Dear Mercedes Orús-Lacort ,

Thank you for correcting the typo (2, -1, -1).

In general, if (a,b,c) is a solution, and the given Diophantine equation is homogeneous in x,y and z, then (na, nb, nc) is a solution for the Diophantine equation

x3 + Ny3 = z3 ∀ n ∈ Z.

So, all mentioned equations are equivalent over Q.

x3 + Ny3 = z3 , x3 + Ny3 = 1 over Q.

We still have the same challenging problem:

x3 + Ny3 = 1 over Q. ( N is not cubic rational)

Best regards

A-3. A method to find all solutions (N, x, y, z)

Answer 3, posted on June 13, 2019.

--------------------------------------------

Here is a method to find all solutions (N, x, y, z)

of the equation x^3 + Ny^3 = z^3 in non-zero integers.

--------------------------------------------

1. Choose any 2 distinct non-zero integers x and z:

For example, z = 7 and x = -2.

--------------------------------------------

2. Compute the difference d = z^3 – x^3:

d = 7^3 – (-2)^3 = 343 + 8 = 351.

--------------------------------------------

3. Choose any cube y^3 that divides d:

d = 351 = 27 times 13, with y = 3.

--------------------------------------------

4. Compute N = d/y^3:

N = 351/27 = 13.

--------------------------------------------

5. Check: x^3 + Ny^3 = z^3:

(-2)^3 + (13)(3^3) = -8 + (13)(27) = - 8 +351 = 343 = 7^3.

--------------------------------------------

For every integers x and z, we have the 2 trivial solutions

y = 1 and N = d,

y = -1 and N = -d.

--------------------------------------------

A big question is:

What are the possible values of N

that give non-trivial solutions.

--------------------------------------------

With best regards, Jean-Claude

Dear's Jean-Claude Evard , Mercedes Orús-Lacort

Look at (1)³ + 7(1)³ = (2)³

(2)³ + 19(1)³ = (3)³

(3)³ + 37(1)³ = (4)³ ,etc.

it is an easy task to generate solutions. (1) N is not arbitrary. N is a give fixed integer. Similar to the famous Pell's Diophantine equation.

Your method doesn't work. If N is variable, then the task is a trivial one. Examples: N is variable.

Solve: x³ + Ny³ = z³

we have (1)³ + 7(1)³ = (2)³

(2)³ + 19(1)³ = (3)³

(3)³ + 37(1)³ = (4)³

we have infinite number of similar solutions.

(2) I have proved the following theorem:

Theorem: If N has the form 3ab³+3a²+b⁶

then the Diophantine equation x³ + Ny³ = z³

has the following solution (x, y, z) = (a, b, a+b3 ).

Proof: By direct substitution.

x³ + Ny³ , we obtain a3 + (3ab³+3a²+b⁶)b3 = (a+b3 )3 .

The true challenge is how to solve

x³ + Ny³ = z³ for a given (non cubic N).

We seek general cases.

Best regards

Dear Issam Kaddoura,

On cubic Pell’s Equations, see Chapter 7 (“The Cubic Analogue of Pell’s Equation”, pp. 92-112) of the book: Edward J. Barbeau, Pell’s Equation, Problem Books in Mathematics, Springer-Verlag, New York, Inc., 2015.

Best regards,

Romeo

Dear Romeo Meštrović ,

What are the main results there?

Have you a (pdf) link to the mentioned textbook?

Best regards

Dear Issam Kaddoura,

Unfortunately, it seems that this book is not avalable (free of charges) at Internet. By searching this book at Google, I found another interesting book of Burbeau (Polynomials) available at http://poincare.matf.bg.ac.rs/~zarkom/Polynomials_EJBarbeau.pdf

and an interesting book on Diophantine Equations (with numerous examples, and in particular, concerning Pell's equations, but not Cubic Pell's Equations) available at https://trungtuan.files.wordpress.com/2015/11/an-introduction-to-diophantine-equations-a-problem-based-approachby-titu-andreescu-dorin-andrica-ion-cucurezeanu.pdf

When I consider what results on cubic Pell's equations are presented in this book, I'll let you know. However, it seems to be only particular cases are considered, rather than a general theory as in the case of the classical Pell's equations.

