Integer Points on Elliptic Curve

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1 Iteger Poits o Elliptic Curve Weipig Zhou School of Mathematics ad Computatioal Sciece, Aqig Normal Uiversity,Aqig46133, Ahui,Chia Abstract Elliptic curve is a importat problem i iteger factorizatio ad cryptography.usig the elemetary umber theory,it provesthat the o-egative iteger poits ofthe elliptic curve x 1qy 1have the structure of 6 1, 3 1 qa a qa where a is a o-egative iteger such that 3qa 1is square, whe q1s 5 is prime where s is a o-egative iteger. This result gives the method of solvig the above elliptic curve. Ad 3 it proves that the elliptic curve y x ( p 4) x p has up to two groups of positive iteger poits besides ( xy, ) (,),where1s 5, 6s 1ad p36s 7 are primes, ad s is a o-egative odd iteger. The two groups of positiveiteger poits ca be take as the key of password if they are exist, ad this result ca avoid the ivalid algorithm of iteger factorizatio usig elliptic curves. Key words:elliptic Curve, Diophatie Equatio, Iteger Poits, Key of Password, Iteger Factorizatio 1. INTRODUCTION The prime test ad the iteger factorizatio algorithm based o the elliptic curve are the hot issue at preset. The researchig of elliptic curve s iteger poits have become a iterestig subject i iteger factorizatio ad cryptography. I 1987, Zagierprovided whether there isoe o-egative iteger poit o the elliptic curve (Zagier, 1987) 3 y x 7x 6 (1) For elliptic curve (1), Zhuad Cheproved there are oly two o-egative iteger poits ( xy, ) (, ) ad ( xy, ) (88444, ) usig algebraic umber theory ad p -adic aalysis (Zhu ad Che, 9). Guaproved the elliptic curve (Guag, 14) 3 y x ( p 4) x p () has iteger poits ( xy, ) (,) ad ( xy, ) (88444, ) whe p 31, ad there is oly oe iteger poit ( xy, ) (,) whe p , where p 36s -5 is prime ad s is a o-egative iteger such that s, 6s 1are primes. Ad the related coclusios are as follows. Lemma 1If D is ot square iteger, the the equatio (Pa ad Pa, 199) x Dy 1 (3) has ifiite may positive iteger solutio. Let ( x, y) deote the basic solutio of (5), the the all iteger solutios ( x, y ) ca be writte as follows x y D x y D, Lemma If D is ot square iteger,the the equatio (Walsh, 1999) 4 x Dy 1 (4) has two sets of o-egative iteger solutios at most, ad which has just two sets of o-egative iteger poits if ad oly if D 1785 or D 865 or x, y are square where ( x, y) is (5) s basic solutio. 1 Lemma 3If the o-egative iteger solutio ( x, y ) of (5) satisfy x y 1,the ( x, y) is the basic solutio of (5) (Luo ad Yua, 1). Lemma 4If the equatio (6) just has oe o-egative iteger solutio ( xy(wu,, ) 1), the Z 9

2 x y D x y D k where x y is the basic solutio of equatio (5), ad k Z, k or k., First,this paper proves the o-egative iteger solutios of the elliptic curve x 1qy 1 (5) are (6qa 1, a 3qa 1),where q1s 5 is prime, s ad a are o-egative itegers, 3qa 1is square. This coclusio gives the method of solvig the elliptic curve (3). Secod, it proves whe p36s 7, q1s 5, r 6s 1are prime ad s is odd, the elliptic curve 3 y x ( p 4) x p (6) has up to two groups of positive iteger solutios if it has other solutios besides ( xy, ) (, ), ad its structure is preseted. Usig this result, the iteger poit of the elliptic curve (4) ca be take as the key of password. Ad this result ca avoid the ivalid algorithm of iteger factorizatio usig elliptic curves.. INTEGER POINTS ON TWO CLASSES OF ELLIPTIC CURVE.1. Iteger poits o Biary Quadratic Elliptic Curve Theorem 1Let s be a iteger such that q1s 5 is prime, the the o-egative iteger solutios of the elliptic curve x 1qy 1 (7) have the structure of 6qa 1, a 3qa 1, where a is a o-egative iteger such that 3qa 1 is square. Proof : (7) is equivalet to x 1 ( x 1)( x 1) 1 qy, q 1s 5 is prime (8) It is obviously that ( 1,) are (8) s solutios. Let ( x, y) be a (8) s solutio,because we cosider (8) i four cases. Case 1 The qa 6b. gcd( x1, x 1), x 1qa, y ab, gcd( qa, b) 1(9) x 1 6b If y is odd, the y 1or9mod 1, a, b are odd, ad qa 6b 4ormod 1,which is cotradictory. If y is eve, the y or4mod 1, oe of a, b is odd ad the other is eve. Because q 1s 5,so qa 6b 6,or1 mod1, which is cotradictory. So (9) has o solutio. Case The 6qy. x 16qa, y ab, gcd( qa, b) 1 x 1 b If y is odd, the abare, odd, ad 6qa b 4or mod1, it is cotradictory. If y is eve, the oe of abis, odd ad the other is eve,ad 6qa b 6or -mod1, which is acotradictory. So (1) has o solutio. Case 3 x16b, y ab, gcd( qa,6 b ) (11) x 1 qa The 6b qa. (1) 1

