Solving $X^2-6Y^2=Z^3$ in positive integers

Question

I’m trying to solve the Diophantine equation $$X^2-6Y^2=Z^3 \tag{$\star$}$$ in positive integers $x,y,z$.

Brute force calculations confirm the naïve intuition that there are many [read: surely infinite!] primitive solutions; numerical observation suggests the solutions have determinable [perhaps even “descent”- or “recurrence”-based?] patterns. For example, one “stream” of solutions is $$(x,y,z) \in \{(5,2,1),\ (49,20,1),\ (485,198,1),\ (4801,1960,1),\dots\},$$ where $z=1$ and the $(x_n,y_n)$ satisfy the recurrence $t_n = 10t_{n-1}-t_{n-2}$. Another such stream exists for $z=19$, another for $z=25$, etc.

Evidently, ($\star$) is related to [but different from] the Pellian equation $$X^2-6Y^2=Z^2,$$ so I’m wondering:

Question #1: Is a complete solution already known, either for the equation or for one of the variables (e.g. characterization of $z$)?

Question #2: If not, what are the most promising ways to attack the problem?

Answer 1

The represented numbers can be described explicitly; this is what becomes possible when there is one class per genus of the discriminant. There are two genera, one form each. We know $x^2 - 6 y^2.$ the other genus has $ -x^2 + 6 y^2.$ This may be unfamiliar; for example, a form and its "negative" are sent to just one ideal in the appropriate quadratic field.

We name three types of (positive) primes. "Good" primes are $3$ along with all primes $p \equiv 1, 19 \pmod {24}$ "Medium" primes are $2$ along with all primes $p \equiv 5, 23 \pmod {24}$ "Bad" primes are all primes $p \equiv 7, 11, 13, 17 \pmod {24}$

A positive number is integrally represented by $x^2 - 6 y^2$ when every bad prime has an even exponent and the sum of all exponents for medium primes is even. The exponents of good primes don't do any harm.