Best regards,

Romeo

Dear Issam Kaddoura,

Chapter 7 of Barbeau;s book mainly deals with the cubic equations of the form

x3 + cy3 + c2z3 – 3cxyz = k , where c is not a perfect cube.

It seems that related results are mainly presented in

https://www.researchgate.net/profile/Thong_Nguyen...Do3/.../cubic+Pell+3.pdf

(available at RG, file in attachment). There is another Burbeau’s 2003 paper https://www.researchgate.net/publication/300871708_The_Cubic_Analogue_of_Pell's_Equation (not available at RG).

Best regards,

Romeo

Dear Romeo Meštrović ,

Thank you for the attached article. This one is available online.

I asked you about the other textbook: (“The Cubic Analogue of Pell’s Equation”, pp. 92-112) of the book: Edward J. Barbeau, Pell’s Equation, Problem Books in Mathematics, Springer-Verlag, New York, Inc., 2015. .

Anyway, I think the problem is still active. And my theorem

Theorem:

If N has the form 3ab³+3a²+b⁶

then the Diophantine equation x³ + Ny³ = z³

has the following solution (x, y, z) = (a, b, a+b3 ), (-a, -b, -a-b3 )

It seems good and very close to some results in the mentioned article

about x³ - Ny³ = 1.

I hope that serious deep discussions may reveal new results.

Best wishes

See my article “On the solution of the cubic Pythagorean Diophantine equation x^3+y^3+z^3=a^3”.

Missouri Journal of Mathematical Sciences, Vol 18,, Number1, (2006), 3-16.

Maybe this would be helpful????

Louis John Scheinman ,

Thank you for sharing your article. It includes nice relations. You can see also the recent work of Sajal Mukherjee

"On the Euler's Diophantine equation for cubic exponent "

In this current question, we have a different task.

We seek solutions to the problem:

x³ + Ny³ = z³ for a given non-cubic integer N.

Best wishes

Dear Issam Kaddoura and Louis John Scheinman,

Please, also see my recent RG questions

https://www.researchgate.net/post/Do_there_exist_infinitely_many_prime_numbers_p_such_that_x3_y3_z3p_3xyz_for_some_integers_x_y_z_such_z_is_less_than_or_equal_to_0 and https://www.researchgate.net/post/Irreducible_super_Pythagorean_triples_Number_Theory., as well as RG question https://www.researchgate.net/post/For_an_arbitrary_positive_integer_kobviously_2_crelaan_we_find_k_perfect_cubes_whose_sum_is_again_a_perfect_cube and related answers.

Best regards

Dear Romeo Meštrović ,

Yes, all results, as well as discussions in the mentioned threads, are related to the Diophantine equations in cubic order.

Thanks for the reminder.

Best regards

A-16. [Non-trivial A3] not of the form [non-trivial A5]

Answer 16, posted on June 14, 2019.

--------------------------------------------

The non-trivial solution of answer 3:

---

(-2)^3 + (13)(3^3) = 7^3.

---

is not of the form of the non-trivial solution of answer 5:

---

a^3 + (3ab^3 + 3a^2 + b^6)b^3 = (a+b^3)^3.

--------------------------------------------

With best regards, Jean-Claude

A-17. Another family of solutions

Answer 17, posted on June 14, 2019.

--------------------------------------------

(3a + 1)^3 + (9a^2 + 33a + 37)3^3 = (3a + 10)^3.

--------------------------------------------

I found it by applying my method of answer 3.

