Multiplicative group law on the folium of Descartes

Transcription

1 Multilicative grou law on the folium of Descartes Steluţa Pricoie and Constantin Udrişte Abstract. The folium of Descartes is still studied and understood today. Not only did it rovide for the roof of some roerties connected to Fermat s Last Theorem, or as Hessian curve associated to an ellitic curve, but it also has a very interesting roerty over it: a multilicative grou law. While for the ellitic curves, the addition oeration is comatible with their geometry additive grou, for the folium of Descartes, the multilication oeration is comatible with its geometry multilicative grou. The original results in this aer include: oints at infinity described by the Legendre symbol, a grou law on folium of Descartes, the fundamental isomorhism and adequate algorithms, algorithms for algebraic comutation. M.S.C. 200: G20, 5E5. Key words: grou of oints on Descartes folium, algebraic grous, Legendre symbol, algebraic comutation. Introduction An ellitic curve is in fact an abelian variety G that is, it has an addition defined algebraically, with resect to which it is a grou G necessarily commutative and 0 serves as the identity element. Ellitic curves are esecially imortant in number theory, and constitute a major area of current research; for examle, they were used in the roof, by Andrew Wiles assisted by Richard Taylor, of Fermat s Last Theorem. They also find alications in crytograhy and integer factorization see, []-[5]. Our aim is to show that the folium of Descartes also has similar roerties and erhas more. Section 2 refers to the folium of Descartes and its rational arametrization. Section 3 underlines to connection between the oints at infinity for this curve and the Legendre symbol. Section 4 gives the geometrical meaning of the multilicative grou law on folium of Descartes, exceting the critical oint. The law was roosed by the second author and to our knowledge is the first time it aears in Balkan Journal of Geometry and Its Alications, Vol. 8, No., 203, c Balkan Society of Geometers, Geometry Balkan Press 203.

2 Multilicative grou law 55 the mathematical literature. Section 5 sotlights an isomorhism between the grou on the folium of Descartes and a multilicative grou of integers congruence classes. Section 6 formulates to conclusions and oen roblems. 2 Folium of Descartes The folium of Descartes is an algebraic curve defined by the equation x 3 + y 3 3axy = 0. Descartes was first to discuss the folium in 638. He discovered it in an attemt to challenge Fermat s extremum-finding techniques. Descartes challenged Fermat to find the tangent straightline at arbitrary oints and Fermat solved the roblem easily, something Descartes was unable to do. In a way, the folium layed a role in the early days of the develoment of calculus. If we write y = tx we get x 3 + t 3 x 3 3atx 2 = 0. Solving this for x and y in terms of t t 3 yields the arametric equations xt = 3at 3at2, yt = + t3 + t 3. The folium is symmetrical about the straightline y = x, forms a loo in the first quadrant with a double oint at the origin and has an asymtote given by x+y+a = 0. The foregoing arametrization of folium is not defined at t =. The left wing is formed when t runs from to 0, the loo as t runs from 0 to +, and the right wing as t runs from to. It makes sense now why this curve is called folium - from the latin word folium which means leaf. It s worth mentioning that, although Descartes found the correct shae of the curve in the ositive quadrant, he believed that this leaf shae was reeated in each quadrant like the four etals of a flower. 3 Points at infinity According to Eves [], infinity has been introduced into geometry by Keler, but it was Desargues, one of the founders of rojective geometry, who actually used this idea. Addition of the oints and the straightline at infinity metamorhoses the Euclidean lane into the rojective lane. Projective geometry allows us to make sense of the fact that arallel straightlines meet at infinity, and therefore to interret the oints at infinity. Even if in rojective geometry no two straightlines are arallel, the essential features of oints and straightlines are inherited from Euclidean geometry: any two distinct oints determine a unique straightline and any two distinct straightlines determine a unique oint. Let K be a field. The two-dimensional affine lane over K is denoted A 2 K = {x, y K K}.

