This is a preprint of: “Spatial convex but non-strictly convex double-pyramidal central configu- rations of the (2n+2)-body problem”, Montserrat Corbera, Jaume Llibre, J. Dyn. Diff. Equat., 2019. DOI: [10.1007/s10884-019-09798-3] Spatial convex but non–strictly convex double–pyramidal central configurations of the (2n + 2)–body problem Montserrat Corbera1 and Jaume Llibre2 1 Departament de Tecnologies Digitals i de la Informació, Universitat de Vic, 08500 Vic, Barcelona, Spain. E–mail: montserrat.corbera@uvic.cat 2 Departament de Matemàtiques, Universitat Autònoma de Barcelona, 08193 Bellaterra, Barcelona, Spain. E–mail: jllibre@mat.uab.cat Abstract A configuration of the N bodies is convex if the convex hull of the positions of all the bodies in R3 does not contain in its interior any of these bodies. And a configuration is strictly convex if the convex hull of every subset of the N bodies is convex. Recently some authors have proved the existence of convex but non–strictly convex central configurations for some N –body problems. In this paper we prove the existence of a new family of spatial convex but non–strictly convex central configurations of the (2n + 2)–body problem. More precisely, we prove that for n > 4 there are spatial convex but non–strictly convex central configurations of the (2n + 2)–body problem consisting of n masses equal to m1 at the vertices of a regu- lar n–gon, n masses equal to m2 at the vertices of the n–gon whose vertices are the midpoints of the edges of the initial n–gon, and two masses equal to m3 on the straight line orthogonal to the plane con- taining the two n–gons passing through their barycenters. Moreover, we show that for n = 3 does not exist such spatial convex but non– strictly convex central configurations. Note that the convex hull of such central configurations is formed by two equal pyramides with 1 the base formed by the big n–gon, glued by their bases. So we call them spatial convex but non–strictly convex double–pyramidal central configurations of the 2n + 2–body problem. 1 Introduction and statement of the main result The equations of motion of the spatial N –body problem are N X qk − qj mk q̈k = − G mk mj , |qk − qj |3 j=1 j 6= k for k = 1, . . . , N , where G is the gravitational constant which will be taken equal to one by choosing conveniently the unit of time, qk ∈ R3 is the position vector of the punctual mass mk in an inertial coordinate system, and the two dots denote the second derivative with respect to the time t. The configuration space of the spatial N –body problem is E = {(q1 , . . . , qN ) ∈ R3N : qk 6= qj , for k 6= j}. Given N positive masses m1 , . . . , mN a configuration (q1 , . . . , qN ) ∈ E is central if there exists a positive constant λ such that q̈k = −λ (qk − cm) , for k = 1, . . . , N , where PN k=1 mk qk cm = P N . k=1 m k is the center of mass of the system. Consequently a central configura- tion (q1 , . . . , qN ) ∈ E of the N –body problem with masses m1 , . . . , mN is a solution of the system of equations N X qk − qj mj = λ (qk − cm) , (1) |qk − qj |3 j=1 j 6= k for k = 1, . . . , N and for some λ. 2 We do not know the first author which shown that N equal masses located at the vertices of a regular N –gon is a central configuration of the N –body problem for all N > 2. If we put equal masses at the vertices of any regular polyhedron, then we obtain a spatial central configuration of the N –body problem with N equal to the number of vertices (see [2]). Other spatial central configurations of the N –body problem are the pyramidal central configurations, which consist of N = n + 1 masses, with n equal masses at the vertices of a regular n–gon and the (n+1)th mass in the orthogonal straight line to the plane containing the n–gon and passing through its barycenter (see for instance [5] and [18]). Another simple spatial central configurations are the double pyra- midal central configurations, which consist of N = n + 2 masses, n equal masses at the vertices of a regular n–gon and the other two equal masses symmetrically localized with respect to the plane con- taining the n–gon on the orthogonal straight line to this plane and passing through its barycenter (see [8, 9, 10, 11, 12, 13, 14, 19, 20]). In [16] the authors study these previous central configurations adding a new mass on the barycenter of the n–gon. In [4] we studied these double pyramidal central configurations but when the two masses on the orthogonal straight line to the plan containing the n–gon through its barycenter are not equal, we called such central configurations by bi–pyramidal central configurations. We recall that a configuration of the N bodies is convex if the convex hull of the positions of all the bodies in R3 does not contain in its interior any of these bodies. And a configuration is strictly convex if the convex hull of every subset of the