--------------------------------------------

With best regards, Jean-Claude

Dear Jean-Claude Evard ,

We can construct many forms for N by applying suitable transformations. The main problem still challenging, if N is fixed and independent of x and y. What are the solutions for x³ - Ny³ = z³, if any? For example, if N = 91. What are the solutions of x³ - 91y³ = 1? Does the equation x³ - 91y³ = 1 admit any solutions?

Computers may produce solutions; we have the question

on how to show or how to generate solutions analytically? Best wishes

93 - 91 ∙ 23 = 1

Dear Romeo Meštrović ,

Recall that:

Computers may produce solutions; we have the question

on how to show or how to generate solutions analytically?

HOW?

Best regards

Dear Issam Kaddoura,

It seems that an analytic approach for generating all (or at most one/some class/classes) of solutions of cubic Pell’s equation x3 – y3 = 1 is generally a very difficult problem. However, there are some results in this direction for the Diophantine equations x3 + cy3 + c2z3 – 3cxyz = 1, where c is not a perfect cube integer.

Best regards

Dear Mercedes Orús-Lacort,

Of course, a mistake, i.e., instead of x3 –y3 = 1 it should be written x3 –Ny3 = 1.

In this moment, a computation up to x ≤ 8000000 in Mathematica 9 shows that the equation x3 – 91y3 = 1 has the only positive solutions (x, y) = (9, 2).

Problem: “Find the analytic expression for a general solution of the Diophantine equation x3 – 19y3 = 1”.

Best regards

Dear Issam Kaddoura

Yes, the equation x³ - 91y³ = 1 has a solution, this by Theorem [*] states that:

(The equation x^3 - d y^3 = 1 has a solution for which y = 2 if and only if

d = u (64u^2 + 24u + 3) for some integer u). So (9, 2) is a solution, since

91= u (64u^2 + 24u + 3) for u= 1.

[*] Chapter The Cubic Analogue of Pell’s Equation

Best regards.

Also, the equation of the form x^3 - d y^3 = 1 has a solution for which y= 3 if and only if d = u(27 u^2 + 9 u + 1) for some integer u, for example; the equation x^3 - 37 y^3 = 1 has a solution in nonzero integers which is (10, 3).

Thus the equation x^3 - d y^3 = 1 is nontrivially solvable for certain values of d.

Rewriting this equation as x^3 + (_1)^3 = d y^3, then equation of the form x^3 + Z^3 = d y^3 are studied in [*] Chapter The Cubic Analogue of Pell’s Equation .

Dear's followers

According to me many of the answers are very close, where N in the Diophantine equation has a specific form.

Sajal Mukherjee

If N = 3, then x3- 3y3 =1 has no solutions. He deduced that the equation is not solvable for some family of N. He wondered, which family?

Adiya K. Hussein discussed partial results if y = 2 or 3 , then

d has exact specific forms in x3- d(y)3 = 1 to guarantee the existence of solutions.

Mercedes Orús-Lacort shew various combinations where

the given equation is solvable. And she raised a question about new forms for N.

Jean-Claude Evard , provided results similar to mine with different forms for N.

Romeo Meštrović , Shew some elegant solutions for special cases using computers and suggest some analytic methods to proceed.

He posted a new particular case x3 - 19y3 = 1, seeking for solutions if any!

Answers are in progress. It is great to see the interaction between many excellent researchers in number theory; each provided his answer based on his proofs or using some published results. This environment, for sure, will end by a fruitful article or at least one can learn many more things about this type of Diophantine equations.

According to me many of the answers are very close, where N in the Diophantine equation x3+ Ny3 = z3, has a specific form.

What about if we consider arbitrary non-cubic integer N, are there any solutions?

Best regards

A computation up to |x| ≤ 107 in Mathematica 9 shows that the Diophanthine equation x3 – 19y3 = 1 has only two integer solutions: (x, y) = (1, 0) and (x, y) = (9, 2).

A-29. Computer solutions are very helpful

Answer 29, posted on June 15, 2019.