3 56 Steluţa Pricoie and Constantin Udrişte The two-dimensional rojective lane P 2 K over K is given by equivalence classes of triles x, y, z with x, y, z K and at least one of x, y, z nonzero. Two triles x, y, z and x 2, y 2, z 2 are said to be equivalent if there exists a nonzero element λ K such that x, y, z = λx 2, λy 2, λz 2. We denote by x : y : z the equivalence class whose reresentative is the trile x, y, z. The oints in P 2 K with z 0 are the finite oints in P2 K because they can be identified with oints in the affine lane. If z = 0 then, considering that dividing by zero gives, we call the oints x : y : 0 oints at infinity in P 2 K. Now let s return to our cubic given by x 3 + y 3 3axy = 0. Its homogeneous form is x 3 + y 3 3axyz = 0. The oints x, y on the affine sace corresond to the oints x : y : in the rojective sace. To see what oints lie at infinity, we set z = 0 and obtain x 3 +y 3 = 0, which is equivalent with x + yx 2 xy + y 2 = 0. We now have two ossibilities:. x + y = 0. In this case we find the oint at infinity = : : x 2 xy + y 2 = 0. We consider this as an equation in the unknown x, with the arameter y, and we comute the discriminant = 3y 2. If there is an element α K such that α 2 = 3, then, rescaling by y, we have the following two oints at infinity 2 = + α : : 0 2, 3 = α : : 0. 2 Remark If we have 3 oints at infinity, the sum of their x coordinates is zero. For examle, when K = R, since : : 0 = : : 0, we can imagine that the oint at infinity lies on the bottom and on the to of every straightline that has sloe. 3. Legendre symbol Of course, we want to know when we have just one oint at infinity and when we have 3 oints at infinity. To decide we need the Legendre symbol which hels us decide whether an integer a is a erfect square modulo a rime number. a For an odd rime and an integer a such that a, the Legendre symbol is defined by { a, if a is a quadratic residue of =, otherwise. The Legendre symbol has the following roerties:. Euler s Criterion Let be an odd rime and a any integer with a. Then a a 2 mod.

4 Multilicative grou law Let be an odd rime, and a and b be any integers with ab. Then a b a If a b mod, then. b c a a 2 b =. = ab. 3. If is an odd rime, then = 4. If is an odd rime, then 2 = {, if a mod 4, if a mod 4. {, if a ± mod 8, if a ±3 mod Law of Quadratic Recirocity Let and q be distinct odd rimes. Then q = q 2 2. q Now we can give the following result. Proosition 3.. Let K be a finite field of rime cardinal 2, 3 and let α 2 = 3 so α might lie in an extension of K.. If = 3r +, then α K. 2. If = 3r + 2, then α / K. Proof. First let us make the following two remarks: when = 3r +, the element r cannot be an odd integer and when = 3r + 2, the element r cannot be an even integer. Now, the roof actually consists in comuting the Legendre symbol 3, in the sense that if this Legendre symbol is, then α K and if it equals, then α / K.. Consider = 3r +, where the element r is an even integer. We first comute the Legendre symbol 3 3 = = = 3r =. 3 Consequently, there is an element α K such that α 2 = 3 mod.

5 58 Steluţa Pricoie and Constantin Udrişte 2. Consider = 3r + 2, where r is an odd integer. We have 3 = 3 = = 3r+2 =. So in this case α / K. 4 The grou law on folium of Descartes We shall define and study a multilicative grou on the non-singular oints of folium of Descartes in an analogous way as the additive grou on an ellitic curve. The grou law is called multilication and it is denoted by the symbol we underline that the oeration over ellitic curves, motivated by geometry, is reresented by the addition sign +. For a better understanding of how the multilicative law works on the folium of Descartes, we describe it for the field R, because in this case we have a clear geometric interretation. Let F be the cubic given by the equation x 3 + y 3 3axy = 0, a R, i.e., F is the folium of Descartes. A first remark is that when K = R there is only one oint at infinity in P 2 K, so we denote it simly by, keeing in mind that this is actually. Start with two oints P = x, y and P 2 = x 2, y 2 on F. Define a new oint S as follows. Draw the straightline l through P and P 2. This straightline intersects F in three oints, namely P, P 2 and one other oint P 3. We take that oint P 3 and consider its symmetric S with resect to the bisectrix y = x. We define P P 2 = S. Of course there are a few subtleties to this law that need to be addressed. First, what haened when P = P 2, i.e., how does P P work? In this case, the straightline l becomes the tangent straightline to F at P. Then l intersects F in one other oint P 3 in some sense, l still intersects F in three oints, but P counts as two af them, and we consider that P P is the reflection of P 3 across the straightline y = x. We define =. Now we can assume that P. The tangent l has the sloe m = x2 + ay y 2 ax. If y ax 2 = 0, then l is the vertical straightline x = x, and this straightline intersects F in two oints P and P 3 that have same abscissa x. So x 3 = x and to find y 3 we consider y 3 3axy + x 3 = 0 as an equation in terms of y. Because y aears as a double root, we find that y 3 = 2y. So in this case P P = 2y, x. If y 2 ax 0, the equation of the straightline l is given by the oint-sloe formulae y = mx x + y. Next we substitute this into the equation for F and get x 3 + mx x + y 3 3axmx x + y = 0.