3 If y is odd, the 6b qa 4or mod1, which is a cotradictory. If y is eve, the oe of abis, odd ad the other is eve. If a is odd ad b is eve, the 6b qa a mod1 we have1 a mod1, so 3b 1 qa 6mod1,but 3b mod1, which is a cotradictio. If b is odd ad a is eve, the 3 3b 1 qa 1or 1mod1, which is cotradictory. So (11) has o solutio Case 4 x1b, y ab, gcd( qa, b) 1 x 1 6qa The b -6qa. b 3qa 1, If y is odd, the abare, odd, ad b -6qa -4or mod1, which is cotradictory. If y is eve, the oe of abis, odd ad the other is eve, ad b 13qa 1or 4mod1, so x b 3qa 6qa 1. So equatio (7) has o--egative iteger solutios with structure 6qa 1, a 3qa 1,where a is ay o-egative iteger such that 3qa 1is square... Iteger Poits o Biary Cubic Elliptic Curve Theorem Let s be a odd umber such that p36s 7, q1s 5 ad r 6s 1are prime. If the elliptic curve 3 y x ( p 4) x p (13) has other o-egative iteger solutio i additio to ( xy, ) (, ), the it has two set of o-egative iteger 4 4 solutio at most with has structure of ( x, y) (3qa,3 qab ),where a fg, b f 4rg, f is odd ad g is eve such that f 1mod1, ad ( f 6 g, g) is the solutio of the elliptic curve i.e. There is a square umber d which ca be divided by 4 such that (1) 4 X 1qY 1 (14) f 6 g, g 6 qd 1, d 3 qd 1 ad 3qd 1is fourth power umber. Proof : The elliptic curve (13) ca be writte as follows y ( x )( x x p ) (15) It is easy to verify that ( xy, ) (,) is a set of solutio. We cosider the case x ad y. Let d gcd( x, x x p ), x dt, t Z, the x x p dt dt p 6 8 gcd( x, x x p) gcd( x, p 8) gcd( x, 36s 15) Because q1s 5, so gcd( x, x x p) gcd( x,3 q ), ad d 1,3, q, or3q. We cosider d i four cases. Case 1 If d 1, let x a, x x p b, y ab,gcd( a, b ) 1 b ( x 1) p 1, x a b a s ( 3) mod8 so ba, 3are both odd or eve, i.e. Oe of bais, odd ad the other is eve. If b is odd ad a is eve, the ad if a is odd ad b is eve, the b ( a 3) 1-1 mod8 11

