show this sequence always is rational number
let $\{a_{n}\}$ such $a_{1}=-8$,and such $$4\sqrt[3]{a_{n}}+5\sqrt[3]{a_{n+1}}=3\sqrt[3]{7(a_{n}+1)(a_{n+1}+1)}$$
show that $$a_{n}\in Q,\forall n\in N^{+}$$
I try let $a_{2}=x$,and for $n=1$, then we have $$-8+5\sqrt[3]{x}=3\sqrt[3]{-49(x+1)}\Longrightarrow x=-1/8$$ and for $n=2$ I get $a_{3}=-\dfrac{389017}{4913}$ and so on
The first observation
i want use this well known identity: $$a^3+b^3+c^3=3abc ~~~~~~~~~~~~~~~~~~if~~~~~ a+b+c=0$$ let $a=4\sqrt[3]{a_{n}},b=5\sqrt[3]{a_{n+1}},c=-3\sqrt[3]{7(a_{n}+1)(a_{n+1}+1)}$,so $$64a_{n}+125a_{n+1}-189(a_{n}+1)(a_{n+1}+1)=-180\sqrt[3]{7a_{n}a_{n+1}(a_{n}+1)(a_{n+1}+1)}$$ ADD it by 2021,11.6.PM.18:05
The second observation
and Now I have found this interesting: if $x,y,p,q,a,b\ge 0$, Hölder's inequality : $$(x^3+y^3)(p^3+q^3)(a^3+b^3)\ge (xpa+yqb)^3$$ $1+\sqrt[3]{e^{2a}}\sqrt[5]{e^{b}}\sqrt[15]{e^{2c}} \leq \sqrt[3]{(1+e^{a})^2}\sqrt[5]{1+e^{b}}\sqrt[15]{(1+e^{c})^2}$ the condition it's $$(a_{n}+1)(1+a_{n+1})(4^3+5^3)= (4\sqrt[3]{a_{n}}+5\sqrt[3]{a_{n+1}})^3$$ or $$4\sqrt[3]{a_{n}}+5\sqrt[3]{a_{n+1}}=3\sqrt[3]{7(a_{n}+1)(a_{n+1}+1)}$$ but this sequence $a_{n}<0$ or $a_{n}>0$,so this Hölder inequality seem can't hold
Solution 1:
We could write a cubic curve in equation$\sum_{i=1}^3\sum_{j=1}^3\sum_{k=1}^3a_{ijk}x_ix_jx_k=0$, where we use $x_1,x_2,x_3$ to replace symbol $x,y,1$ here. And the equation is symmetric so that $a_{123}=a_{321}$ etc, similar as a symmetric matrix could be used to represent a quadratic curve. And we could use symbol $A=(a_{ijk})$ to represent the 3-D tensor.
For any point in homogenous coordinates $p=(p_1,p_2,p_3), q=(q_1,q_2,q_3),r=(r_1,r_2,r_3)$, define $A_p=\sum_{i=1}^3\sum_{j=1}^3\sum_{k=1}^3a_{ijk}p_ix_jx_k, A_{pq}=\sum_{i=1}^3\sum_{j=1}^3\sum_{k=1}^3a_{ijk}p_iq_jx_k, A_{pqr}=\sum_{i=1}^3\sum_{j=1}^3\sum_{k=1}^3a_{ijk}p_iq_jr_k$.
If any point p in the cubic curve, we have $A_{ppp}=0$.
If a line intersect with the cubic curve in three different points p,q,r, we have $A_{pqr}=0$.
It is because for any p and q in the curve, we have $A_{ppp}=A_{qqq}=0$, using homogenous coordinates, for any point r in the line passing p&q, we have $r=p+\lambda q$, where $\lambda$ is a real number.
If r is also in the curve, we could have $A_{qqq}\lambda^3+3A_{qqp}\lambda^2+3A_{qpp}\lambda+A_{ppp}=0$.
Since $A_{qqq}=A_{ppp}=0$, we have $A_{qqp}\lambda+A_{qpp}=0$, or $ r=A_{qqp} p - A_{qpp} q$.
So $A_{pqr}=A_{qqp} A_{pqp} - A_{qpp}A_{pqq}$.
Since $A_{qqp}=A_{pqq}, A_{pqp}=A_{qpp}$, we got $A_{pqr}=0$.
Calvin has constructed cubic curve $x^3+1-7y^3=0$ so that it means that if a line intersect with the curve in $(x_1,y_1), (x_2,y_2), (x_3,y_3)$, we must have $x_1x_2x_3+1-7y_1y_2y_3=0$.