1 =   1   
1 =   1   
3 =  3   3:  good  
4 =  2^2   2:  med  
4 =  2^2   2:  med  
9 =  3^2   3:  good  
9 =  3^2   3:  good  
10 =  2 5   2:  med   5:  med  
10 =  2 5   2:  med   5:  med  
12 =  2^2 3   2:  med   3:  good  
16 =  2^4   2:  med  
16 =  2^4   2:  med  
19 =  19   19:  good  
19 =  19   19:  good  
25 =  5^2   5:  med  
25 =  5^2   5:  med  
25 =  5^2   5:  med  
25 =  5^2   5:  med  
27 =  3^3   3:  good  
30 =  2 3 5   2:  med   3:  good   5:  med  
30 =  2 3 5   2:  med   3:  good   5:  med  
36 =  2^2 3^2   2:  med   3:  good  
36 =  2^2 3^2   2:  med   3:  good  
40 =  2^3 5   2:  med   5:  med  
40 =  2^3 5   2:  med   5:  med  
43 =  43   43:  good  
43 =  43   43:  good  
46 =  2 23   2:  med   23:  med  
46 =  2 23   2:  med   23:  med  
48 =  2^4 3   2:  med   3:  good  
49 =  7^2   7:  bad  
49 =  7^2   7:  bad  
57 =  3 19   3:  good   19:  good  
57 =  3 19   3:  good   19:  good  
58 =  2 29   2:  med   29:  med  
58 =  2 29   2:  med   29:  med  
64 =  2^6   2:  med  
64 =  2^6   2:  med  
67 =  67   67:  good  
67 =  67   67:  good  
73 =  73   73:  good  
73 =  73   73:  good  
75 =  3 5^2   3:  good   5:  med  
75 =  3 5^2   3:  good   5:  med  
75 =  3 5^2   3:  good   5:  med  
76 =  2^2 19   2:  med   19:  good  
76 =  2^2 19   2:  med   19:  good  
81 =  3^4   3:  good  
81 =  3^4   3:  good  
90 =  2 3^2 5   2:  med   3:  good   5:  med  
90 =  2 3^2 5   2:  med   3:  good   5:  med  
94 =  2 47   2:  med   47:  med  
94 =  2 47   2:  med   47:  med  
97 =  97   97:  good  
97 =  97   97:  good  
100 =  2^2 5^2   2:  med   5:  med  
100 =  2^2 5^2   2:  med   5:  med  
100 =  2^2 5^2   2:  med   5:  med  
100 =  2^2 5^2   2:  med   5:  med  
106 =  2 53   2:  med   53:  med  
106 =  2 53   2:  med   53:  med  
108 =  2^2 3^3   2:  med   3:  good  
115 =  5 23   5:  med   23:  med  
115 =  5 23   5:  med   23:  med  
115 =  5 23   5:  med   23:  med  
115 =  5 23   5:  med   23:  med  
120 =  2^3 3 5   2:  med   3:  good   5:  med  
120 =  2^3 3 5   2:  med   3:  good   5:  med  
121 =  11^2   11:  bad  
121 =  11^2   11:  bad  
129 =  3 43   3:  good   43:  good  
129 =  3 43   3:  good   43:  good  
138 =  2 3 23   2:  med   3:  good   23:  med  
138 =  2 3 23   2:  med   3:  good   23:  med  
139 =  139   139:  good  
139 =  139   139:  good  
142 =  2 71   2:  med   71:  med  
142 =  2 71   2:  med   71:  med  
144 =  2^4 3^2   2:  med   3:  good  
144 =  2^4 3^2   2:  med   3:  good  
145 =  5 29   5:  med   29:  med  
145 =  5 29   5:  med   29:  med  
145 =  5 29   5:  med   29:  med  
145 =  5 29   5:  med   29:  med  
147 =  3 7^2   3:  good   7:  bad  
160 =  2^5 5   2:  med   5:  med  
160 =  2^5 5   2:  med   5:  med  
163 =  163   163:  good  
163 =  163   163:  good  
169 =  13^2   13:  bad  
169 =  13^2   13:  bad  
171 =  3^2 19   3:  good   19:  good  
171 =  3^2 19   3:  good   19:  good  
172 =  2^2 43   2:  med   43:  good  
172 =  2^2 43   2:  med   43:  good  
174 =  2 3 29   2:  med   3:  good   29:  med  
174 =  2 3 29   2:  med   3:  good   29:  med  
184 =  2^3 23   2:  med   23:  med  
184 =  2^3 23   2:  med   23:  med  
190 =  2 5 19   2:  med   5:  med   19:  good  
190 =  2 5 19   2:  med   5:  med   19:  good  
190 =  2 5 19   2:  med   5:  med   19:  good  
190 =  2 5 19   2:  med   5:  med   19:  good  
192 =  2^6 3   2:  med   3:  good  
193 =  193   193:  good  
193 =  193   193:  good  
196 =  2^2 7^2   2:  med   7:  bad  
196 =  2^2 7^2   2:  med   7:  bad  
201 =  3 67   3:  good   67:  good  
201 =  3 67   3:  good   67:  good  
202 =  2 101   2:  med   101:  med  
202 =  2 101   2:  med   101:  med  
211 =  211   211:  good  
211 =  211   211:  good  
219 =  3 73   3:  good   73:  good  
219 =  3 73   3:  good   73:  good  
225 =  3^2 5^2   3:  good   5:  med  
225 =  3^2 5^2   3:  good   5:  med  
225 =  3^2 5^2   3:  good   5:  med  
225 =  3^2 5^2   3:  good   5:  med  
228 =  2^2 3 19   2:  med   3:  good   19:  good  
228 =  2^2 3 19   2:  med   3:  good   19:  good  
232 =  2^3 29   2:  med   29:  med  
232 =  2^3 29   2:  med   29:  med  
235 =  5 47   5:  med   47:  med  
235 =  5 47   5:  med   47:  med  
235 =  5 47   5:  med   47:  med  

Answer 2

For fixed $Z$ you have a Pell-type equation in $X$ and $Y$; by the theory of such equations, if there is one solution to this there are infinitely many. In fact, if $(X,Y,Z)$ is one solution then $(5X+12Y,2X+5Y,Z)$ is another. The real question is for what values of $Z$ are there solutions?

EDIT: It appears that $X^2 - 6 Y^2 = Z^3$ has integer solutions if and only if $X^2 - 6 Y^2 = Z$ has solutions. One direction is easy: if $X^2-6 Y^2 = Z$ then $(XZ)^2 - 6 (YZ)^2 = Z^3$.