--------------------------------------------

At the beginning of a research work on a Diophantine equation,

computer solutions are very helpful.

A striking example is the following beautiful article:

---

On the solution of the cubic Pythagorean

Diophantine equation x^3 +y^3 +z^3 =a^3

Louis John Scheinman

Etobicoke General Hospital

Department of surgery

Missouri Journal of Mathematical Sciences

Vol 18, Number 1, (2006), 3--16

December 2005

https://www.researchgate.net/publication/267478581_On_the_solution_of_the_cubic_Pythagorean_Diophantine_equation_x_3_y_3_z_3_a_3

--------------------------------------------

With best regards, Jean-Claude

Dear Jean-Claude Evard ,

Thank you for the reminder about Louis answer 2 days ago.

Let me remind you of my reply, it was:

>

Best wishes

It seems that the solutions of the Diophanthine equation x3 – Ny3 = 1 are “very rare”. Namely, a computation in Mathematica 9 up to |x| ≤ 106 and non-square positive integers N ≤ 21 give only the following solutions (x, y) with y ≠ 0: (-1, -1) for N = 2; (2, 1) for N = 7; (-2, -1) for N = 9; (18, 7) for N = 17; (-8, -3) for N = 19 and (-19, -7) for N = 20.

I believe x^3-19y^3=1 is a misprint.

It should be x^3-91y^3=1. This equation has the solutions (1,0) and (9,2).

However, is it not true that (1,0) is a trivial solution since it would be a solution of any equation of the form x^3-Ny^3=1 regardless of the value of N?

Dear Louis John Scheinman ,

Any solution of a Diophantine equation where one of its components is zero called a trivial solution.

Best regards

A-35. The general solution in the case y = 2 and z = 3.

Answer 35, posted on June 16, 2019.

--------------------------------------------

In answer 25, Adiya Hussein has given the general

solution of the Diophantine equation

x^3 + N*y^3 = z^3

in the special case where y = 2 and z = 1,

that is to say, has given the set of all integer values

of x and N such that x^3 + 8N = 1.

--------------------------------------------

Here is the general solution in the case y = 2 and z = 3,

that is to say, the set of all integer values of x and N

such that x^3 + 8N = 27.

--------------------------------------------

Theorem. Let x and N be integers.

Then x and N satisfy the equation x^3 + 8N = 27

if and only if there exists an integer n such that

x = 8n + 3 and N = -n(64n^2 + 72n + 27).

---

Proof. Suppose that x^3 + 8N = 27.

This implies that x^3 is congruent to 27 modulo 8,

that is to say, congruent to 3 modulo 8.

By checking all cubes satisfying this condition,

we find that this condition is satisfied

if and only if x is congruent to 3 modulo 8,

that is, if and only if there exists an integer n

such that x = 8n + 3.

Substituting this into the equation x^3 + 8N = 27

gives (8n + 3)^3 + 8N = 27,

that is,

(8n)^3 + (3)((8n)^2)(3) + (3)(8n)(3^2) + 3^3 + 8N = 27,

that is,

(8^3)(n^3) + (9)(8^2)(n^2) + (3)(8n)(9) + 27 + 8N = 27,

that is

(8^3)(n^3) + (9)(8^2)(n^2) + (27)(8n) + 8N = 0,

that is

8[(8^2)(n^3) + (9)(8)(n^2) + 27n + N] = 0,

that is,

(8^2)(n^3) + (9)(8)(n^2) + 27n + N = 0,

that is,

(64)(n^3) + (72)(n^2) + 27n + N = 0,

that is,

N = -[(64)(n^3) + (72)(n^2) + 27n],

that is,

N = -n(64n^2 + 72n + 27).

---

It is easy to check that conversely,

if x = 8n + 3 and N = -n(64n^2 + 72n + 27),

then x^3 + 8N = 27.

--------------------------------------------

We obtain the smallest solution when n = -1.

