppliction of Theorem : Theorem Theorem below is direct ppliction of Theorem. Consider the cses where p is prime. b c, p, or p Since, in Theorem, we hve b c ; then d d b c d d, with d d. It follows tht when When b c p ; then d nd d p. When b c p ; then either d nd d p ; or d d p. We stte Theorem without further elborting. Theorem Let bc,, be integers nd consider the function (, ). f( x) x bx c x with domin (, ) U ( ) If, then the grph of y f ( x) contins exctly four integrl points. These re b c (, b ), (, b ), (, b ), nd (, b ) (b) If, then the grph of y f ( x) contins exctly two integrl points. These re: b c (, b ), (, b ) (c) If b c p, p prime. When the grph of y f ( x) contins exctly four integrl points: 3
24 ( p, b ( p )), (, b ( p )), (, b ( p )), ( p, b ( p )) (d) If b c p, p prime. When the grph of y f ( x) contins exctly four integrl points: ( p, b ( p )), (, b ( p )), (, b ( p )), ( p, b ( p )) (e) If b c p, p prime. When the grph of y f ( x) contins exctly eight integrl points: ( p, b ( p )), (, b ( p )), (, b ( p )), ( p, b ( p )), ( p, b p), ( p, b p), ( p, b p), ( p, b p) (f) If b c p, p prime. When the grph of y f ( x) contins exctly six integrl points. These re: ( p, b ( p )), (, b ( p )), (, b ( p )), ( p, b ( p )), ( p, b ), ( p, b ) 4
25 References [] L.E. Dickson, History of the Theory of Numbers, Vol II, 803 pp., Chelse Publishing Compny 99. ISBN: For Pythgoren tringles see pp For Diophntine equtions of degree, see pp [] Konstntine Zeltor, Integrl points on hyperbols: specil cse, rxiv: , August 009, 7 pges, no figures. [3] Konstntine Zeltor, Integer roots of qudrtic nd cubic polynomils with integer coefficients, rxiv: 0.60, October 0, 4 pges. [4] W. Sierpinski, Elementry Theory of Numbers, 480 pp. Wrsw, 964. ISBN: For Euclid s lemm (Lemm ) see Theorem 5 on pge 4. For Pythgoren triples, see pp Also, see Theorem 8 on pge 7. 5
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