EDIT: The nonnegative integers $z$ for which $X^2 - 6 Y^2 = z$ has integer solutions are OEIS sequence A242661. On the other hand, you don't always have solutions with $\gcd(X,Y,z)=1$. The sequence of $z$ for which $X^2-6Y^2=z^3$ has such solutions is $1, 19, 25, 43, 67, 73, 97, 115, 139, 145, 163, 193, 211, 235, 241, 265, 283, \ldots$, which is not (yet) in OEIS.

Answer 3

This answer does not solve the problem. It only makes an observation at the elementary level. The constructed solution set is only a subset of the countable infinite solution set.

---

Let,

$$X=Ym, ~Z=Yn$$

$$\begin{align}&\implies Y^2(m^2-6)=Y^3n^3 \\ &\implies m^2-6=Yn^3 \\ &\implies Y=\frac{m^2-6}{n^3}\end{align}$$

If $n=1,~m≥3$, then

$$X=m(m^2-6),~Y=m^2-6, \\ Z=m^2-6 $$

$$X=m^3-6m,~ Y=m^2-6, \\ Z=m^2-6.$$

Answer 4

Setting (with Euler):

$x+y\sqrt{6}=(p+q\sqrt{6})^{3}$,

We get:

$x=p^{3}+18p.q^{2}$

$y=3p^{2}q+6q^{3}$

therefore

$z=p^{2}-6q^{2}$

with $(p,q)∈N$ and $p>q\sqrt{6}$

Ex.:

$(p,q)=(3,1)$, $(x,y,z)=(81,33,3)$;

$(p,q)=(4,1)$, $(x,y,z)=(136,54,10)$;

$(p,q)=(5,1)$, $(x,y,z)=(215,81,19)$;

$(p,q)=(5,2)$, $(x,y,z)=(485,198,1)$;

$(p,q)=( 2450,1000)$, $(x,y,z)=( 58806125000, 24007500000,2500)$.

Answer 5

Better looking output, I did some cutting and pasting to illustrate the double sequences..... Taking $w=1,$ the recipes are (A) $ (p+q \sqrt 6) = \left(x+y \sqrt 6\right)^3 \; , \; \;$ then (B) $ (p+q \sqrt 6) = (5+2 \sqrt 6) \left(x+y \sqrt 6\right)^3 \; . \; \;$

Show $\mathbb{Z}[\sqrt{6}]$ is a Euclidean domain

Recipe A: \begin{align} p &= w^3 (x^3 + 18xy^2) \\ q &= w^3 (3x^2y + 6y^3) \\ r &= w^2 (x^2 - 6y^2) \end{align}

Recipe B: \begin{align} p &= w^3 \color{blue}{(5x^3 + 36x^2y + 90xy^2 + 72y^3)} \\ q &= w^3 \color{green}{(2x^3 + 15x^2y + 36xy^2 + 30y^3)} \\ r &= w^2 (x^2 - 6y^2) \end{align}

Output for $r$ up to $27$ and then $817 = 19 \cdot 43.$ The thing we demand is double direction sequences obeying $p_{j+2} = 10 p_{j+1} - p_j.$ These begin large, shrink down then begin growing again. For some $z$ such as $1,$ the growing part merely duplicates the shrinking part, note how the numbers $485, 49, 5, 1, 5, 49, 485$ obey $p_{j+2} = 10 p_{j+1} - p_j.$

Let's see, $$ p^2 - 6 q^2 = r^3 \; . \; \; $$ The first number in each line is $r,$ I left off $q$ to save space. In this output all $w=1.$ There is no real need to have any other choice of $w,$ since we are already allowing $\gcd(x,y)$ to be arbitrary.

I mentioned $p_{j+2} = 10 p_{j+1} - p_j.$ Now that I have gotten rid of most repetition, notice how the same is true of every third value of $x,$ also every third value of $|y|$