N bodies is convex. As far as we know the first example of a convex but non–strictly convex central configuration appeared in 2010 in the paper [7] of Gidea and Llibre for the planar 5–body problem. In 2012 Chen and Hsiao [3] reproved this previous result of the planar 5–body problem and provided a spatial convex but not strictly convex central configuration for the 7–body problem. In these two first examples of convex but non–strictly convex central configurations three of the bodies are on a straight line. More recently in 2018 Fernandes, Garcia and Mello provided convex but non–strictly convex central configuration for the planar 8–body problem, and for the spatial 10– and 20–body problem. We think that these are all the know convex but non–strictly convex central configuration which appear until now in the literature. Inspired in the convex but non–strictly convex central configura- 3 tion for the spatial 10–body problem found by Fernandes, Garcia and Mello in this paper we shall prove the existence of convex but non– strictly convex central configuration for the spatial (2n+2)–body prob- lem for all n > 4. More precisely, we prove that for n > 4 there are spatial convex but non–strictly convex central configurations of the (2n + 2)–body problem consisting of n masses equal to m1 = 1 (we choose the unit of mass equal to m1 ) at the vertices of a regular n–gon, n masses equal to m at the vertices of the n–gon whose vertices are the midpoints of the edges of the initial n–gon, and two masses equal to µ on the straight line orthogonal to the plane containing the two n–gons passing through their barycenters. Moreover, we show that for n = 3 does not exist such spatial convex but non–strictly convex central configurations. Note that the convex hull of such central con- figurations is formed by two equal pyramides with the base formed by the big n–gon, glued by their bases. So we call them spatial convex but non–strictly convex double–pyramidal central configurations of the 2n + 2–body problem. Thus our main result is the following one. Theorem 1 The following statements hold for the central configura- tions of the (2n + 2)–body problem consisting of n masses equal to 1 at the vertices of a regular n–gon inscribed in a circle of radius 1, n masses equal to m at the vertices of the n–gon whose vertices are the midpoints of the edges of the initial n–gon, and two masses equal to µ on the straight line orthogonal to the plane containing the two n-gons passing through their barycenters and at a distance a of it. (a) For n = 3 there does not exist central configurations of this type. (b) For each n > 4 there exists at least one family of central config- urations of this type depending on the parameter a. Statements (b) and (a) of Theorem 1 are proved in sections 3 and 4, respectively. We note that the spatial convex but non–strictly convex central configuration found by Fernandes, Garcia and Mello for the 10–body problem is the central configuration of Theorem 1 for n = 4. In section 2 we provide the explicit equations for the double– pyramidal central configurations of the 2n + 2–body problem. 4 2 Equations of the central configura- tions We consider N = 2n+2 with n ∈ N, n > 3, n equal masses m1 = · · · = mn at the vertices of a (regular) n–gon, mn+1 = · · · = m2n = m at the midpoint (regular) n–gon (the n–gon whose vertices are the midpoints of the edges of the initial n–gon), and 2 masses m2n+1 = m2n+2 = µ located symmetrically on the straight line orthogonal to the plane containing the n–gons passing through their center. Without loss of generality we can choose the unit of mass so that m1 = · · · = mn = 1, and we can take the unit of length so that the radius of the circle containing the initial n-gon be one. By using complex coordinates in the plane that contains the regular n–gons, the positions of the vertices of the initial n-gon can be written as qj = (eiϕj , 0) ∈ C × R with ϕj = 2πj/n for j = 1, . . . , n and the vertices of the midpoint n–gon can be written as qj+n = α(ei(ϕj −π/n) , 0) ∈ C × R with α = cos(π/n) for j = 1, . . . , n. Let the positions of the masses m2n+1 and m2n+2 be q2n+1 = (0, a) and q2n+2 = (0, −a), respectively, with a > 0. It is easy to check that the center of mass of the system is at the origin. The first n equations of (1) become n X n X qk − qj+n 2 X qk − q2n+ℓ qk − qj 3 +m 3 +µ = λqk , |qk − qj | |qk − qj+n | |qk − q2n+ℓ |3 j=1 j=1 ℓ=1 j 6= k (2) for k = 1, . . . , n, the following n equations become n X n X 2 X qk+n − qj qk+n − qj+n qk+n − q2n+ℓ 3 +m 3 +µ = λqk+n , |qk+n − qj | |qk+n − qj+n | |qk+n − q2n+ℓ |3 j=1 j=1 ℓ=1 j 6= k (3) for k = 1, . . . , n, and the last 2 equations of (1) are Xn Xn q2n+1 − qj q2n+1 − qj+n q2n+1 − q2n+2 3 + m 3 +µ = λq2n+1 , |q − qj | |q2n+1 − qj+n | |q2n+1 − q2n+2 |3 j=1 2n+1 j=1 Xn Xn q2n+2 − qj q2n+2 − qj+n q2n+2 − q2n+1 + m +µ = λq2n+2 . |q2n+2 − qj |3 |q2n+2 − qj+n |3 |q2n+2 − q2n+1 |3 j=1 j=1 (4) 5 It is easy to check that for j, k = 1, . . . , n and ℓ = 1, 2 s   √ 2π(j − k) |qk − qj | = |eiϕk − eiϕj | = 2 1 − cos , n s   iϕk i(ϕj −π/n) 2π(j − k) π |qk − qj+n | = |e − αe | = 1 + α2 − 2α cos − , n n s   i(ϕk −π/n) iϕj 2π(j − k) π |qk+n − qj | = |αe − e | = 1 + α2 − 2α cos + , n n |qk+n − qj+n | = α|ei(ϕk −π/n) − ei(ϕj −π/n) | = α|eiϕk − eiϕj |, p |qk − q2n+ℓ | = 1 + a2 , p |qk+n − q2n+ℓ | = α2 + a2 , |q2n+1 − q2n+2 | = 2a. The first two components of the n vectorial equations (2) are n X n X eiϕk − αei(ϕj −π/n) eiϕk − eiϕj eiϕk +m +2µ = λeiϕk , |eiϕk − eiϕj |3 |eiϕk − αei(ϕj −π/n) |3 (1 + a2 )3/2 j=1 j=1 j 6= k (5) for k = 1, . . . , n. By dividing each equation of (5) by eiϕk we get n X Xn 1 − ei(ϕj −ϕk ) 1 − αei(ϕj −ϕk −π/n) 1 iϕ iϕ 3 +m i(ϕ −π/n) +2µ = λ, |e − e | k j j=1 iϕ |e k − αe j |3 (1 + a2 )3/2 j=1 j 6= k (6) for k = 1, . . . , n. The first two components of the n vectorial equations (3) are Xn n X αei(ϕk −π/n) − eiϕj 1 ei(ϕk −π/n) − ei(ϕj −π/n) αei(ϕk −π/n) i(ϕ −π/n) iϕ 3 +m 2 iϕ iϕ 3 +2µ 2 2 3/2 = λαei(ϕk −π/n) , j=1 |αe k −e | j α |e k − e j | (α + a ) j=1 j 6= k (7) for k = 1, . . . , n. By dividing each equation of (7) by ei(ϕk −π/n) we get Xn n X α − ei(ϕj −ϕk +π/n) 1 1 − ei(ϕj −ϕk ) α i(ϕ −π/n) iϕ +m 2 iϕ iϕ 3 +2µ 2 = λα, j=1 |αe k −e | j 3 α |e − e | k j (α + a2 )3/2 j=1 j 6= k (8) 6 for k = 1, . . . , n. The first two components of the two vectorial equa- tions (4) become n X Xn −eiϕj −αei(ϕj −π/n) + m = 0. (9) j=1 (1 + a2 )3/2 j=1 (α2 + a2 )3/2 P P Since nj=1 eiϕj = 0 and consequently nj=1 ei(ϕj −π/n) = 0, equation (9) is always satisfied. It is easy to check that the equations corresponding to the third component of the n vectorial equations (2) and (3) are always satisfied. The third component of the two vectorial equations (4), after dividing by a, is equivalent to n n µ 2 3/2 +m 2 2 3/2 + 3 = λ. (10) (1 + a ) (α + a ) 4a Since for all k = 1, . . . , n the set {ϕj − ϕk }j=1,...,n modulus 2π is equal to the set {2πj/n}j=1,...,n and the set {ϕj − ϕk ± π/n}j,k=1,...,n modulus 2π is equal to the set {2πj/n±π/n}j=1,...,n , the equations (6) (respectively, (8)) are independent of k. So it is not restrictive to take k = n. On the other hand, since sin(2πj/n) = − sin(2π − 2πj/n), cos(2πj/n) = cos(2π − 2πj/n) and sin(2πj/n ± π/n) = − sin(2π − (2πj/n ± π/n)), cos(2πj/n ± π/n) = cos(2π − (2πj/n ± π/n)), the equations corresponding to the imaginary parts of (6) and (8) are always satisfied. The equations corresponding to the real parts of (6) and (8) are equivalent to     n−1 X 1 − cos 2πjn Xn 1 − α cos 2πj n − π n 1 r   3 +m r  3 +2µ (1 + a2 )3/2 = λ, j=1 2 1 − cos 2πj n j=1 1 + α2 − 2α cos 2πj n − n π (11) and     2πj π 2πj n X α − cos n + n n−1 X 1 1 − cos n α r +m +2µ 2 = λα,  3 α2 r    3 (α + a2 )3/2 j=1 2πj π j=1 2πj 1 + α2 − 2α cos n + n 2 1 − cos n (12) 7 respectively. Notice that     2πj π 2πj π Xn 1 − α cos n − n Xn 1 − α cos n + n r  3 = r  3 . j=1 2πj π j=1 2πj π 1 + α2 − 2α cos n − n 1 + α2 − 2α cos n + n (13) We isolate λ from equation (10) and we substitute it into equations (11) and (12). Then, taking into account (13), the equations (1) for the central configurations of our problem are equivalent to the following system of two equations      a11 a12 m −b1 = , (14) a21 a22 µ −b2 where   2πj π n X 1 − α cos n + n n a11 = r − ,  3 (α2 + a2 )3/2 j=1 2πj π 1 + α2 − 2α cos n + n 2 1 a12 = 2 3/2 − 3, (1 + a ) 4a   2πj n−1 X 1 1 − cos nnα a21 = − , α2 r   3 (α2 + a2 )3/2 j=1 2 1 − cos 2πj n 2α α a22 = − 3, (α2 +a )2 3/2 4a   2πj n−1 X 1 − cos n n b1 = r  − ,  3 (1 + a2 )3/2 j=1 2πj 2 1 − cos n   Xn α − cos 2πj n + π n αn b2 = r  3 − (1 + a2 )3/2 . j=1 1 + α2 − 2α cos 2πj n + n π 8 3 Proof of statement (b) of Theorem 1 We start with some auxiliary lemmas. We define   n−1 X 1 − cos 2πj n−1 X n−1   n 1 1X πj βn = r  = r = csc .  3    4 n j=1 2πj j=1 2 2 1 − cos 2πj j=1 2 1 − cos n n Lemma 2 Let s  2/3 ∗ n a = − 1, βn then the following statements hold. (a) a∗ is not defined for n > 473, (b) for n < 473, b1 > 0 when a > a∗ , b1 = 0 when a = a∗ , and b1 < 0 when a < a∗ , (c) b1 > 0 for n > 473, √ √ (d) a12 > 0√ for a > 1/ 3, a12 = 0 for a = 1/ 3, and a12 < 0 for a < 1/ 3, √ √ (e) a22 > 0√for a > α/ 3, a12 = 0 for a = α/ 3, and a12 < 0 for a < α/ 3, √ √ (f ) a∗ > 1/ 3 for 3 6 n < 53, 0 < a∗ < 1/ 3 for 53 6 n 6 472. Proof. In [17] it is proved that βn /n is increasing with n and that βn /n > 1 for n > 473. Therefore statement (a) follows directly from the results in [17]. Moreover for n > 473 b1 βn 1 = − > 0, n n (1 + a2 )3/2 because βn /n > 1 (see again [17]). This proves statement (c). State- ments (b), (d) and (e) follow from simple computations. Finally, com- puting the values of a∗ for n = 3, . . . , 472 we get statement (f). Lemma 3 The function H(n) = α b1 − b2 is positive for all n > 473. 