This gives x = 8n + 3 = -8 + 3 = -5, so that x = -5,

N = 64 – 72 + 27 = -8 + 27 = 19, so that N = 19.

and the equation x^3 + 8N = 27 becomes

(-5)^3 + (8)(19) = 27.

--------------------------------------------

Thus, N = 19 is a possible value for non-trivial

solutions of x^3 + N*y^3 = z^3,

and we have the above solution

x = - 5, y = 2, and z = 3,

and the solution obtained by Adiya Hussein:

x = -7, y = 2, and z = 1,

and the solution obtained by Romeo Meštrović:

x = -8, y = 3, and z = 1.

--------------------------------------------

With best regards, Jean-Claude

A-36. A first theorem about Modulo 9.

Answer 36, posted on June 17, 2019.

--------------------------------------------

The modulus 9 has the following magic property:

---

Lemma. Let n be an integer.

Then n^3 is congruent to -1 or 0 or 1 modulo 9.

Furthermore, n^3 is congruent to 0 modulo 9

if and only if n is a multiple of 3.

---

Proof: It is sufficient to check the conclusion for the nine

integers in the interval [-4, 4], because any other integer

is congruent modulo 9 to one of these 9 integers.

We have:

(-4)^3 = -64 = -63 – 1 = -(9)(7) -1 is congruent to -1 modulo 9.

(-3)^3 = -27 = -(9)(3) is congruent to 0 modulo 9.

(-2)^3 = -8 = -9 + 1 = is congruent to 1 modulo 9.

(-1)^3 = -1 is congruent to -1 modulo 9.

0^3 = 0 is congruent to 0 modulo 9.

1^3 = 1 is congruent to 1 modulo 9.

2^3 = 8 = 9 – 1 = is congruent to -1 modulo 9.

3^3 = 27 = (9)(3) is congruent to 0 modulo 9.

4^3 = 64 = 63 + 1 = (9)(7) + 1 is congruent to 1 modulo 9.

---

Thus, in all of the 9 cases, the two conclusions of the lemma hold.

--------------------------------------------

Theorem. Let N be an integer

that is congruent to -4 or -3 or 3 or 4 modulo 9.

Let x, y, and z be integers such that x^3 + Ny^3 = z^3.

Then y is a multiple of 3.

---

Proof. The equation x^3 + Ny^3 = z^3

is equivalent to Ny^3 = z^3 – x^3.

By the above lemma, we have that both z^3 and x^3

are congruent to -1 or 0 or 1 modulo 9.

This implies that (z^3 – x^3) is congruent

to -2 or -1 or 0 or 1 or 2 modulo 9.

This implies that (z^3 – x^3) is never congruent

to -4 or -3 or 3 or 4 modulo 9.

Let us prove by contradiction that y is a multiple of 3.

Suppose that y is not a multiple of 3.

Then by the above lemma, we have that

y^3 is congruent to -1 or 1 modulo 9.

This, together with the hypothesis

that N is congruent to -4 or -3 or 3 or 4 modulo 9, implies

that Ny^3 is congruent to -4 or -3 or 3 or 4 modulo 9.

This, together with the hypothesis that Ny^3 = z^3 – x^3,

implies that (z^3 – x^3) is congruent to -4 or -3 or 3 or 4

modulo 9.

This contradicts our above result that (z^3 – x^3)

is never congruent to -4 or -3 or 3 or 4 modulo 9.

--------------------------------------------

With best regards, Jean-Claude

Dear Jean-Claude Evard ,

It is more convenient to write the proof of the lemma in this form:

Lemma: For nay integer n, then n3 is congruent mod(9) to 0,1, or -1.

Proof: Any integer n has one of the following forms 3k, 3k+1,3k+2

If n=3k, then n³=27k³ ≡ 0mod9

If n=3k+1,then n³=(3k+1)³= 27k³+27k²+9k+1 ≡ 1 mod9.