1      p:  4517251249 B    x: 485  y:  198  gcd 1
1      p:  456335045 A    x: 485  y:  198  gcd 1
1      p:  46099201 B    x: 485  y:  -198  gcd 1
1      p:  4656965 B    x: 49  y:  20  gcd 1
1      p:  470449 A    x: 49  y:  20  gcd 1
1      p:  47525 B    x: 49  y:  -20  gcd 1
1      p:  4801 B    x: 5  y:  2  gcd 1
1      p:  485 A    x: 5  y:  2  gcd 1
1      p:  49 B    x: 5  y:  -2  gcd 1
1      p:  5 B    x: 1  y:  0  gcd 1
1      p:  1 A    x: 1  y:  0  gcd 1
3      p:  753651369 B    x: 267  y:  109  gcd 1
3      p:  76134249 A    x: 267  y:  109  gcd 1
3      p:  7691121 B    x: 267  y:  -109  gcd 1
3      p:  776961 B    x: 27  y:  11  gcd 1
3      p:  78489 A    x: 27  y:  11  gcd 1
3      p:  7929 B    x: 27  y:  -11  gcd 1
3      p:  801 B    x: 3  y:  1  gcd 1
3      p:  81 A    x: 3  y:  1  gcd 1
3      p:  9 B    x: 3  y:  -1  gcd 1
4      p:  37255720 B    x: 98  y:  40  gcd 2
4      p:  3763592 A    x: 98  y:  40  gcd 2
4      p:  380200 B    x: 98  y:  -40  gcd 2
4      p:  38408 B    x: 10  y:  4  gcd 2
4      p:  3880 A    x: 10  y:  4  gcd 2
4      p:  392 B    x: 10  y:  -4  gcd 2
4      p:  40 B    x: 2  y:  0  gcd 2
4      p:  8 A    x: 2  y:  0  gcd 2
9      p:  125738055 B    x: 147  y:  60  gcd 3
9      p:  12702123 A    x: 147  y:  60  gcd 3
9      p:  1283175 B    x: 147  y:  -60  gcd 3
9      p:  129627 B    x: 15  y:  6  gcd 3
9      p:  13095 A    x: 15  y:  6  gcd 3
9      p:  1323 B    x: 15  y:  -6  gcd 3
9      p:  135 B    x: 3  y:  0  gcd 3
9      p:  27 A    x: 3  y:  0  gcd 3
10      p:  1249335392 B    x: 316  y:  129  gcd 1
10      p:  126208504 A    x: 316  y:  129  gcd 1
10      p:  12749648 B    x: 316  y:  -129  gcd 1
10      p:  1287976 B    x: 32  y:  13  gcd 1
10      p:  130112 A    x: 32  y:  13  gcd 1
10      p:  13144 B    x: 32  y:  -13  gcd 1
10      p:  1328 B    x: 4  y:  1  gcd 1
10      p:  136 A    x: 4  y:  1  gcd 1
10      p:  32 B    x: 4  y:  -1  gcd 1
10      p:  184 B    x: 8  y:  -3  gcd 1
10      p:  1808 A    x: 8  y:  3  gcd 1
10      p:  17896 B    x: 8  y:  3  gcd 1
10      p:  177152 B    x: 76  y:  -31  gcd 1
10      p:  1753624 A    x: 76  y:  31  gcd 1
10      p:  17359088 B    x: 76  y:  31  gcd 1
12      p:  6029210952 B    x: 534  y:  218  gcd 2
12      p:  609073992 A    x: 534  y:  218  gcd 2
12      p:  61528968 B    x: 534  y:  -218  gcd 2
12      p:  6215688 B    x: 54  y:  22  gcd 2
12      p:  627912 A    x: 54  y:  22  gcd 2
12      p:  63432 B    x: 54  y:  -22  gcd 2
12      p:  6408 B    x: 6  y:  2  gcd 2
12      p:  648 A    x: 6  y:  2  gcd 2
12      p:  72 B    x: 6  y:  -2  gcd 2
16      p:  298045760 B    x: 196  y:  80  gcd 4
16      p:  30108736 A    x: 196  y:  80  gcd 4
16      p:  3041600 B    x: 196  y:  -80  gcd 4
16      p:  307264 B    x: 20  y:  8  gcd 4
16      p:  31040 A    x: 20  y:  8  gcd 4
16      p:  3136 B    x: 20  y:  -8  gcd 4
16      p:  320 B    x: 4  y:  0  gcd 4
16      p:  64 A    x: 4  y:  0  gcd 4