9 Independently of us and almost simultaneously Barrabés and Cors in [1] have proved that H(n) > 0 for all n > 5. Here we give our proof for n > 473 which is different from their proof. Proof. We define α(1 − cos x) α − cos x f (x) = , F (x) = . (2(1 − cos x))3/2 (1 + α2 − 2α cos x)3/2 Then H(n) can be written as n−1 X   n X   2πj 2πj π H(n) = f n − F n + n . j=1 j=1 Since α = cos(π/n), we have that F (2πn/n ± π/n) = 0. Therefore Xh n−1  2πj   2πj i π H(n) = f n −F n + n . j=1 On the other hand, since f (x) = f (2π − x) and F (x) = F (2π − x) we have that H(n) = 2A + Ao when n is odd and H(n) = 2A + Ae when n is even, where [n ]−1 h    i X2 2πj 2πj π A= f n −F n + n , j=1 [ n2 ] denotes the integer part of n2 and      2π[ n2 ] 2π[ n2 ] π π Ao = 2f −F + = 2f π − − F (π), n n n n   2π[ n2 ] Ae = f = f (π). n Next we want to see which terms in the sum A are positive and which are negative when n > 473. The sum A can be written as [n ]−1   X2 2πj A= G n , j=1 where G(x) = f (x) − F (x + π/n). Notice that the first term in the sum A corresponds to x = 2π/n and the last term corresponds to x = π − 3π/n when n is odd, and x = π − 2π/n when n is even. Let 10 Ix = [2π/n, π − 3π/n] when n is odd, and Ix = [2π/n, π − 2π/n] when n is even. We claim that the function G has a unique zero for x ∈ Ix and this zero belongs to the interval (π − 5π/n, π − 4π/n). Next we prove the claim. To simplify the computations we introduce the change of variables e z = x + π/n and consider the equation G(z) = 0 with e α2 (α − cos z)2 G(z) = (f (z−π/n))2 −(F (z))2 = − . 8(1 − sin πn sin z − cos πn cos z) (1 + α2 − 2α cos z)3 Let g(z) be the numerator of the √ e factorization of G(z), after doing the π π √ substitutions cos n = α, sin n = 1 − α , and sin z = 1 − cos2 z we 2 get that equation g(z) = 0 is equivalent to equation   (1 − α)(1 + α) α2 α4 + 4α2 + 7 − 2α 3α4 + 9α2 + 8 cos z+    4 3α4 + 6α2 + 2 cos2 z − 8α α2 + 1 cos3 z p p = 8 1 − α2 1 − cos2 z (α − cos z)2 . Squaring both sides of this equation to drop off the square roots and doing the substitution c = cos z we get an equation equivalent to g(z) = 0 given by    g(c) = −(1 − α2 ) 64 α8 + α6 − α4 − α2 − 1 c6 − 64α 3α8 + 6α6 − α4 − 6α2 − 6 c5 + e  16α2 15α8 + 45α6 + 22α4 − 42α2 − 60 c4 −  16(α − 1)α3 (α + 1) 10α6 + 50α4 + 95α2 + 73 c3 +  4α2 15α10 + 75α8 + 133α6 + 21α4 − 168α2 + 4 c2 −  4α3 3α10 + 18α8 + 44α6 + 30α4 − 39α2 + 8 c +   α4 α2 + 3 α8 + 4α6 + 10α4 − 4α2 + 5 = 0. Note that all solutions of equation G(x) = G(z−π/n) = 0 are solutions of ge(c) = 0 with c = cos z. The interval Ix in the variable c is given by Ic = [cos(3π/n), cos(π− 2π/n)] when n is odd and Ic = [cos(3π/n), cos(π − π/n)] when n is even. In what follows we will work with α = cos(π/n) instead of n, so we will not distinguish between n odd and n even and we take Ic = [cos(3π/n), cos(π − π/n)]. Since n = π/ arccos α, after some sim- plifications Ic can be written as Iα = [4α3 −3α, −α]. We are interested in solutions of ge(c) = 0 with c ∈ Iα and α ∈ [cos(π/473), 1) ⊂ D with D = [999/1000, 1). 11 First we prove that the number of solutions of ge(c) = 0 does not change for α ∈ D. The number of solutions of e g(c) = 0 can change when e ′ g(c) = 0 and ge (c) = 0 simultaneously. We compute Res[eg (c), ge′ (c), c] and we get the following polynomial in the variable α  T (α) = −4398046511104(α2 − 1)31 α16 α8 + α6 − α4 − α2 − 1  729α10 − 2187α8 + 459α6 + 33831α4 + 13824α2 − 65536 . Applying Sturm’s algorithm we see that there are no zeroes of T for α ∈ D. Therefore, using the properties of resultants, we get that the number of real solutions of e g(c) = 0 does not change for α ∈ D. The number of solutions of ge(c) = 0 with c ∈ Iα can change when α is such that either c = 4α3 − 3α or c = −α. Applying Sturm’s algorithm we see that the polynomials ge(4α3 − 3α) and ge(−α) have no zeroes for α ∈ D. So the number of solutions of e g(c) = 0 with c ∈ Iα does not change for α ∈ D. Applying Sturm’s algorithm again we see that for instance for α = 99999/100000 ∈ [cos(π/473), 1) there is a unique real root c∗ of ge(c) = 0 with c∗ ∈ Iα . Next we see that this real root satisfies that c∗ ∈ (cos(π − 4π/n), cos(π − 3π/n)). After some simplifications and taking n = π/ arccos α we get cos(π − 3π/n) = 3α − 4α3 and cos(π − 4π/n) = −8α4 + 8α2 − 1. By doing the change of variables γ = (α − 99/100)/(1 − α) we get (100γ + 99)4 (200γ + 199)2 ge(3α − 4α3 ) = B1 , 3814697265625 · 1038 (γ + 1)28 where B1 is a polynomial of degree 20 in the variable γ with positive coefficients that is given by  3125000000000000000γ 10 + 30078125000000000000γ 9 + 130192968750000000000γ 8 + 333726022656250000000γ 7 + 560991664380859375000γ 6 + 646166008553593750000γ 5 + 516451807542245312500γ 4 + 282813389281368250000γ 3 + 101545293194930191250γ 2 +  21586058638174381500γ + 2062870261351502751 40625000000000000000γ 10 + 401078125000000000000γ 9 + 1781852968750000000000γ 8 + 4691003022656250000000γ 7 + 8104450691880859375000γ 6 + 9601063173433593750000γ 5 + 7898547219442445312500γ 4 + 4455668938082168250000γ 3 + 1649462704496130191250γ 2 + 361845223038974381500γ +  35719897641551502751 . 