The Somos-5 sequence and diophantine equation $9p^4-10p^2q^2+17q^4=r^2$?

Question

I. Sequence

The Somos-$k$ sequences have the special property that, even though their recurrence involves division, for certain $k$ and certain initial starting values, then all entries are integers. One example is the generalized Somos-5 sequence,

$a(n)=\color{blue}1, \color{red}1, \color{blue}3, \color{red}{13}, \color{blue}{113}, \color{red}{1525},\dots$ (A360187)

It is associated with the elliptic curve $x^3 - 2x = y^2$, and its bisection turn out to be solutions to $p^4-2q^4=\pm r^2$, namely,

$\color{blue}1^4-2\times0^4 = 1^2\\ \color{blue}3^4-2\times2^4 = 7^2\\ \color{blue}{113}^4-2\times84^4 = 7967^2$

$1^4-2\times\color{red}1^4 = -1^2\\ 1^4-2\times\color{red}{13}^4 = -239^2\\ 1343^4-2\times\color{red}{1525}^4 = -2750257^2$

and so on. A question then: Can we find other Somos-k sequences which solve $ap^4+bp^2q^2+cq^4 = \pm r^2$?

---

II. The Somos 5-sequence

As discussed here, this is given by,

$a(n) = \color{blue}1, 1, \color{blue}1, 1, \color{blue}1, 2, \color{blue}3, 5, \color{blue}{11}, 37, \color{blue}{83}, 274, \color{blue}{1217}, 6161, \color{blue}{22833}, 165713, \color{blue}{1249441},\dots$ (A006721)

starting with $n=0$, hence the blue terms have even index $n$ (A097495) while the black terms have odd index (A097496). Both are associated with the elliptic curve $4x^3 - (121/12)x + 845/216 = y^2$. After some experimentation and transforming it from $\text{cubic} = z_1^2\,$ to $\text{quartic} = z_2^2$, the $a(2n)$ seem to solve the Diophantine equation,

$$9p^4-10p^2q^2+17q^4=(12r)^2$$

with solutions (discarding the repeated $1$'s),

$p = 1,\, 1,\, 4,\, 19,\, 13,\, 254,\, 1291,\dots$

$q = 1,\, 3,\; 2,\; 9,\; 35,\; 264,\; 37,\dots$

$r = \dfrac{\color{blue}1}3,\, \dfrac{\color{blue}3}1, \dfrac{\color{blue}{11}}3, \dfrac{\color{blue}{83}}1, \dfrac{\color{blue}{1217}}3, \dfrac{\color{blue}{22833}}1, \dfrac{\color{blue}{1249441}}3,\dots$

and denominators repeating $(3,1,3,1,\dots)$ though $12r$ remains an integer.

---

III. Question

1. Does this pattern continue for all $a(2n)$?
2. And what's the corresponding Diophantine equation for $a(2n+1)$?

Answer 1

We know $9p^4-10p^2q^2+17q^4=(12r)^2$ and $y^2 = 4x^3 - (121/12)x + 845/216$ are reduced to minimal Weierstrass form, $Y^2+XY = X^3+X^2-2X.$

The value of j-invariant of $Y^2+XY = X^3+X^2-2X\,$ is $j=\frac{1771561}{612}=\frac{11^6}{17\times36}.$

Hence, we search the quartic equation $ap^4+bp^2q^2+cq^4=r^2$ which has the same j-invariant.

The quartic equation $ap^4+bp^2q^2+cq^4=r^2$ can be transformed to $Y^2 = X^3+bX^2+acX\,$ with $X=aU^2,\,$ $Y=arU,\,$ and $U=\frac{p}{q}$.

After depressing the cubic (making the coefficient of $X^2$ vanish) we get

$Y^2=X^3+(-1/3b^2+ac)X+2/27b^3-1/3abc$

$Y^2=X^3+AX+B$

Searching for $(a,b,c)$ such that j-invariant = $12^3\frac{4A^3}{4A^3+27B^2}= \frac{1771561}{612},$ I found the equation $8p^4+5p^2q^2-4q^4=r^2$.