Let $(x_3,y_3)=(\frac45,\frac35)$, we have $4x_1x_2+5=21y_1y_2$.
And according to Calvin's another construct, so that we could replace $(x_2,y_2)$ by $(\frac{1}{x_4},\frac{y_4}{x_4})$ and get formula that $4x_1+5x_4=21y_1y_4$.
continue to use condition that 3 points in same line and finally we could get formula:
$\begin{cases} u_1=-2,v_1=1,w_1=-1\\ u_{h+1}=-20u_h^2 + 16u_hv_h -63v_hw_h + 105w_h^2\\ v_{h+1}=25u_hv_h - 63u_hw_h-20v_h^2 + 84w_h^2\\ w_{h+1}=-12u_h^2 + 16u_hw_h -15v_h^2 + 25v_hw_h\\ a_h=(\frac{u_h}{v_h})^3 \end{cases}$
In the picture above, slope of $OA_h$ gives $\sqrt[3]{a_h}$, where $B_h, A_{h-1}, S$ is in one straight line and $A_h$ is the point of symmetry of $B_h$ about y=x. The orange curve is $x^3+y^3=7$. $S(\frac53, \frac43), A_1=B_2(-1,2)$
Weierstrass normal form $U^2=V^3-\frac2{6^37^4}$. $E_0: U_0=\frac1{9*6*7^2}, V_0=\frac{2}{3*6*7}, E_1: U_1=-\frac3{6*7^2}, V_1=\frac2{6*7}$ $x_h=\frac1{42V_h}+\frac{7U_h}{V_h}, y_h=\frac1{42V_h}-\frac{7U_h}{V_h}, a_h=\left(\frac{y_h}{x_h}\right)^3$. $E_h=(h-1)E_0+E_1$(Add in elliptic cure group)
pari/gp code
getslope(n)={
local(E,S0,S1);
local(R,V,U,x,y);
E=ellinit([0,-2/(6^3*7^4)]);
S0=[2/(3*6*7), 1/(9*6*7^2)];
S1=[2/(6*7), -3/(6*7^2)];
R=ellmul(E,S0,n-1);
R=elladd(E,R,S1);
V=R[1];
U=R[2];
x=1/(42*V)+7*U/V;
y=1/(42*V)-7*U/V;
y/x
}
? getslope(1)
%27 = -2
? getslope(2)
%28 = -1/2
? getslope(3)
%29 = -73/17
? getslope(4)
%30 = -65882/90271
? getslope(5)
%31 = 4309182809/191114642
Solution 2:
By looking at the initial values, we hypothesize that
- $a_n = x_n^3, x_n \in \mathbb{Q}$
- $a_n + 1 = 7y_n^3, y_n \in \mathbb{Q}$
This suggests to study rational points on $7y^3 = x^3 + 1$.
We map $ a_i$ to the point $ ( \sqrt[3]{a_i}, \sqrt[3]{\frac{a_i + 1}{7} } )$.
We have the starting points $a_1: (-2, -1), a_2: (-1/2, 1/2)$.
Notice that we also have the rational points $ (-1, 0), (4/5, 3/5)$ on the curve. The latter was inspired by the coefficients in the question, and you will see how this is applicable in a moment.
We apply the usual tricks with group addition to determine the sequence.
Given a solution $(x,y)$, divide throughout by $x^3$ to get $ 7 (\frac{y}{x} ) ^3 = ( \frac{1}{x} ) ^3 + 1 $, so observe that $(\frac{1}{x}, \frac{y}{x} )$ is also a solution.
Given a solution $(x,y)$, we can construct the line that passes through $(x,y)$ and $(\frac{4}{5}, \frac{3}{5} )$ which intersects the cubic again at $(x', y')$.
Furthermore, if $(x,y)$ is a rational point, then so is $(x', y')$. (Prove this.)
We combine this with the previous observation and send $(x,y)$ (to $(x',y')$ and then ) to $(\frac{1}{x'} , \frac{y'}{x'} ) $.
Claim: $(x,y)$ and $(\frac{1}{x'} , \frac{y'}{x'} )$ satisfy $4x + \frac{5}{x'} = 21 y \frac{y'}{x'}$.
See Zhaohui's solution for a beautiful proof that if the cubic intersects a line at 3 points, then $x_1x_2x_3 + 1 = 7 y_1y_2y_3$.
Applying it to the 3 points on the line, we get $\frac{4}{5} x x' + 1 = 7 \times \frac{3}{5} y y'$, and hence the claim follows.