6 Multilicative grou law 59 Figure : The grou law on Folium of Descartes main idea Now we can assume that P. The tangent l has the sloe m = x2 + ay y 2 ax. If y 2 ax = 0, then l is the vertical straightline x = x, and this straightline intersects F in two oints P and P 3 that have same abscissa x. So x 3 = x and to find y 3 we consider y 3 3axy + x 3 = 0 as an equation in terms of y. Because y aears as a double root, we find that y 3 = 2y. So in this case P P = 2y, x. If y 2 ax 0, the equation of the straightline l is given by the oint-sloe formulae y = mx x + y. Next we substitute this into the equation for F and get x 3 + mx x + y 3 3axmx x + y = 0. This can be rearranged to the form m 3 + x 3 + 3m 3 x + 3m 2 y 3max 2 + = 0. The three roots of this cubic equation corresond to the three oints of intersection of l with F as mentioned before, P is counted twice. If m = this haens when x = y, then P P =. Otherwise, denote s = 3m3 x 3m 2 y + 3ma m 3. + We can now say that x 3 = s 2x and, substituting in the equation for l, y 3 = ms 3x + y. We thus obtain that P P = ms 3x + y, s 2x. Next we consider the case when P P 2. A secial case here is when x = x 2 see Figure 5, left. Then l is the vertical straightline x = x and it intersects F in the oint P 3, with x 3 = x and y 3 = y y 2. So P P 2 = y y 2, x. If x x 2, then the straightline l has the sloe m = y 2 y x 2 x.

7 60 Steluţa Pricoie and Constantin Udrişte Figure 2: Multilying a oint P to itself and equation y = mx x + y. To find the intersection with F, substitute to get x 3 + mx x + y 3 3axmx x + y = 0. As seen before, this can be rearranged to the form 0 = m 3 + x 3 + 3x m 3 + 3y m 2 3amx 2 + 3x 2 m 3 6y x m ax + 3y 2 m 3ay x + x 3 m 3 + 3y x 2 m 2 3y 2 x m + y 3. If m =, then P P 2 = this haens when x = y 2 and x 2 = y - see Figure 3, right. Figure 3: Secial cases of oint multilication on Folium of Descartes

8 Multilicative grou law 6 If m, then we can consider s = 3x m 3 3y m 2 + 3am m 3. + Keeing in mind that s = x + x 2 + x 3, we can recover x 3 = s x x 2 and then we find substituting as we did many times before y 3 = ms 2x x 2 + y. Therefore, we have P P 2 = ms 2x x 2 + y, s x x 2. Theorem 4.. Let F a R = {x, y R R x 3 + y 3 3axy = 0} { }. Then F a R, is a commutative grou. Proof. The commutativity is obvious, the identity element is and the inverse of a oint is its reflection across the straightline y = x. In fact all of the grou roerties are trivial to check excet for the associative law. The associative law can be verified by a lengthy comutation using exlicit formulas, or by using more advanced algebraic or analytic methods we do not resent here a rigorous roof because we will do this later, in a different aroach. We consider next the case when the field K is a finite field with elements, i.e. K = F. Although we do not have a geometric interretation, the ideas resented before remain the same. 4. Case = 3r + 2 In this case, because m 3 = has the unique solution m =, we have only one oint at infinity. So ractically, all the results resented in the case K = R are available here too. We now give two algorithms, the first one being for the multilication of a oint P with itself, and the second one is for the multilication of two oints P and P 2 in general. In all that follows, the comutations are made modulo. Algorithm: StarSquareP. If P =, then P P =. 2. Otherwise, if y 2 ax = 0, then P P = 2y, x. 3. Otherwise, define m by m = x2 + ay y 2 ax. 4. If m 3 + = 0, then P P =. 5. Otherwise, define s = 3m3 x 3m 2 y + 3am m 3. + Then P P = ms 3x + y, s 2x. Algorithm: StarP roductp, P 2. If P =, then P P 2 = P Otherwise, if P 2 =, then P P 2 = P.