4 b or4mod8,( a 3) mod8 b ( a 3) or 4mod8 But b ( a 3) 36s 6 6mod8, which is cotradictory. Case. If d 3, let x a x x p b y ab a b 3, 3, 3,gcd(, ) 1 we have x3a ad p36s 7, so x x p 3b is equivalet to 3( a 1) 1s b b a s -3( 1) 1 mod 4 Thus ba, 1 are both odd or eve, i.e. Oe of bais, odd ad the other is eve. If b is odd ad a is eve, the If a is odd ad b is eve,the b ( a 3) 1-1 mod 4 b mod4,( a 3) mod4 b a but b -3( a 1) 1s mod 4, which is cotradictory. Case 3. If d q, let ( 3) mod 4 x qa x x p qb y qab a b,,,gcd(, ) 1 so x qa ad p 3q -8, thus x x p qb is equivalet to ( qa 3) 3q 9 qb 3( a 1) ( q 3) a b 4 By q1s 5 ad kow that ba, 1 are both eve or both odd, i.e. Oe of bais, odd ad the other is eve. If b is odd ad a is eve, the 4 4 ( q 3) a b -3( a 1) ( q 3) a 1-3 6mod8 which is cotradictory. If a is odd ad b is eve, the 4 4 ( q 3) a b -3( a 1) ( q 3) a mod 4 which is cotradictory. Case 4. If d 3q, let x qa x x p qb y qab a b 3, 3, 3,gcd(, ) 1 so x 3qa ad p 3q -8, thus x x p 3qb is equivalet to we have (3qa 3) 3q 9 3qb (3a 1) 3( q 3) a b 4 By q1s 5 we kow that b,3a 1 are both odd ad both eve, i.e. oe of bais, odd ad the other is eve. if b is eve ad a is odd, the 4 b mod 4,(3a 1) 3( q 3) a 3 mod 4 which is cotradictory. So b is odd ad a is eve. Let a t, wheret is a o-egative iteger. The ( 3) t q t 4 b (16) Let r 6s 1be prime, where s is o-egative odd. There is qr 3, hece (16) is equivalet to Let k gcd( b 1t 1, b 1t 1),the k, t 4t 196rt b ( b 1t 1)( b 1t 1) 96rt 4, 4 b t If k,let p1 deote oe prime divisor k of, the p b, p 1t 1, ad 1 1 p 96rt 4. By 1 4 gcd(1 t 1,96 rt ) gcd(1 t 1, r ) 1

5 We kow p 1 r,so p 1 r, ad r b, r 1t 1,as well as t ad 96 ca t be divided by r, but r a cotradictio. So p 1 1, i.e. k. Let t fg,gcd( f, g ) 1, the we ca deote where c,6,16,48, r,6 r,48r If c,the 3r b 1t 1 f, b 1t 1 cg c b 4 rf g, 1t 1 4rf g Because 3 ca t divide1t 1,so 3 do t divide g,hece which is cotradictory. If c 6,the t 1 4rf g -1mod3 b 8rf 3 g, 1t 1 8rf 3g Because of t fg, so the secod equatio of the above is equivalet to the followig 1 f g 1 8rf 3g 4 4 (r 3)( f ) 3( f g ) 1 Because of r 6s 1, so the above equatio is equivalet to the followig (1 s 5)( f ) 3( f g ) 1 we ca get the equatio (1 s 5) x 3y 1 has iteger poit ( x, y) ( f, f g ). But It is cotradictory. If c 16, the 1 (1 s 5) x 3y 5x or mod3 b 3rf 8 g, 1t 1 3rf 8g rt, it is By the secod equatio of the above,we kow f is odd, so 1t 8g 3rf 1 mod 4, which is cotradictory. If c 48, the b rf 4 g, 1t 1 rf 4g By the secod equatio of the above, it is easy to verify f ca t be divided by 3 ad f is odd. Because s is odd, so t 4g rf 1 (6s 1) f 1 (6 1) f 1 6or mod1 which is cotradictory. If c r, the b 4 f rg, 1t 1 4 f rg By the secod equatio of the above, it is easy to verify g is odd, so which is cotradictory. If c 6r, the so g is odd. If f is odd, the rf 1t 1 rg 1 g or 4mod 6 b 8 f 3 rg, 1t 1 8 f 3rg f 1t 13rg mod8 It is a cotradictio. If f is eve, because of t fg, we has It is cotradictory. If c16r, the f 1t 13rg mod b 3 f 8 rg, 1t 1 3 f 8rg 13