Corollary: This shows that we send $ a_n : (x_n, y_n) $ to $a_{n+1} : (x_{n+1}, y_{n+1}) $ via the above description.
Since the starting point $a_1$ is rational, hence all of these are rational points.
So $ x_n \in \mathbb{Q}$ and $ a_n \in \mathbb{Q}$.
Notes
- Someone who is more familiar with elliptic curves might be able to manipulate the equations better. (IE See Zhaohui's solution)
- This approaches illustrates the theory behind the question, and why the question isn't just "magic".
- If one knew the fact that Zhaohui showed, then this approach would have felt natural. My guess is that similar questions like this can be dealt with in a similar way, especially for quadratic (and cubic) terms. Previously, I mainly dealt with similar questions previously via heavy-handed induction + guessing what the sequence is (EG Zhaohui's formulas at the end). I'm excited to apply this to future questions.
- For $a_1$, the line is actually tangential at $a_1$. So the 3rd point of intersection is $a_1$ (which explain why $a_1, a_2$ are related via $ (\frac{1}{x}, \frac{y}{x} )$.
Uncompleted attempt at proof: The line that passes through $(x,y)$ and $(\frac{4}{5}, \frac{3}{5} )$ is $ (Y' - y) ( \frac{4}{5} - x ) = (X' - x) ( \frac{3}{5} - y ) $, or that $Y' = X'\frac{5y-3}{5x-4} + \frac{3x-4y}{5x-4} $.
Substituting this into $7Y^3 = X^3 + 1$, and applying vieta's formula to find the sum of roots, we get that $\frac{4}{5} + x + x' = \frac{3 (\frac{5y-3}{5x-4})^2 (\frac{3x-4y}{5x-4}) } {1 - (\frac{5y-3}{5x-4})^3 } $, so $ x' = -\frac{625 x^4 - 1000 x^3 - 625 x y^3 + 675 x y + 370 x + 1000 y^3 - 900 y^2 - 148}{5 (5 x - 5 y - 1) (25 x^2 + 25 x y - 55 x + 25 y^2 - 50 y + 37))}$.
I'm unwilling to continue this tedious calculations, but one can find $y'$ and then verify that the equation holds.
Solution 3:
This problem is about the sequence $\,\{a_n\}\,$ where$\,a_ 1 = -8\,$ and it satisfies the recurrence relation $$ 4\sqrt[3]{a_{n}}+5\sqrt[3]{a_{n+1}} = 3\sqrt[3]{7(a_{n}+1)(a_{n+1}+1)}. \tag{1} $$ Notice that the real cube root is a one-to-one and onto function. Thus, define the related real sequence $$ t_n := \sqrt[3]{a_{n}} \quad \text{ for all }\quad n>0. \tag{2} $$ Also, define the function $$ f(u,v) := (4u + 5v)^3 - 3^3\,7(u^3+1)(v^3+1). \tag{3} $$ Notice that $$ (u\,v)^3\, f(1/v,1/u) = f (u,v) \tag{4} $$ for all nonzero real $\,u,v.\,$ Now redefine $\,t_n\,$ as the unique real solution to the recurrence relation $$ f(t_{n-1}, t_n) = 0 \quad \text{ for all }\quad n>1. \tag{5} $$ Also, define $\,t_n\,$ for $n<1$ as the real solution to $\,f(t_n,t_{n+1}) = 0\,$ and notice that equation $(4)$ with equation $(5)$ together imply that $$ t_n = 1/t_{3-n}\quad \text{ for all }\quad n\in \mathbb{Z}. \tag{6} $$
The question asks for a proof that $\,\{a_n\}\,$ is a rational sequence. Since $\,a_n = t_n^3,\,$ It is sufficient to prove that $\,t_n=x_n/y_n\,$ for some rational sequences $\,x_n,y_n.\,$ The proof of this requires a lemma as follows.