9 62 Steluţa Pricoie and Constantin Udrişte 3. Otherwise, if P = P 2, then P P 2 = starsquarep. 4. Otherwise, if x = x 2 and y y 2, then P P 2 = y y 2, x. 5. Otherwise, define m by m = y 2 y x 2 x. 6. If m 3 + = 0, then P P 2 = [, 0]. 7. Otherwise let s = 3m3 x 3m 2 y + 3am m 3. + Then P P 2 = ms 2x x 2 + y, s x x Case = 3r + In this case, we know that we have 3 oints at infinity, each of them corresonding in a way that we will see in the next section to an element ω K with ω 3 =. We described oint multilication of non-singular oints of folium of Descartes for all finite oints. Because in this case we have more than one oint at infinity, it is referable to resent this grou law in the rojective version. If we look at all oints like oints in rojective lane, then we can treat all of them just in the same way. The idea is the same. Let us start with two distinct oints P = P : P 2 : P 3 and Q = Q : Q 2 : Q 3. They determine a unique straightline l given by the following equation P P 2 P 3 l : Q Q 2 Q 3 x y z = 0 or, equivalently l : P 2 Q 3 P 3 Q 2 x + P 3 Q P Q 3 y + P Q 2 P 2 Q z = 0. A first remark is that if both P 3 and Q 3 equals 0, so they are both oints at infinity, l is the ideal straightline. So, when we intersect l with our curve we also get a oint at infinity. More recisely, we get the oint R at infinity that is different from both P and Q. Using the observation made earlier we have that R = P Q : : 0, so P Q = : P P 2 : 0, which is the same as P Q = : : 0. P + P 2 If one of the oint is finite, say P so P 3 =, and the other one, Q, is a oint at infinity so Q 3 =, Q 2 = and Q 3 = 0, then P Q will be a finite oint. The equation of l becomes P + Q y x P 2 Q = 0. Denote m = and n = P 2Q P and relace y = mx+n in x 3 +y 3 3axy = 0. Q Q We will get a quadratic equation in x, a 2 x 2 + a x + a 0 = 0 with a known solution P. So the x coordinate of the oint R is just the other solution on this equation, R = a P and R 2 = mr + n. a 2 Therefore, in this case P Q = m a a 2 P + n : a a 2 P :.

10 Multilicative grou law 63 In the same way, if P is a oint at infinity and Q is a finite oint, we have P Q = m a Q + n : a Q :, where m =, n = P Q 2 Q a 2 a 2 P P and a, a 2 are coefficients obtained the same way as before. In all the cases that remain, both oints are finite, so we can aly the same rules as in the affine version from Algorithm 2. Algorithm : StarSquareP. If P 3 = 0, then P P = P 2 : : Otherwise, if P 2 2 ap = 0, then P P = 2P 2 : P :. 3. Otherwise, define m by m = P 2 + ap 2 P2 2 ap. 4. If m 3 + = 0, then P P = m : : Otherwise, define s by s = 3m3 P 3m 2 P 2 + 3am m 3. + Then P P = ms 3P + P 2 : s 2P :. Algorithm 2: StarP roductp, Q. If P = Q, then P Q = StarSquareP. 2. Otherwise, if P 3 = 0 and Q 3 = 0, then P Q = : : 0. P + Q 3. Otherwise, if P 3 = and Q 3 = and P = Q, then P Q = P 2 Q 2 : P :. 4. Otherwise, if P 3 = and Q 3 = 0, denote m = Q and n = Q P 2 P Q. Relace y = mx + n in x 3 + y 3 3axy = 0 and obtain a 2 x 2 + a x + a 0 = 0. Let s = a. Then P Q = ms P + n : s P :. a 2 5. Otherwise, if P 3 = 0 and Q 3 =, denote m = P and n = P Q 2 Q P. Relace y = mx + n in x 3 + y 3 3axy = 0 and obtain a 2 x 2 + a x + a 0 = 0. Let s = a. Then P Q = ms Q + n : s Q :. a 2 6. Otherwise, comute m = Q 2 P 2 Q P. 7. If m 3 + = 0, then P Q = m : : 0.