6 4 4 It is easy to verify that f is odd, ad 1 1t 1 3 f 8rg 3mod 4, which is cotradictory. If c 48r, the b f 4 rg, 1t 1 f 4rg 4 so f is odd, f 1mod1 f 1mod1, f 1mod 4,ad t fg is eve, i.e. g is eve. By the secod equatio of the above we ca get 4 ( f 6 g ) 1(r 3) g 1 Because of r 6s 1, the the above is equivalet to the follows 4 ( f 6 g ) 1(1 s 5) g 1 so ( f 6 g, g ) is a set of iteger solutio of the equatio 4 X (144 s 6) Y 1 By lemma we kow the above equatio has two set of o-egative iteger solutio at most, so (13) has o more tha two solutios i additio to (,). Ad ( f 6 g, g ) is oe solutio of the equatio X (144 s 6) Y 1 By theorem 1, we kow the equatio 6 1, 3 1 X 1qY 1, where q1s 5 is primehas the structure qd d qd of the o-egative iteger solutio, where d is ay o-egative iteger such that 3 1 qd is square, so there is some iteger d such that f 6g 6qd 1, g d 3qd 1 Because of f g qd d, gcd( d,3qd 1) 1, g d 3qd mod1 so d is square which ca be divided by 4. By the above aalysis, we kow if the elliptic curve 3 y x ( p 4) x p has other o-egative iteger poit except of (,), the it has two set of o-egative iteger solutio at most, 4 4 ad it s solutio is ( x, y) (3qa,3 qab ), where a fg, b f 4rg, f is odd, g is eve which satisfy : f 1mod1, ad ( f 6 g, g ) is oe solutio of the elliptic curve X (144 s 6) Y 1 i.e. there is a square iteger d which ca be divide by 4 such that 6, 6 1, 3 1 3qd 1 is a fourth power iteger. f g g qd d qd, ad 3.CONCLUSIONS For theorem 1,we gave the followig example to describe it was feasible for solvig the equatio (7). 7 Example 1 If q 113,we could fid oly oe value 453 for a i (,1 ), ad the (7) s solutio is , For theorem, it was easy to prove that r 6s 1, q1s 5, p36s 7 were primes whe s take the followig 4 values ad there values of the above were odd. Example if s 1, the r 6s 1 7, q1s 5 17, p 36s 7 43 are prime, we could fid eight values of d such as 7, 7, 69993, 69986, ,849731, , i (,1 ) such that 3qd 1 are square. For the first fourth values of d, we could give the (7) s solutios 14

7 ad It was easy to verify that 4999,35, 49981,34993, , , ( , ) But 4999,35 was t the (7) s basic solutio. Betwee i this values of d,oly d 7, 6986 could be divided by 4 but which were ot square. So 8 there was t square iteger d i (,1 ) which ca be divided by 4 such that f 6 g, g 6 qd 1, d 3 qd 1 8 i.e. we ca t fid (13) iteger poit for d i (,1 ),or i other words, the equatio (13) had o iteger poits i 4 (,1 ). Example showed that if the elliptic curve (13) has a solutio had oe iteger solutio, the the iteger was very large whe we take large eough prime umber prime q which satisfied the give coditio. Ad it was relatively safe if the solutio of the elliptic curve was take as the key of password. O the other had, because of the scarcity of the elliptic curve (13) solutio, it is ot feasible to use the form of (13) to factor iteger. Ackowledgemets This work was supported by the key projects of Provicial Natural Sciece Research of Ahui Provice(No. KJ14A144). REFERENCES Zagier,D. (1987) Large itegral poit o elliptic curves, Math Comp,48(177), pp Zhu,H.L., Che,J.H. (9) Itegral poits o y x 7x 6, J. Math. Study,4(), pp Gua Xugui (14) Poits o the Elliptic Curve y x ( p 4) x p, Advaces i Mathematics, 43(4), pp Pa,C.D., PaC.B.(199)Elemetary Number Theory,BeijigUiversityPress: Beijig. 4 Walsh,G.(1999) A ote o a theorem of Ljuggre ad the Diophatieequatios x kxy y 1or4,Arch. Math,73(), pp Luo, G., YuaP.Z.(1) O the Diophatie equatio x Dy 1,J.Sichua Uiv.(Natural Sciece Editio),38(1), pp Wu,H.M.(1) Poits o y x 7x 6, Acta Math. Siica,53(1), pp