19      p:  1925229703 B    x: 365  y:  149  gcd 1
19      p:  194487695 A    x: 365  y:  149  gcd 1
19      p:  19647247 B    x: 365  y:  -149  gcd 1
19      p:  1984775 B    x: 37  y:  15  gcd 1
19      p:  200503 A    x: 37  y:  15  gcd 1
19      p:  20255 B    x: 37  y:  -15  gcd 1
19      p:  2047 B    x: 5  y:  1  gcd 1
19      p:  215 A    x: 5  y:  1  gcd 1
19      p:  103 B    x: 5  y:  -1  gcd 1
19      p:  815 B    x: 13  y:  -5  gcd 1
19      p:  8047 A    x: 13  y:  5  gcd 1
19      p:  79655 B    x: 13  y:  5  gcd 1
19      p:  788503 B    x: 125  y:  -51  gcd 1
19      p:  7805375 A    x: 125  y:  51  gcd 1
19      p:  77265247 B    x: 125  y:  51  gcd 1
25      p:  582120625 B    x: 245  y:  100  gcd 5
25      p:  58806125 A    x: 245  y:  100  gcd 5
25      p:  5940625 B    x: 245  y:  -100  gcd 5
25      p:  600125 B    x: 25  y:  10  gcd 5
25      p:  60625 A    x: 25  y:  10  gcd 5
25      p:  6125 B    x: 25  y:  -10  gcd 5
25      p:  625 B    x: 5  y:  0  gcd 5
25      p:  125 A    x: 5  y:  0  gcd 5
25      p:  43191059 B    x: 103  y:  42  gcd 1
25      p:  4363183 A    x: 103  y:  42  gcd 1
25      p:  440771 B    x: 103  y:  -42  gcd 1
25      p:  44527 B    x: 11  y:  4  gcd 1
25      p:  4499 A    x: 11  y:  4  gcd 1
25      p:  463 B    x: 11  y:  -4  gcd 1
25      p:  131 B    x: 7  y:  -2  gcd 1
25      p:  847 A    x: 7  y:  2  gcd 1
25      p:  8339 B    x: 7  y:  2  gcd 1
25      p:  82543 B    x: 59  y:  -24  gcd 1
25      p:  817091 A    x: 59  y:  24  gcd 1
25      p:  8088367 B    x: 59  y:  24  gcd 1
25      p:  80066579 B    x: 583  y:  -238  gcd 1
25      p:  792577423 A    x: 583  y:  238  gcd 1
25      p:  7845707651 B    x: 583  y:  238  gcd 1
27      p:  20977947 B    x: 81  y:  33  gcd 3
27      p:  2119203 A    x: 81  y:  33  gcd 3
27      p:  214083 B    x: 81  y:  -33  gcd 3
27      p:  21627 B    x: 9  y:  3  gcd 3
27      p:  2187 A    x: 9  y:  3  gcd 3
27      p:  243 B    x: 9  y:  -3  gcd 3
817      p:  1674697433 B    x: 349  y:  142  gcd 1
817      p:  169178797 A    x: 349  y:  142  gcd 1
817      p:  17090537 B    x: 349  y:  -142  gcd 1
817      p:  1726573 B    x: 41  y:  12  gcd 1
817      p:  175193 A    x: 41  y:  12  gcd 1
817      p:  25357 B    x: 41  y:  -12  gcd 1
817      p:  78377 B    x: 61  y:  -22  gcd 1
817      p:  758413 A    x: 61  y:  22  gcd 1
817      p:  7505753 B    x: 61  y:  22  gcd 1
817      p:  74299117 B    x: 569  y:  -232  gcd 1
817      p:  735485417 A    x: 569  y:  232  gcd 1
817      p:  7280555053 B    x: 569  y:  232  gcd 1
817      p:  187021517 B    x: 169  y:  68  gcd 1
817      p:  18893017 A    x: 169  y:  68  gcd 1
817      p:  1908653 B    x: 169  y:  -68  gcd 1
817      p:  193513 B    x: 29  y:  2  gcd 1
817      p:  26477 A    x: 29  y:  2  gcd 1
817      p:  71257 B    x: 29  y:  -2  gcd 1
817      p:  686093 B    x: 121  y:  -48  gcd 1
817      p:  6789673 A    x: 121  y:  48  gcd 1
817      p:  67210637 B    x: 121  y:  48  gcd 1