11 64 Steluţa Pricoie and Constantin Udrişte 8. Otherwise, let n = P 2Q P Q 2 and relace y = mx + n in x 3 + y 3 3axy = 0 Q P to obtain a 3 x 3 + a 2 x 2 + a x + a 0 = 0. Let s = a 2 a 3. Then P Q = ms P Q + n : s P Q : 5 The fundamental isomorhism Like grou, the folium of Descartes is one of the most imortant objects in mathematics. Let us sotlight an isomorhism between the grou on the folium of Descartes and a multilicative grou of integers congruence classes. In what follows, we denote by Z the set of integers modulo the rime number and we consider the multilicative grou Z,, where denotes the usual multilication. Also, let F a, = {x, y F F x 3 + y 3 3axy = 0} { } and consider the oeration that endows this set whith a grou structure as seen before. 5. Case 2 mod 3 Theorem 5.. There is an isomorhism between Z, and F a,,, given by ϕ : Z 3at F a, ; ; t P t = t 3, 3at 2 t 3, t. Proof. We first rove that P t P t = P t 2 following the stes in Algorithm. If P t =, i.e., t =, then from Algorithm we have that P t P t = and obviously, because t 2 =, we find that also P t 2 =, so P t P t = P t 2. 2 If y 2 t ax t = 0, then 6t4 +3ta 2 t 6 2t 3 + = 0 3t2t3 +a 2 t 3 = 0 2t 3 + = t 2 a From Algorithm, we obtain P t P t = 2y t, x t = t 3, 3ta t 3. 3t 2 a We want to show that P t P t = P t 2 = t 6, 3t 4 a t 6, i.e., 6t2 a t 3 = 3t2 a t 6 and 3ta t 3 = 3t4 a t 6. Both of them hold because of the relation 2t3 + = 0. 3 If y t 2 ax t 0, we can comute m = x2 t + at t y 2 t ax t = t4 4t 2t 3 +. If m 3 + = 0, i.e., if m =, from Algorithm, we find P t P t =. We want to show that P t 2 =, i.e., t 2 =. From m = we get x 2 t ay t = yt 2 ax t, so x t y t x t + y t + a = 0. If x t y t = 0, we get 3at t 3 = 3at2 t 3, so t2 + t = 0. Because t 0, it remains t =, so t 2 = and P t 2 =. If x t + y t + a = 0, we get t2 2t + a 2 t 2 = 0 so t =, meaning P t is actually, + t + so from the first case, we know that P t 2 =.