If n=3k+2,then n³ =(3k+2)³ = 27k³+54k²+36k+8 ≡-1mod9.

Best regards

Can you send your problem at [email protected]

I will share information regarding your problem.

Amin Asgari ,

You are welcome to share your p0int of view. Your request is a strange one!!! The problem is evident as posted. If you have an answer, you can show it and share others your ideas. Otherwise, you can learn from the experts and their replies. Best regards

A-41. Checking answer 40.

Answer 41, posted on June 18, 2019.

--------------------------------------------

In answer 40, just after equation (1),

the term that does not depend on y

is not -1, it is (t^3 – 1).

--------------------------------------------

With best regards, Jean-Claude

A-42. Recall answer 23.

Answer 42, posted on June 18, 2019.

--------------------------------------------

Recall answer 23 from Sajal Mukherjee:

---

… there are values of N, such that the given equation

does not have integral solutions.

For example, a theorem of Legendre says that, for N=3,

the equation is not solvable in integers.

…

--------------------------------------------

With best regards, Jean-Claude

A-43. The cases y = 1 and y = -1 are trivial

Answer 43, posted on June 18, 2019.

--------------------------------------------

The cases y = 1 and y = -1 give trivial solutions (N, x, z)

of the equation x^3 + Ny^3 = z^3:

--------------------------------------------

This equation is equivalent to Ny^3 = z^3 – x^3.

--------------------------------------------

When y = 1, this equation becomes N = z^3 – x^3.

Therefore, for all integers x and z,

this equation is satisfied with N = z^3 – x^3.

--------------------------------------------

In particular, when y = 1, z = 1, and N = -d,

the cubic Pell equation x^3 – dy^3 = 1

becomes x^3 – d = 1, and it is satisfied

for all integers x with d = x^3 – 1.

--------------------------------------------

When y = -1, the equation Ny^3 = z^3 – x^3

becomes -N = z^3 – x^3,

that is, N = x^3 – z^3.

Therefore, for all integers x and z,

this equation is satisfied with N = x^3 – z^3.

--------------------------------------------

In particular, when y = -1, z = 1, and N = -d,

the cubic Pell equation x^3 – dy^3 = 1

becomes x^3 + d = 1, and it is satisfied

for all integers x with d = 1 – x^3.

--------------------------------------------

With best regards, Jean-Claude

Cubic Pell’s equations would be very useful for approximations of cube roots of positive integers. More precisely, large positive solutions of the Diophantine equations x3 – Ny3 = k with relatively small values |k| lead to the best rational approximation of cube root N1/3. Namely, this immediately yields from the fact that the previous equality can be written as

x/y - N1/3 = k/(x2y + xy2N1/3 + y3N2/3).

For example, (x, y) = (467, 257) is a solution of the equation x3 – 6y3 = 5, which substituting into previous equality gives

467/257 –61/3 = 5/(56048873+30844883*61/3 +16974593*62/3)

Since 61/3 > 1.8, and hence 62/3 > 3.24, from the above equality we find that

0 < 467/257 –61/3 < 5/(56048873+30844883*1.8 +16974593*3.24)

≈ 5/166567343.7 ≈ 0.00000003.

From the above estimate we see that the rational number 467/257 is is an upper approximation/bound of cube root 61/3 with accuracy up to seven decimals. This together with fact that 467/257 ≈ 1.817120622 yields 61/3 ≈ 1.8171206.

In fact, a computation in Mathematica 9 gives 61/3 ≈ 1.8171205928321463.

Dear Mercedes Orús-Lacort ,

Thank you for the detailed answer(6 hours ago):

you mentioned:

Dear Mercedes Orús-Lacort ,

I forgot to include a counterexample:

Notice that, (9)3 +91(-2)3 =1.

That mentioned by others in this thread.

x > 0, N = 91 > 0, y=-2 neither 1 nor -1.