Answer 6

We can always construct a linear relation between x and y such as $x=ay+b; a, b \in \mathbb z$ to convert pell equations $x^2-Dy^2=1$ to a single unknown equation. We can find numerous families of solutions by this method. For Pell like equations we use rational solutions. I try to show this bellow by an example:

Let $x=y+a$ we have:

$x^2-6y^2=1\Rightarrow 5y^2-2ay-a^2+1=0$

$\Delta'=6a^2-5$

which have number of solutions:

$(a, \Delta')=(1, 1), (3, 7^2), (7, 17^2), \cdot \cdot\cdot$

which gives:

$(x, y)=(\frac 25, 1), (\frac 4 5, 1), (8, 5), (\frac {11}5, -\frac 45)\cdot\cdot\cdot $

For example take $(x, y)=(\frac {11}5, -\frac 45)$ we have:

$\big(\frac {11}5\big)^2-6\big(\frac 45\big)^2=1\Rightarrow\big(\frac {11^2}{25}\big)-6\big(\frac{4^2} {25}\big)=1$

We can rewrit this relation as:

$(11\times 25)^2-6(4\times 25)^2=25^3$

So: $(x, y, z)=(675, 100, 25) $

Similarly we can find more solutions where $z=25$ as a family of solutions.

Generally there can alway be a solution for equation $x^2-Dy^2=z^{2k+1}$ . for example for $x^2-6x^2=z^5$ we can write:

$(11\times625)^2-6(4\times 625)^2=25^5$

For D=19 we have:

$x=\frac{35} 3$ , $y=\frac 83$ and we have:

$(35\times 9)^2-19(6\times 9)^2=9^3$

$(35\times 81)^2-19(8\times 81)^2=9^5$

Answer 7

I’m taking individ’s comment and spinning it out into an answer, in part to invite others to help parse it.

The formulas given on this AoPS page do indeed yield an identity solving ($\star$) in my original question.

There is some simplification which can be done in order to clarify the formulas. Writing $b=r-s$, and “compressing” Pell equations [wherever I found them, anyway!] yields \begin{align} x &= \bigl(p^2+qr^2\bigr)\bigl(p^2(p^2+3qr(r-2s))-q(3p^2+qr(r-2s))(r-2s)^2\bigr), \\[0.5em] y &= 2p\bigl(p^2+qr^2\bigr)\bigl(p^2(r-3s)+q(r+s)(r-2s)^2\bigr), \\[0.5em] z &= \bigl(p^2+qr^2\bigr)\bigl(p^2+q(r-2s)^2\bigr). \end{align} Substitution into ($\star$) verifies that these formulas hold identically.

Evidently, solving $p^2+qr^2=\pm 1$ — always possible when $q$ is a nonsquare negative number — will eliminate the [algebraic] common factor $p^2+qr^2$; of course, using the minus sign will introduce a sign change in $z^3$ [but not $x^2$ or $y^2$]; furthermore, if one can simultaneously solve $p^2+q(r-2s)^2=\pm 1$ [using the same sign as the first choice], then we recover $z=1$.

I’m making this a community wiki answer, so that others will contribute. I’m hoping we can (a) find optimal formulas for $x,y,z$; (b) prove or disprove that this is a complete solution; and (c) perhaps determine how to use this technique [which individ appears unwilling to illuminate] to find similar “parameterizations” for other Diophantine equations.

I am writing $p^2 - 6 q^2 = r^3.$ It is important to allow the variables to be positive or negative in recipe B, as needed. Note that we may take absolute values of $p,q,r$ without changing anything. That allows the natural sequences, Fibonacci type. For example $r=1$ uses the sequence of (positive) $p$ values $1, 5, 49, 485, 4801, $ made with rule $r_{k+2} = 10 r_{k+1} - r_k$ Ummm. Taking $x,y$ coprime and possible conditions mod(2,3) will cause $\gcd(p,q,r) = | u^2 - 6 v^2 |$

Progress: Getting three recipes by the original, then multiplication by units $5 + 2 \sqrt 6$ or by $49 + 20 \sqrt 6,$ appears to allow us to use only $x,y \geq 0$. There is considerable repetition. The concept is that: we demand $x,y$ coprime; then $u,v$ coprime; and finally $w$, if larger than one, a prime $\equiv 7,11,13,17 \pmod{24}$ or the product of such primes.

Recipe A: \begin{align} p &= w^3 (x^3 + 18xy^2) \\ q &= w^3 (3x^2y + 6y^3) \\ r &= w^2 (x^2 - 6y^2) \end{align}

Recipe B: \begin{align} p &= w^3 \color{blue}{(5x^3 + 36x^2y + 90xy^2 + 72y^3)} \\ q &= w^3 \color{green}{(2x^3 + 15x^2y + 36xy^2 + 30y^3)} \\ r &= w^2 (x^2 - 6y^2) \end{align}

Answer 8

For the first equation: $X^2-6Y^2=Z^3$ the following produces solutions in ascending $(z,y)$.