Its minimal Weierstrass form is $y^2+xy=x^3+x^2+8x+10.$

---

Summary:

I. Given the Somos-5 sequence (A006721, with the first two $1$'s dropped),

$a(n) = 1, 1,\, 1,\, 2,\, 3,\, 5,\; 11,\; 37,\; 83,\; 274,\; 1217,\; 6161,\; 22833,\dots$

and the generalized Somos-5 sequence (A360381),

$b(n) = 0, 1, -1, 1, 1, -7,\, 8, -1, -57, 391, -455, -2729, 22352,\dots$

both starting at $\color{red}{n=0}$. OEIS says $a(n)$ and $b(n)$ have the same recurrence but with different initial values. Then the two related Diophantine equations,

$$4p^4+5p^2q^2-8q^4=y_1^2$$ $$8r^4+5r^2s^2-4s^4=y_2^2$$

(where the LHS factors over $\sqrt{17}\,$) are solved by,

$$p=a(2m),\quad q=b(2m)$$

$$r=a(2m+1),\quad s=b(2m+1)$$

For example, let $m=2$, then $(p,q) = (3,1)$ and $(r,s)=(5,-7)$.

---

II. The above is similar to the odd-even alternation for $x^4-2y^4 = \pm z^2$ cited in the post. Assume another generalized Somos-5 sequence (A360187),

$c(n) = 1,\; 1,\; 3,\; 13, 113, 1525, 57123,\, 2165017,\, 262621633,\dots$

and (no OEIS entry yet)

$d(n) = 0, -1, 2, -1,\, 84,\, 1343,\, 6214,\, 2372159, -151245528,\dots$

both starting at $\color{red}{n=0}$. Both $c(n)$ and $d(n)$ have the same recurrence but different initial values. Then,

$$p^4-2q^4=y_1^2$$ $$2r^4-s^4=y_2^2$$

are solved by,

$$p=c(2m),\quad q=d(2m)$$

$$r=c(2m+1),\quad s=d(2m+1)$$

For example, let $m=1$ then $(p,q) = (3,2)$ and $(r,s)=(13,1)$.

Answer 2

My "WXYZ Math Project" is partly motivated to give solutions to Diophantine equations such as the congruent number problem. The main construction is four polynomial sequences that correspond to the four Jacobi theta functions or the four Weierstrass sigma functions.

These four sequences I denote by $w(n),x(n),y(n),z(n)$. The difference of the squares of the last three sequences is a multiple of the square of the first with a factor that does not depend on n. All four of these sequences are generalized Somos-4 sequences with the same parameters.

Note that for all integer $n$ $$w(-n)=-w(n),\,\,x(-n)=x(n),\,\,y(-n)=y(n),\,\,z(-n)=z(n).$$

As a numerical example, let the sequence $w(n)$ be defined by the initial values and recurrence

$$ w(0) = 0, w(1) = 1, w(2) = 12, w(3) = -385, w(4) = -701592, \\ w(n) = (144 w(n-1)w(n-3) + 385 w(n)^2)/w(n-4). $$

The other three sequences have initial values

$$ x(0) \!=\! y(0) \!=\! z(0) \!=\! x(1) \!=\! 1, x(2) \!=\! -23,\,\,y(1) \!=\! 2, y(2) \!=\! 31,\,\, z(1) \!=\! 3, z(2) \!=\! 41 $$

and the same recurrence as $w(n)$. An important relation between the four sequences is

$$ x(n)^2 + 3w(n)^2 = y(n)^2,\quad x(n)^2 + 8w(n)^2 = z(n)^2,\quad y(n)^2 +5w(n)^2 = z(n)^2. $$

The product of the first two relations is

$$ x(n)^4 + 11x(n)^2w(n)^2 + 24w(n)^4 = (y(n)z(n))^2. $$

This gives an affirmative answer to the question

A question then: Can we find other Somos-k sequences which solve $ap^4+bp^2q^2+cq^4 = \pm r^2$?