12 Multilicative grou law 65 4 Otherwise we comute s = 3m3 x t 3m 2 y t + 3am m 3 = 3t4 + 6ta + t 6 and using 3at Algorithm, we deduce P t P t = ms 2 3x t + y t, s 2x t = t 6, 3at 4 t 6. 3at 2 Because P t 2 = t 2 3, 3at 2 2 t 2 3, we obtain P t P t = P t 2. We next rove the relation P t P u = P tu following the stes in Algorithm 2. Let P t =, i.e., t =. Then from Algorithm 2 we have P t P u = P u, and obviously P tu = P u = P u, so in this case P t P u = P tu. 2 In the same way, if P u =, then from Algorithm 2 we have P t P u = P t, and because P tu = P t = P t we also get P t P u = P tu. 3 If P t = P u, i.e., t = u, then we are in the settings of Algorithm, and we roved that in this case P t P u = P tu. 4 If x t = x u, but y t y u, then we obtain the following relation between t and u 3at t 3 = 3au u 3 t tu3 = u ut 3 t u + t 2 u + tu 2 = 0. Because t u, it remains + t 2 u + tu 2 = 0. We want to rove that the following two equalities hold tu x tu = y t y u, i.e., tu 3 = t2 t 3 u2 u 3 and y tu 2 tu = x t, i.e. tu 3 = t t 3. We rove first that y tu = x t. From tu2 tu 3 = t t 3 we get that tu2 t 4 u 2 t 3 u 3 + must be 0. Using + t 2 u + tu 2 = 0, we we can write tu 2 = t 2 u, so relacing tu 2 in tu 2 t 4 u 2 t 3 u 3 + we get t 2 u t 4 u 2 t 3 u 3 +, which is equal to t 2 u + t 2 u + tu 2 = 0. What is left to rove is t 2 u t 4 u 2 t 3 u 3 = 0. From 3at t 3 = 3au u 3 we can say that u 2 t 3 = tu, and now we relace it t3 tu in tu 3 = t2 t 3 u2 u 3 and obtain tu tu 3 = t2 t 3 tu t 3, i.e., tu tu 3 = t2 u 2 t 3. So the equality ut 3 + t t 4 u 3 t 3 u 4 = 0 must hold. Using again the relation +t 2 u+tu 2 = 0, but this time relacing t 2 u = tu 2, we get tu 2 t 3 u 3 t 2 u 4 = 0, which is equivalent to tu 2 + t 2 u + tu 2 = 0. Because tu 2 0 and we know that + t 2 u + tu 2 = 0, we are done roving also that in this case P t P u = P tu. 5 If x t x u, we can comute m = y u y t = u2 t 3 t 2 u 3 x u x t u t 3 + t u 3 = u2 t 2 + u 3 t 2 u 2 t 3 u + t + ut 3 tu 3 = u tu + t + u2 t 2 u t u t utu tu + t = u + t + u2 t 2 u 2 t ut 2.

13 66 Steluţa Pricoie and Constantin Udrişte 6 If m 3 + = 0, i.e., m =, then u + t + u 2 t 2 = + utu + t. We denote u + t = v and ut = w. This yields w 2 vw + v = 0. One solution is w =. From this we get ut =, so P tu =. Solution w 2 = v yields ut = u + t so u = or t =, but this cannot be the case. It remains, in this case, P t P u =, as in Algorithm 2. 7 If m, we can comute where s = 3m3 x t 3m 2 y t + 3am + m 3 = 3aA B, A = u 4 t 6 + u 5 u 2 t 5 + u 6 + u 3 t 4 + u 4 + ut 3 + u 5 + u 2 t 2 + +u 3 t u, B = u 6 + u 3 t 6 + u 6 t 3 + u 3 +. From Algorithm 2, we have that x tu should be equal to ms 2x t x u + y t and y tu should be equal to s x t x u. Therefore we comute ms 2x t x u + y t and obtain 3atu t 3 u 3 which is indeed x tu, and also comuting s x t x u, we get 3at2 u 2 t 3 u 3, which is equal to y tu. 5.2 Case mod 3 In what follows, we also denote by Z, the multilicative grou of integers modulo n. This time, let P F a, = {x : y : z F F {0, } x 3 + y 3 3axyz = 0} and consider the oeration that endows this set whith a grou structure as seen before, in the rojective version. Theorem 5.2. There is an isomorfism between Z, and P F a,,, given by t P t = 3at t 3 : 3at 2 t 3 :, if t 3 t 2 : : 0, otherwise. Proof. Algorithm 3: If z t = 0, then P t = t 2 : : 0 and t 3 =. We want to show that P t P t = t 4 : : 0. But this is exactly what Algorithm 2 says. 2 If z t =, then yt 2 ax t = 0. From this condition we find 2t 3 + = 0. From the foregoing Algorithm 3, we have 6at 2 P t P t = 2y t : x t : = t 3 : 3at t 3 :. So we want to show that 6at2 t 3 = 3at2 3at and t6 t 3 = 3at4 t 6.