12 g(3α − 4α3 ) > 0 for all α ∈ D. In a similar Since γ > 0 for all α ∈ D, e way, by doing the change of variables γ = α − 9/10 we get (20γ + 19)6 g(−8α4 + 8α2 − 1) = − e B2 , 3814697265625 · 1016 (γ + 1)34 where B2 is a polynomial of degree 26 in the variable γ with positive coefficients. Since γ > 0 for all α ∈ D, ge(−8α4 + 8α2 − 1) < 0 for all α ∈ D. Therefore the unique solution of ge(c) with c∗ ∈ Iα always satisfies that c∗ ∈ (cos(π −4π/n), cos(π −3π/n)). This solution provides a solution x∗ of G(x) = 0 with x∗ ∈ (π − 5π/n, π − 4π/n). This proves the claim. Now we prove that G(x) > 0 when x ∈ Ix and x < x∗ , and G(x) < 0 when x ∈ Ix and x > x∗ . Since f (x) > 0 and F (x + π/n) > 0 for x ∈ e Ix it is sufficient to analyze the sign of G(x) = (f (x))2 −(F (x+π/n))2 . After some simplifications and using that α = cos(π/n) we get   e 2π α2 g1 (α) G = , n 16 (1 − α2 ) (8α2 + 1)3    2 1 − α2 g (α) e π− 4π α 2 G = − , n 16 (2α − 1) (8α − 5α2 + 1)3 2 2 4 where g1 (α) = 512α6 + 192α4 + 24α2 − 255, g2 (α) = 512α10 − 448α8 − 680α6 + 1003α4 − 434α2 + 63. After doing the change of variables γ = α − 99/100 in the functions g1 and g2 respectively we get that g1 (γ + 99/100) and g2 (γ + 99/100) are polynomials in the variable γ of degrees 6 and 10 respectively having positive coefficients, so g1 (γ+99/100) > 0 and g2 (γ+99/100) > 0 in D because γ > 0 in D. Moreover 8α4 − 5α2 + 1 > 0 for all α. Therefore e G(2π/n) > 0 and G(π e − 4π/n) < 0 for all α ∈ [cos(π/473), 1). In short, G(x) > 0 when x ∈ Ix and x < x∗ , and G(x) < 0 when x ∈ Ix and x > x∗ We have just seen that A can be written as A = Ap + Am with 13 Ap > 0 and Am < 0 where [n ]−2     X2 2πj 2πj π Ap = Apo = f −F + , n n n j=1     2π([ n2 ] − 1) 2π([ n2 ] − 1) π Am = Amo = f −F + n n n     3π 2π = f π− −F π− , n n when n is odd, and [n ]−3      X2 2πj 2πj π Ap = Ape = f + −F n n n j=1     2π([ n2 ] − 1) 2π([ n2 ] − 1) π Am = Ame = f −F + + n n n     2π([ n2 ] − 2) 2π([ n2 ] − 2) π f −F + n n n         2π π 4π 3π = f π− −F π− +f π− −F π− , n n n n when m is even. In short, H(n) = 2Apo + 2Amo + Ao when n is odd and H(n) = 2Ape + 2Ame + Ae when m is even, where the terms Apo and Ape are positive. Next we see that the terms 2Amo + Ao and 2Ame + Ae are also positive. We start with the case n odd. Since α+1 > 4α2 −3α+1, after some simplifications we see that for all α ∈ [cos(π/473), 1)    2Amo + Ao = 2f π − 3π 2π π n − 2F π − n + 2f π − n − F (π) √ 2 ! 1 2α 2(2α − 1) 1 = √ − − α + 1 2α − 1 (4α2 − 3α + 1)3/2 (α + 1)3/2 √ 2 ! 1 2α 4α − 1 > √ − . α + 1 2α − 1 (4α2 − 3α + 1)3/2 After doing the change of variables γ = α − 9/10 √ !2 !2 2α2 4α − 1 B3 − = , 2α − 1 (4α2 − 3α + 1) 3/2 800(5γ + 2)2 (200γ 2 + 210γ + 77)3 14 where B3 is a polynomial of degree 10 in the variable γ with positive coefficients, so it is positive because γ > 0 for α ∈ [cos(π/473), 1). In short 2Amo + Ao > 0 for all α ∈ [cos(π/473), 1) and consequently if n is odd, then H(n) > 0 for all n > 473. In the case n even     2Ame + Ae = 2f π − 2π π 4π n − 2F π − n + 2f π − n − 2F π − n + f (π). 3π It is easy to check that f (x), F (x) > 0 for all x ∈ (2π/n, π). After doing the change of variables γ = (α − 9/10)/(1 − α) we have 4π 2 2 α2 16α2 2f π − n − 2F (π − πn ) = − 4 (2α2 − 1)2 (3α2 + 1)3 25(γ + 1)2 (10γ + 9)2 (20γ + 19) = 4 (50γ 2 + 80γ + 31)2 (400γ 2 + 740γ + 343)3  1120000γ 4 + 4184000γ 3 + 5836000γ 2 + 3601080γ + 829053 . and 2 2π  2 1 3π 2  2 16α2 2α2 − 1 2f π − n + f (π) − 2F (π − = (α + 2) − n ) 16 (8α4 − 5α2 + 1)3 B4 = , 1600(γ + 1) (10000γ + 34500γ 3 + 44575γ 2 + 25570γ + 5497)3 2 4 where B4 is a polynomial of degree 14 in γ with positive coefficients, so they are positive for α ∈ [cos(π/473), 1) because γ is positive in [cos(π/473), 1). In short if n is even, H(n) > 0 for all n > 473. This completes the proof. Lemma 4 Let Nm (a, n) = a12 b2 − a22 b1 , then the following state- ments hold √ √ (a) Nm (1/ 3, n) > 0 for 3 6 n < 53 and Nm (1/ 3, n) < 0 for n > 53, (b) Nm (a∗ , n) > 0 for n = 3, 4, Nm (a∗ , n) < 0 for 5 6 n < 53 and Nm (a∗ , n) > 0 for 53 6 n < 473, (c) lima→0+ Nm (a, n) = +∞ for n > 473. Proof. √ If a√= 1/ 3, then from Lemma 2 (d) a12 = 0, and consequently Nm (1/ 3, n) = −a22 b1 . Since α ∈ [1/2, 1) (recall that α = cos(π/n)), 15 √ √ we get 1/ 3 > α/ 3; then from Lemma 2 (e) a22 > 0. Moreover from Lemma 2 (b) and (f) b1 < 0 for n < 53 and b1 > 0 for 53 6 n 6 473 and from Lemma 2 (c) b1 > 0 for n > 473. This proves statement (a). Statement (b) can be proved by computing the values of Nm (a∗ , n) for n = 3, . . . , 472. Now we prove statement (c). If follows easily that 1 lim Nm (a, n) = lim (α b1 − b2 ), a→0+ a→0+ 4a3 Then the proof follows from the fact that α b1 − b2 > 0 for all n > 473 (see Lemma 3). 