 110 print "Enter H1" : input h1 : print
 120 for z1 = 1 to h1
 130    for y1 = 1 to h1
 140       x1 = sqr(6*y1^2+z1^3)
 150       if x1 = int(x1)
 160          print "(" x1 "," y1 "," z1 ")\quad "
 170       endif
 180    next y1
 190 next z1

$$(y,z) \le100\longrightarrow\quad (5 ,2 ,1 )\quad (49 ,20 ,1 )\quad (9 ,3 ,3 )\quad (81 ,33 ,3 )\quad (40 ,16 ,4 )\\ (135 ,54 ,9 )\quad (32 ,2 ,10 )\quad (40 ,10 ,10 )\quad (80 ,30 ,10 )\quad (136 ,54 ,10 )\quad (184 ,74 ,10 )\\ (72 ,24 ,12 )\quad (95 ,19 ,19 )\quad (103 ,25 ,19 )\quad (215 ,81 ,19 )\quad (247 ,95 ,19 )\\ (131 ,16 ,25 )\quad (145 ,30 ,25 )\quad (175 ,50 ,25 )\quad (275 ,100 ,25 )\quad (243 ,81 ,27 )\\ (180 ,30 ,30 )\quad (252 ,78 ,30 )\quad (256 ,16 ,40 )\quad (320 ,80 ,40 )\quad (301 ,43 ,43 )\\ (356 ,70 ,46 )\quad (477 ,84 ,57 )\quad (464 ,58 ,58 )\quad (553 ,29 ,67 )\quad (675 ,75 ,75 )\\ (864 ,54 ,90 )\quad (940 ,94 ,94 )\quad $$

From this sample, we can infer that, if we fix the $z$-value, we will find streams for $z\in\big\{1,3,4,10,12,19,25,27,30,40,43,46,57,58,67,75,90,94\big\}$ and more.

For examples $$(9 ,3 ,3 )\quad (81 ,33 ,3 )\quad (801 ,327 ,3 )\quad (7929 ,3237 ,3 )\quad (78489 ,32043 ,3 )\quad \\ (40 ,16 ,4 )\quad (392 ,160 ,4 )\quad (3880 ,1584 ,4 )\quad (38408 ,15680 ,4 )\quad (380200 ,155216 ,4 )\\\ (32 ,2 ,10 )\quad (40 ,10 ,10 )\quad (80 ,30 ,10 )\quad (136 ,54 ,10 )\quad (184 ,74 ,10 )\\ (72 ,24 ,12 )\quad (648 ,264 ,12 )\quad (6408 ,2616 ,12 )\quad (63432 ,25896 ,12 )\\ (95 ,19 ,19 )\quad (103 ,25 ,19 )\quad (215 ,81 ,19 )\quad (247 ,95 ,19 )\quad (703 ,285 ,19 )\\ (131 ,16 ,25 )\quad (145 ,30 ,25 )\quad (175 ,50 ,25 )\quad (275 ,100 ,25 )\quad (365 ,140 ,25 )\\ (243 ,81 ,27 )\quad (2187 ,891 ,27 )\quad (21627 ,8829 ,27 )\quad (214083 ,87399 ,27 )\\ (180 ,30 ,30 )\quad (252 ,78 ,30 )\quad (324 ,114 ,30 )\quad (540 ,210 ,30 )\quad (1260 ,510 ,30 ) $$

To find another stream, you could, for example, change line $120$

 120 for z1 = 40 to 40

and get $$(256 ,16 ,40 )\quad (320 ,80 ,40 )\quad (640 ,240 ,40 )\quad (1088 ,432 ,40 )\quad (1472 ,592 ,40 )$$

Of course, none of this is a complete solution but the "streams" are generated as fast as the monitor can display them and perhaps the patterns within steams can provide insight. What I find interesting are those cases whee $y=z$: $$(9 ,3 ,3 )\quad (40 ,10 ,10 )\quad (95 ,19 ,19 )\quad (301 ,43 ,43 )\quad (464 ,58 ,58 )\quad (940 ,94 ,94 )$$

BTW, if you have interest in the second equation (with $z^2$), there $is$ a general solution for that.