14 Multilicative grou law 67 For both equalities we need 2t 6 t 3 = 0 and this holds because 2t 3 + = 0. 3 If z t = and yt 2 ax t 0, we can comute m = x2 t + ay t yt 2 ax t = t4 + 2t 2t If m 3 + = 0, then t2 2t 9 + 2t 3 8t 9 + 2t 6 + 6t 3 + = 0, i.e., t3 3 t 3 + 2t = 0. Because we are in a case where z t =, we know that t 3, so it remains t 3 + = 0. Therefore t 6 =, so t 2 3 = 0, which means that P t 2 = t 4 : : 0. It remains to show t4 + 2t 2t 3 + = t4. This reduces to t 7 = t, and because t 0, to t 6 =, which is a true equality, as mentioned before. 5 Otherwise, we can comute s = 3m3 x t 3m 2 y t + 3am m 3 = 3at4 + 6at + t 6. Algorithm 3 says that P t P t = ms 3x t + y t : s 2x t :. 3at 2 Relacing m, s, x t and y t, we obtain that P t P t = t 6 : 3at 4 t 6 : and because P t 2 = 3at 2 t 2 3 : 3at 2 2 t 2 3 : we have that P t P t = P t 2. Algorithm 4: If P t = P u, then we roved P t P u = P tu. 2 If z t = z u = 0, it means that t 3 = u 3 =, from where we have that tu 3 =, so P tu = tu 2 : : 0. From Algorithm 4, we have P t P u = : : 0, where x t = t 2 and x t + x u x u = u 2. So we need to show t 2 + u 2 = t2 u 2, i.e., t 4 u 2 +t 2 u 4 + = 0. Because t 3 = u 3 =, this is equivalent to tu 2 + t 2 u + = 0. We also have t 3 u 3 = 0, i.e., t ut 2 + tu + u 2 = 0 and, because t u, it remains that t 2 + tu + u 2 = 0. If we multily this by t 2 u 2 and kee in mind that t 3 = u 3 =, we get that tu 2 +t 2 u+ = 0, which roves that in this case P t P u = P tu. 3 If z t = z u = and x t = x u but y t y u, so P t P u, then, from Algorithm 4, tu we find P t P u = y t y u : x t :, so what we have to show is that tu 3 = t2 t 3 u2 u 3 and tu 2 tu 3 = t t 3. We rove first that y tu = x t, using x t = x u From tu2 tu 3 = t t 3 we get tu2 t 4 u 2 t 3 u 3 + must be 0. Using the equality + t 2 u + tu 2 = 0, i.e., tu 2 = t 2 u, and relacing tu 2 in tu 2 t 4 u 2 t 3 u 3 +, we find t 2 u t 4 u 2 t 3 u 3 +, which is equal to t 2 u + t 2 u + tu 2 = 0. What is left to rove is t 2 u t 4 u 2 t 3 u 3. From 3at t 3 = 3au u 3, we can say that u 2 t 3 = tu, and now we relace it t3

15 68 Steluţa Pricoie and Constantin Udrişte tu in tu 3 = t2 t 3 u2 u 3 and obtain tu tu 3 = t2 t 3 tu t 3, i.e., tu tu 3 = t2 u 2 t 3. So the equality ut 3 + t t 4 u 3 t 3 u 4 = 0 must hold. Using again the relation +t 2 u+tu 2 = 0, but this time relacing t 2 u = tu 2, we get tu 2 t 3 u 3 t 2 u 4 = 0, which is equivalent to tu 2 + t 2 u + tu 2 = 0. Because tu 2 0 and we know that + t 2 u + tu 2 = 0, we are done roving also that in this case P t P u = P tu. 4 If z t = x t = 3at t 3, y t = 3at2 t 3 and z u = 0 x u = u 2, y u =, then let m = = x u u 2 and n = x uy t x t = 3u2 t 2 a 3ta x u u 2 u 2 t 3. We next relace y = mx+n in x 3 +y 3 3axy = 0 and obtain a 3 x 3 +a 2 x 2 +a x+a 0 = 0, where which is 0, because u 3 = a 3 = u6 u 6 a 2 = 3u4 t 3 + 9u 2 t 2 9t + 3u 4 a u 6 t 3 + u 6 a = 9u6 t 5 36u 4 t u 2 t 3 + 9u 6 27t 2 + 9u 4 ta 2 u 6 t 6 2u 6 t 3 + u 6 a 0 = 27u6 t 6 8u 4 t 5 + 8u 2 t 4 27t 3 a 3 u 6 t 9 + 3u 6 t 6 3u 6 t 3 + u 6. It remains a quadratic equation in x, a 2 x 2 + a x + a 0 = 0, and we know that one solution is x t. To find the other solution, we comute s = a = 3u2 t 2 3ta a 2 t 3. So the second solution is s x t = 3u2 t 2 a, so the equality 3 3u 2 t 2 a t 3 t 3. This should be equal to y tu = 3atu2 tu = 3atu2 tu 3 must hold, which is true because u3 =. It remains to show that x tu = ms x t + n, i.e., 3atu tu 3 = 3ta u 2 t 3. This is u2 also true, for the same reason that u 3 =. 5 The case z t = 0 and z u = are treated the same way, and we also get that P t P u = P tu. 6 Otherwise if z t = z u = and x t x u, we can comute 7 If m 3 + = 0, then m = y t y u x t x u = u2 t 2 t u ut 2 + u 2 t +. u 6 + u 3 t 6 + u 6 t 3 u 3 + u 3 t 6 + 3u 4 t 5 + 3u 5 + 3u 2 t 4 + u 6 + 6u 3 t 3 + 3u 4 + 3ut 2 + 3u 2 t + = 0