Lemma: Define three sequences by recursion $$ x_{n+1} := 12y_n^2 - 25x_nz_n,\;\; y_{n+1} := 15x_n^2 - 16y_nz_n,\;\; z_{n+1} := 20z_n^2 - 9x_ny_n. \tag{7} $$ using these sequences, define the special sequence $$ e_n := x_n^3 + y_n^3 - 7z_n^3. \tag{8} $$ Verify using algebra that $$ 4x_ny_{n+1} + 5y_nx_{n+1} - 21z_nz_{n+1} = 60e_n. \tag{9} $$ Also, verify that $$ e_{n+1} \!=\! e_n (15x_n \!+\!12y_n \!+\!20z_n) ((15x_n)^2 \!+\!(12y_n)^2 \!+\!(20z_n)^2 - 180(3x_ny_n \!+\!5x_nz_\!+\!4y_nz_n). \tag{10} $$ This implies that if $\,e_1 = 0\,$ then $e_n = 0\,$ for all $n>0.\,$ Using equation $(9)$ this implies that $\, 4x_ny_{n+1} + 5y_nx_{n+1} - 21z_nz_{n+1} = 0\,$ for all $n>0.\,$ Also, verify that if $\,t_n:=x_n/y_n,\,$ then $$ f(t_n,t_{n+1}) = e_n\frac{-27((5x_n - 4y_n) (25x_n^2 + 20x_ny_n + 16y_n^2))^2}{(y_ny_{n+1})^3}. \tag{11} $$ This implies that if $\,e_1=0\,$ then $\,f(t_n,t_{n+1}) = 0\,$ for all $n>0.\,$ Note that if the initial values $\,x_1,y_1,z_1\,$ are integers, then $\,x_n,y_n,z_n\,$ are integers and hence $\,t_n\,$ are rational for all $n>0.\,$ In the application of the lemma, let $\,x_1 = -2,\, y_1 = 1, \, z_1 = -1\,$ which gives $\,e_1 = 0\,$ and hence $$ x_n^3 + y_n^3 - 7z_n^3 = 0 \quad \text{ for all } n>0. \tag{12} $$ and also $$ f(t_n,t_{n+1}) = 0 \quad \text{ for all } n>0. \tag{13} $$
As a check, here is a small table of sequence values: $$\begin{array}{|c|c|c|c|} \hline n & x_n & y_n & z_n \\ \hline 1 & -2 & 1 & -1 \\ \hline 2 & -38 & 76 & 38 \\ \hline 3 & 105412 & -24548 & 54872 \\ \hline 4 & -137372929952 & 188227311856 & 83507611664 \\ \hline \end{array}$$ Note that $\,x:y:z = u:v:w\,$ from the Zhaohui Du answer.
Note that I used Mathematica to verify these computations.
My original answer is below
This answer uses generalized Somos-$4$ sequences, which are sequences that satisfy $$ s_{n+2}s_{n-2} = p_1 s_{n+1}s_{n-1} + p_2 s_ns_n, \quad \text{ for all }\quad n\in \mathbb{Z} $$ where $\,p_1,\,p_2\,$ are two parameters. In this question we have $\,p_1 = -3721,\; p_2 = 1663260.\,$ Using these two parameters, define the sequences $\,x_n,y_n,z_n\,$ with this recursion and initial values: $$ x_0 = -17,\; x_1 = -2,\; x_2 = -1,\; x_3 = -73,\; z_2 = 1,\; z_3 = -38. $$ They satisfy the equations $\, z_n = z_{3-n}\,$ and $\, y_n = -x_{3-n} \,$ for all $\, n\in \mathbb{Z}.\,$ Now define $$ t_n := x_n/y_n,\qquad a_n := t_n^3. $$ The three sequences $\,x_n,y_n,z_n\,$ satisfy $$ x_n^3 + y_n^3 = 7z_n^3,\quad 4x_ny_{n+1} + 5x_{n+1}y_n = 21z_nz_{n+1}. $$ These equations imply $$ 4\sqrt[3]{a_{n}}+5\sqrt[3]{a_{n+1}}=3\sqrt[3]{7(a_{n}+1)(a_{n+1}+1)}.$$ Note that all sequences here except $\,t_n\,$ and $\,a_n\,$ are integer sequences, although it is sufficient that $\,x_n,y_n,z_n\,$ are rational sequences which implies that $\,t_n\,$ and hence $\,a_n\,$ are rational sequences also.
As a check, here is a small table of sequence values: $$\begin{array}{|c|c|c|c|} \hline n & x_n & y_n & z_n \\ \hline 1 & -2 & 1 & -1 \\ \hline 2 & -1 & 2 & 1 \\ \hline 3 & -73 & 17 & -38 \\ \hline 4 & -65882 & 90271 & 40049 \\ \hline 5 & -4309182809 & -191114642 & -2252725111 \\ \hline \end{array}$$ Note that $\,x:y:z = u:v:w\,$ from the Zhaohui Du answer.
L. E. Dickson, History of the Theory of Numbers, Volume II, Chapter XXI, section "Numbers the sum of two rational cubes: $x^3+y^3=Az^3$, pp. $572$-$578$ has several solutions for $\,A=7.$