I noticed that (6 hours ago, A-41) Jean-Claude Evard mentioned the same observation.

Best regards

Dear Romeo Meštrović ,

Thank you for the remark about the approximations of the cube roots of integers( positive or negative).

I prefer to stay within the target of the Diophantine equations.

I appreciate if you devote your experience to discuss the detailed answers posted by Mercedes Orús-Lacort , Jean-Claude Evard ,

and others. Some of them seem to be new.

Jean proved that: If N ≡ 3,4,5,6mod(9), then y ≡ 0 mod(3).

What about the other cases N ≡ 0,1,2,7,8 mod(9)?

Best wishes

Good News about the progress of the problem.

I have proved the following theorem, according to the best of my knowledge, it seems new and indeed the same as in the quadratic Pell's Diophantine equation:

Theorem: If the Diophantine equation x3 - Ny3 = z3

has a non trivial primitive solution (x,y,z), then it has

infinite number of non trivial primitive solutions.

Example: we mentioned that ( 2, 1, -1) satisfies

the Diophantine equation x3 - 9y3 = z3

you can see that

x = - 487 267171714 352336560

y = - 609 623835676 137297449

z = 1243 617733990 094836481

is a new primitive solution.

we can continue to generate infinite number of huge primitive solutions.

The work is in progress.

I hope to hear from , @Thong Nguyen Quang Do, Romeo Meštrović , Mercedes Orús-Lacort , Jean-Claude Evard , Adiya K. Hussein , Sajal Mukherjee

if the encountered any similar results in the literature or any Journal article.

Appreciate your feedback.

Best regards

A-53. Answer 50 is quite a major theorem

Answer 53, posted on June 21, 2019.

--------------------------------------------

The theorem found by Issam Kaddoura

and announced in answer 50

is quite a major theorem on the cubic Pell equation.

---

It was not obvious at all to find a proof.

---

I present my very sincere congratulations to Issam.

---

I think that it is better that Issam submits it for publication

immediately, without any additional results,

because referees and editors are overloaded

with too much work, and strongly prefer

to receive short articles with one major theorem

at a time.

---

Of course, the question asked by Issam

whether this theorem has already been published

is important.

---

We all can try to help by searching for references

about the cubic Pell equation.

--------------------------------------------

With best regards, Jean-Claude

A-54. A major reference on the cubic Pell equation

Answer 54, posted on June 21, 2019.

--------------------------------------------

The following very recent book is a major reference

on the cubic Pell equation:

---

Cubic Fields with Geometry

Part 5: Cubic Pell Equations

Samuel A. Hambleton and Hugh C. Williams

CMS Books in Mathematics

DOI: 10.1007/978-3-030-01404-9

Hardcover ISBN-13: 9783030014025

493 + xix pages, Springer, 2018

26 b/w illustrations, 27 illustrations in colour

https://www.springer.com/gp/book/9783030014025

https://www.researchgate.net/publication/328796669_Cubic_Pell_Equations_Ouvrages_de_mathematiques_de_la_SMC

--------------------------------------------

With best regards, Jean-Claude

A-55. The top expert on the cubic Pell equation

Answer 55, posted on June 21, 2019.

--------------------------------------------

Hugh Williams is the most highest top expert,

and therefore referee, on the cubic Pell equation.

---

He is co-author of the major book that I posted

in answer 54.

---

He is also co-author of the following other book:

---

Solving the Pell Equation

Michael Jacobson and Hugh Williams

CMS Books in Mathematics

DOI: 10.1007/978-0-387-84923-2

Hardcover ISBN-13: 9780387849225

Softcover ISBN-13: 9781441927477

eBook ISBN-13: 9780387849232

495 + xx pages, Springer, 2009

https://www.springer.com/us/book/9780387849225

60 citations

https://www.researchgate.net/publication/267121962_Solving_the_Pell_Equation

---

He has obtained 2482 citations for his work:

https://www.researchgate.net/profile/Hugh_Williams4

--------------------------------------------

With best regards, Jean-Claude

A-56. Does every chain start with y = 1?

