TITLE: show this sequence always is rational number
QUESTION [26 upvotes]: 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

REPLY [10 votes]: 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