3.1 Proof of Theorem 1 for n = 4 We are interested in the solutions of system (14) with m, µ > 0. When n = 4 we claim that we can find a solution of system (14) with µ = 0 and m > 0 that can be continued to a family of solutions of (14) with µ > 0 small and m > 0. Now we prove the claim. When n = 4 the coefficients of system (14) are √ √ 8 2 √ 6 2 a11 = − +2 2+ √ , (2a2 + 1)3/2 5 5 2 1 a12 = 3/2 − 3, (a2 + 1) 4a 8 √ 1 a21 = − 3/2 + 2+ , 2 (2a + 1) 2 4 1 a22 = 3/2 − √ , (2a2 + 1) 4 2a3 4 1 1 b1 = − 3/2 +√ + , 2 (a + 1) 2 4 √ 8 2 2 b2 = √ − . 5 5 (a2 + 1)3/2 We assume that a11 6= 0 and we look for solutions of (14) with µ = 0. Under these assumptions, from (14) we get m = −b1 /a11 . Straightforward q p computations show that a11 > 0 for a > a⋆,1 = 5 3 √ 58 3886 − 870 5 − 12 ≈ 0.7584732493, a11 < 0 for a < a⋆,1 , and 16 q 2/3 p √ 3 a11 = 0 for a = a⋆,1 ; and b1 > 0 for a > = 4 27 a⋆,2 9−4 2−1 ≈ 1.262765225, is b1 < 0 for a < a⋆,2 , and b1 = 0 for a = a⋆,2 . Therefore m > 0 when a ∈ (a⋆,1 , a⋆,2 ). On the other hand, from (14) again, µ = 0 when Nµ (a) = a21 b1 − a11 b2 = 0. We see that Nµ (a⋆,1 ) > 0 and Nµ (a⋆,2 ) < 0. Therefore, since Nµ is defined for all a ∈ R+ , there exists at least a zero a = aµ of Nµ with aµ ∈ (a⋆,1 , a⋆,2 ). If this zero does not satisfy condition D(aµ ) = a11 a22 − a21 a12 = 0, then this solution can be continued to a family of solutions of (14) with m > 0 and µ > 0 small, which is given by Nm (a) a12 b2 − a22 b1 Nµ (a) a21 b1 − a11 b2 m = m(a) = = , µ = µ(a) == . D(a) a11 a22 − a12 a21 D(a) a11 a22 − a12 a21 (15) If the value aµ is such that D(aµ ) = 0, then either system (14) has no solutions when Nm (aµ ) = a12 b2 − a22 b1 6= 0, or it has infinitely many solutions when Nm (aµ ) = 0. We note that the first case is not possible because, since a11 6= 0, from condition D(aµ ) = 0 we get a22 = a21 a12 /a11 and from condition Nµ (aµ ) = 0 we get b2 = b1 a21 /a11 , then Nm (aµ ) = 0. The infinitely many solutions that we can have when Nµ (aµ ) = 0, D(aµ ) = 0 and Nm (aµ ) = 0 are −b1 − a12 µ m= , µ ∈ R. a11 In this last case, since −b1 /a11 > 0 there exists infinitely many families of solutions of (14) with m > 0 and µ > 0 small. Notice that if a11 = 0, that is a = a⋆,1 , then the solution is m = m(a⋆,1 ) ≈ 28614318, 2.4304510854826634604 and µ = µ(a⋆,1 ) ≈ 2.4304510854 6= 0 (see (15)). This completes the proof of the claim, so Theorem 1 is proved for n = 4. Remark: We can eliminate the square roots of equation Nµ (a) = 0 by squaring conveniently as many times as we need, then dropping off the denominators we get a polynomial equation of degree 24 in the variable a whose set of solutions contains at least all solutions of equation Nµ (a) = 0. This polynomial equation has two positive real solutions that can be computed numerically but only a = aµ ≈ 1.1778268479 is a solution of Nµ (a) = 0. Proceeding in the same way with the equation D(a) = 0 we arrive to a polynomial equation of degree 36 in the variable a that has also two positive real solutions but only a = ad ≈ 0.7111336024 is a solution of D(a) = 0. Therefore Nµ and D are not zero simultaneously, so for all positive a with a 6= ad 17 system (14) has a unique solution m = m(a), µ = µ(a) which is given by (15), and for a = ad system (14) has no solution. The solution m = m(a), µ = µ(a) provides a central configurations when m(a) > 0 and µ(a) > 0. Now we find the set where these conditions are satisfied. As above we eliminate the square roots of equation Nm (a) = 0, we drop off the denominators we get a polynomial equation of degree 36 in the variable a that has 6 real solutions but only the solutions a = am,1 ≈ 0.1092983002, a = am,2 ≈ 1.3587461717 and a = am,3 = 3.7016660737 are solutions of Nm (a) = 0. Analyzing the signs of Nµ , D and Nm we get that Nµ (a) > 0 for a ∈ (0, aµ ) and Nµ (a) < 0 for a ∈ (aµ , +∞); D(a) > 0 for a ∈ (ad , +∞) and D(a) < 0 for a ∈ (0, ad ); and Nm (a) > 0 for a ∈ (am,1 , am,2 ) ∪ (am,3 , +∞) and Nm (a) < 0 for a ∈ (0, am,1 ) ∪ (am,2 , am,3 ). So m(a) > 0 and µ(a) > 0 for a ∈ (ad , aµ ) = (0.7111336024, 1.1778268479) and the configuration is central. 