16 Multilicative grou law 69 u 6 + u 3 t 6 + u 6 t 3 u 3 + = 0 u 3 t 3 t 3 u 3 = 0. Because t 3 and u 3, it remains that tu 3 =, so P tu = t 2 u 2 : : 0. From Algorithm 4, we have that P t P u = m : : 0, so we have to show that u 2 t 2 u t ut 2 + tu 2 + = t2 u 2, which yields to t 4 u 3 t 3 u 4 = t u and this is true because t 3 u 3 =. 8 And finally, if m 3, then comute n = x uy t x t y u and relace y = mx + n x u x t in x 3 + y 3 3axy = 0. We get a cubic equation a 3 x 3 + a 2 x 2 + a x + a 0 = 0 in the unknown x, with the coefficients where a 3 = P N, a 2 = Q N, a = R N, a 0 = S N, P = u 6 + u 3 t 6 + u 6 t 3 + u 3 +, Q = 3au 4 t 6 + u 5 + u 2 t 5 + u 6 u 3 t 4 + u 4 ut 3 + u 5 u 2 t 2 + u 3 + t + u, R = 9u 3 a 2 t 5 9u 4 a 2 t 4 + 9u 5 9u 2 a 2 t 3 9u 3 a 2 t 2 + 9ua 2 t, S = 27u 3 a 3 t 3, N = u 3 t 6 + 3u 4 t 5 + 3u 5 + u 2 t 4 + u 6 + 6u 3 t 3 + 3u 4 + ut 2 + 3u 2 t +. For this cubic equation, we know two solutions, namely x t and x u, so we can easily find out the third solution. 6 Conclusions and oen roblems The folium of Descartes can have oints with coordinates in any field, such as F, Q, R, or C. The folium of Descartes with oints in F is a finite grou. There are still many alications to be found for folium of Descartes, related to the fact that there is a multilicative grou law over the curve. For examle, folium of Descartes crytograhy. Oen roblems i Folium of Descartes Discrete Logarithm Problem FDDLP is the discrete logarithm roblem for the grou of oints on folium of Descartes over a finite field. ii The best algorithm to solve the FDDLP is exonential. That is why the folium of Descartes grou can be used for crytograhy. More recisely, we estimate that the best way to solve FDDLP for folium of Descartes over F takes time O. The goal of our future aers is about such tye of roblems, with an emhasis on those asects which are of interest in crytograhy. Acknowledgements Partially suorted by University Politehnica of Bucharest, and by Academy of Romanian Scientists, Bucharest, Romania.

17 70 Steluţa Pricoie and Constantin Udrişte References [] H. W. Eves, A Survey of Geometry, Allyn and Bacon, Inc., 972. [2] N. Koblitz, A Course in Number Theory and Crytograhy, Graduate Texts in Mathematics, 4, Sringer, 995. [3] J. H. Silverman, Advanced Toics in the Arithmetic of Ellitic Curves, Sringer, 994. [4] N. Smart, Crytograhy: an Introduction, McGraw-Hill, [5] L. C. Washington, Ellitic Curves. Number Theory and Crytograhy, Chaman & Hall, Authors address: Constantin Udrişte and Steluţa Pricoie, University Politehnica of Bucharest, Faculty of Alied Sciences, Deartment of Mathematics-Informatics, 33 Slaiul Indeendentei, RO Bucharest, Romania. udriste@mathem.ub.ro, anet.udri@yahoo.com ; maty star@yahoo.com