3.2 Proof of Theorem 1 for n > 5 Proceeding in a similar way than in the case n = 4, we claim that for all n > 5 we can find at least a value of a for which there exist a solution of (14) with m = 0 and µ > 0 that can be continued to a family of solutions of (14) with m > 0 and µ > 0. This family of solutions will provide a family of central configuration of our problem, this proves Theorem 1 for n > 5. Next we prove the claim. Let Nm (a, n) = a12 b2 −a22 b1 , Nµ (a, n) = a21 b1 −a11 b2 , D(a, n) = a11 a22 −a21 a12 . We assume that a12 6= 0 and we look for solutions of (14) with m = 0. If m = 0 and a12 6= 0, then from (14) we get µ = −b1 /a12 . We analyze the sign of −b1 /a√12 . From Lemma 2, if 5 6 n < 53, then −b1 /a12 > 0 for √ a ∈ (1/ 3, a∗ ), if 53 6 n 6 472, then −b1 /a12 > 0 for √a ∈ (a∗ , 1/ 3); and if n > 473, then −b1 /a12 > 0 for a ∈ (0, 1/ 3). Now we prove that there exists at least a value of a such that m = 0 in the region where µ > 0. If m = 0, then Nm (a, n) must be √ zero. From Lemma 4 we have that if 5 6 n 6 472, then ∗ Nm (1/ 3, n) and Nm (a , n) have different √ signs, and if n > 473 then lima→0+ Nm (a, n) = +∞ and Nm (1/ 3, n) < 0. Since Nm (a, n) is well defined for all a ∈ R+ we get that for all n > 5 there exist at least a value of a, a = am , such √ that Nm (am , n) = 0 and µ√> 0. Notice that if a12 = 0 (i.e. a = 1/ 3, see Lemma 2) then Nm (1/ 3, n) is not zero, 18 so the assumption a12 6= 0 is not restrictive. If D(am , n) 6= 0, then the solution of (14) wit a = am is unique and it can be continued to a family of solutions of (14) in a with m > 0 and µ > 0. As in (15), this family is given by m = Nm (a, n)/D(a, n) and µ = Nµ (a, n)/D(a, n). If the value am is such that D(am , n) = 0, then either system (14) has no solutions when Nµ (am , n) 6= 0, or it has infinitely many solutions when Nµ (am , n) = 0. We note that the first case is not possible. This is proved following the same arguments than in the case n = 4. The infinitely many solutions that we have when Nm (am , n) = 0, D(am , n) = 0 and Nµ (am , n) = 0 are −b1 − a11 m m ∈ R, µ= . a12 In this last case since −b1 /a12 > 0, for all n > 5 there exists infinitely many families of solutions of (14) with m > 0 and µ > 0. This proves the claim, so Theorem 1 is proved for n > 5. Remark: We have computed numerically the values of a for which Nm (a, n) = 0 and the values of a for which D(a, n) = 0 for 5 6 n 6 600 and they never coincide. Moreover for 5 6 n 6 600 we have found that both functions, Nm (a, n) and D(a, n), have a unique positive zero. Therefore we have strong numerical evidences that for each n > 5 system (14) cannot have infinitely many solutions and that there exists a unique value of a such that the solution of system (14) satisfies that m = 0 and µ > 0. Proceeding as in the case n = 4 we have computed for 5 6 n 6 600 the set where the solution of system (14) provides a central configu- ration; that is, m > 0 and µ > 0. Let S denote this set. We see that for all 5 6 n 6 600 the set S is an interval whose endpoints are the solutions of Nm (a, n) = 0 and D(a, n) = 0, respectively. In particular, if a = am (n) is the solution of Nm (a, n) = 0 and a = ad (n) is the solution of D(a, n) = 0, then S = (ad (n), am (n)) when n 6 128 and S = (am (n), ad (n)) when n > 128. The functions am (n) and ad (n) for 5 6 n 6 600 are plotted in Figure 1 (a). In Figure 1 (b) we plot the length of the interval S for 5 6 n 6 600. In Table 1 we give the numerical value of S form some values of n. 19 0.00008 0.5776 0.5774 0.00006 0.5772 0.00004 0.5770 0.00002 0.5768 100 200 300 400 500 600 100 200 300 400 500 600 (a) (b) Figure 1: (a) The plot of the solution of Nm (a, n) = 0, am (n), (the function having a minimum) and the plot of the solution of D(a, n) = 0, ad (n), (the increasing function). (b) The plot of the length of the interval S. n D n D 5 (0.6054508601, 1.047660644) 20 (0.5722706230, 0.5876432515) 6 (0.5735628097, 0.8966193846) 30 (0.5746983627, 0.5798882468) 7 (0.5649527372, 0.7993129461) 40 (0.5757077735, 0.5780266218) 8 (0.5631917539, 0.7341595533) 50 (0.5762252724, 0.5774202950) 9 (0.5635643266, 0.6904238044) 60 (0.5765274024, 0.5771954491) 10 (0.5645832304, 0.6608539922) 70 (0.5767199392, 0.5771108814) 11 (0.5657382437, 0.6405062783) 80 (0.5768506065, 0.5770839197) 12 (0.5668499893, 0.6261710814) 90 (0.5769435911, 0.5770820288) 13 (0.5678603234, 0.6158181455) 100 (0.5770122548, 0.5770911638) n D 128 (0.5771316623, 0.5771323567) 129 (0.5771338677, 0.5771346567) 200 (0.5772516235, 0.5772167740) 300 (0.5773027503, 0.5772730689) 400 (0.5772997096, 0.5773220724) 500 (0.5773143684, 0.5773314948) 600 (0.5773233276, 0.5773368180) Table 1: The interval S providing central configurations for some n > 5. 20 4 Proof of statement (a) of Theorem 1 When n = 3 the coefficients of system (14) are 3 4 4 a11 = −  3/2 +√ + , a2 + 14 3 9 2 1 a12 = 3/2 − 3, (a2 + 1) 4a 4 3 a21 = √ −  , 3 2 a2 + 1 3/2 4 1 1 a22 =  3/2 − 3, a2 + 14 8a 1 3 b1 = √ − , 3 (a2 + 1)3/2 4 3 b2 = − . 9 2 (a2 + 1)3/2 Recall that Nm (a) = a12 b2 −a22 b1 , Nµ (a) = a21 b1 −a11 b2 , D(a) = a11 a22 −a12 a21 . Assuming that D(a) 6= 0, the solutions of system (14) are m = m(a) = Nm (a)/D(a) and µ = µ(a) = Nµ (a)/D(a). Next we see that there are no solutions with m(a) > 0 and µ(a) > 0, therefore there are not central configurations for our problem when n = 3. Let a = am denote a solution of Nm (a) = 0, a = aµ a solution of Nµ (a) = 0 and a = ad a solution of D(a) = 0. The set of a’s where m(a) > 0 and µ(a) > 0 is formed by an interval or a union of intervals whose endpoints are zeroes of either Nm , Nµ or D. In particular, we have the following possibilities for the endpoints of each interval: two zeroes of Nm ; two zeroes of Nµ ; a zero of Nm and a zero of D; a zero of Nµ and a zero of D; and two zeroes of D. First we prove that there is no a such that m = 0 and µ > 0. If m = 0 then b1 b2 µ=− =− . a12 a22 √ Straightforward √ computations show that√ b1 > 0 for a > 2, b1 = √0 for a = 2, and b1 < 0 for 0 < a < 2; a12 > 0 for a > 1/ 3, 21 √ √ a12 = 0 for a = 1/ 3,√and a12 < 0 for a < 1/ 3; √ √ b2 > 0 for a > 5/2, b2 = 0 for a = √ 5/2, and b 2 < 0 for a < √ 5/2; and finally a22 > 0 √for a > 1/(2 3), a22 = 0 for a = 1/(2 3), and a22 < 0 for a < 1/(2 3). Therefore √ µ > 0 (that√ is, −b1 /a12 > 0 and −b2 /a22 > 0 simultaneously) when 1/ 3 < a < 5/2. Now √ we√ prove that equation Nm (a) = 0 has no solutions with a ∈ (1/ 3, 5/2). We eliminate the square roots of equation Nm (a) = 0 by squaring conveniently as many times as we need, then dropping off the denominators we get a polynomial equation Pm (a) = 0 of degree 36 in the variable a. We do the change of variables γ = −(5(2a − 1))/(2(5a √ − √6)) in the polynomial Pm (a) (notice that γ > 0 for all a ∈ (1/ 3, 5/2)) and we get a rational function in the variable γ having all the coefficients positive, therefore there is no solution of system (14) with m = 0 and µ > 0. In a similar way, if µ = 0 then b1 b2 m=− =− . a11 a21 Straightforward computations show that a11 < 0 for a < a with s  2/3 1 2 a= 9 √ − 1 ≈ 0.8992976476, 2 1+3 3 √ a11 = 0 for a √ = a, and a11 < 0 for a √ < a; b2 > 0 for a > 5/2, b2 = 0 √ for a = 5/2, and b2 < √ 0 for a < 5/2; and finally √ a21 > 0 for a > 1/ 2, a22 = 0 for a = 1/ 2, and a22 < 0 for a < 1/ 2. Therefore √ is, −b1 /a11 > 0 and −b2 /a21 > 0 simultaneously) when m > 0 (that a < a < 5/2. By eliminating the square roots of equation Nµ (a) = 0 and dropping off the denominators we get a polynomial equation Pµ (a) = 0 of degree 24 in the variable a. Doing the change of variables γ = (4 −√5a)/(5a − 6) in the polynomial Pµ (a) (notice that γ > 0 in a < a < 5/2) as above we see that √ this polynomial equation has no solutions in the interval a < a < 5/2. Therefore there is no solution of system(14) with µ = 0 and m > 0. Now we prove that D and Nµ (respectively, Nm ) are not zero simultaneously. We eliminate the square roots of equation Nµ (a) = 0 (respectively, Nm (a) = 0) by squaring conveniently as many times as we need, then dropping off the denominators we get a polynomial equation Pµ (a) = 0 of degree 24 (respectively, Pm (a) = 0 of degree 36). Doing the same with equation D(a) = 0 we get a polynomial equation 22 PD (a) = 0 of degree 36. We do the resultant of the polynomials PD and Pµ , and of PD and Pm with respect to a, and we get two numbers different from zero. Then by the properties of the resultants these two pairs of polynomials cannot have a common zero. Finally we prove that the set of a’s where m(a) > 0 and µ(a) > 0 cannot be formed √ by an interval whose endpoints are both zeroes D. Since D(1/(2 3)) is different from zero we can assume that a22 6= 0, then from D(a) = 0 we obtain a11 = a12 a21 /a22 . Substituting this value of a11 into the expressions of Nm (a) and Nµ (a) we get a21 Nm (a) = −a22 b1 + a12 b2 , Nµ (a) = (a22 b1 − a12 b2 ). a22 If a21 /a22 > 0 then Nm and Nµ have different signs, so in this case the zeroes of D cannot be the endpoints of an interval providing central configurations. Straightforward √ computations √ show that a21 /a22 > 0 when either 0 < a < 1/(2 3) or a > 1/ 2. On the other hand, by doing the change of variables γ = we √ see that √ the polynomial PD has no zeroes in the interval a ∈ [1/(2 3), 1/ 2], and since this interval is the unique set where Nm and Nµ have the same sign, it follows that two zeroes of D never can be the endpoints of an interval providing central configurations. Remark. Computing numerically the positive real solutions of Nm (a) = 0, Nµ (a) = 0 and D(a) = 0 as we have done for n = 4 we see that Nm (a) = 0 has a unique real solution with a > 0, a = am ≈ 0.13809757; Nµ (a) = 0 has two real solutions with a > 0, a = aµ,1 ≈ 0.3019817428 and a = aµ,2 ≈ 2.5508746364; and D(a) = 0 has a unique real solution with a > 0 a = ad ≈ 0.1235465602. Then Analyzing the signs of Nm , Bµ and D we get that Nm (a) < 0 for a ∈ (0, am ) and Nm (a) > 0 for a ∈ (am , +∞); Nµ (a) < 0 for a ∈ (aµ,1 , aµ,2 ) and Nµ (a) > 0 for a ∈ (0, aµ,1 ) ∪ (aµ,2 , +∞); and D(a) < 0 for a ∈ (ad , +∞) and D(a) > 0 for a ∈ (0, ad ). Therefore m > 0 for a ∈ (ad , am ), and µ > 0 for a ∈ (0, ad ) ∪ (aµ,1 , aµ,2 ). In short, there is no positive a such that µ > 0 and m > 0. Acknowledgements Both authors are partially supported by the Ministerio de Ciencia, Innovación y Universidades, Agencia Estatal de Investigación grants 23 MTM2016-77278-P (FEDER). 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