Answer 56, posted on June 21, 2019.

--------------------------------------------

By the theorem posted by Issam Kaddoura in answer 50,

for every value of N for which there exists a solution

to the extended cubic Pell equation x^3 + Ny^3 = z^3,

there exists an infinite non-trivial chain of solutions

(x, y, z), where non-trivial means that if (a, b, c)

and (d, e, f) are in the chain, none of these two

is a multiple of the other.

---

For example,

(a, b, c), (2a, 2b, 2c), (3a, 3b, 3c) is a trivial chain.

--------------------------------------------

Does every chain start with y = 1?

---

There exists a solution with y = 1 if and only if

N is a sum or a difference of two cubes.

--------------------------------------------

For example, in answer 19, Romeo Meštrović

posted the solution 9^3 + (91)((-2)^3) = 1.

In this solution, N = 91 and y = -2.

---

Is there a solution with same N = 91 and y = 1?

---

Yes: (-3)^3 + (91)(1^3) = 4^3.

--------------------------------------------

In answer 25, Adiya Hussein posted the solution

10^3 + (37)(-3^3) = 1.

In this solution, N = 37 and y = -3.

---

Is there a solution with same N = 37 and y = 1?

---

Yes: 3^3 + (37)(1^3) = 4^3.

--------------------------------------------

With best regards, Jean-Claude

A-57. Update: Crash after crash after crash

Answer 57, posted on June 21, 2019.

--------------------------------------------

In answer 23, Sajal Mukherjee said the following:

---

Trying to generate solutions for arbitrary N would be a next to impossible task, bcoz, there are values of N, such that the given equation does not have integral solutions. For example, a theorem of Legendre says that, for N=3, the equation is not solvable in integers. The proof uses very similar idea to that of the proof of Fermat's last theorem for cubics. See, for example, see Number Theory by Hardy and Wright. So, the first task should be to determine the values of N, for which the equation is solvable in integers. This should be an enormously challenging problem, of what I don't have any clue at present. It is truly a brilliant question due to Issam Kaddoura .

---

I have had several ideas for proving that for N = 3,

the only solution of the cubic Pell equation x^3 – 3y^3 = 1

is the trivial solution x = 1 and y = 0.

---

Each of these ideas was worth trying,

and so far, each ended with a crash.

---

So, during the last few days, I have had crash after crash

after crash after crash.

---

I sincerely admire the hard and courageous work

of Mercedes Orús-Lacort and Issam Kaddoura

on the terribly hard and frustrating subject

of the cubic Pell equation x^3 – dy^3 = 1

and the extended cubic Pell equation x^3 + Ny^3 = z^3.

--------------------------------------------

With best regards, Jean-Claude

Dear Sajal Mukherjee

Thank you for your kind answer. The main point is to share knowledge. One needs to be patient and to do more investigations about all possible available result in this domain, the value, as well as originality, should be convenient to say it deserves to be submitted as an article. Best regards

Jean-Claude Evard , Mercedes Orús-Lacort

Thank you for your supporting words. The real greatness is the sharing knowledge, ideas, thoughts and references, and scientific discussions. This stands behind any progress of research. I think it is the main goal of the RG platform.

Also, I am sure that you can do something as many as other followers.

All of your ideas were inspiring.

PS. Jean, I can't reach the mentioned chapter.

Best regards

The primitive solution (9, 2, 1) of the cubic Diophantine equation

x3 - 91y3 = z3 generates the new primitive solution:

x = 10242000 209300723 381783092 544497244 283071509 385558977 824812799

y= 2882159 022487264 411417571 566350714 465266639 799663405 749447120

z = - 10336289 774054504 045535897 816726623 868487582 480309309 250003201

Best wishes

A-61. Table of contents of the major book

Answer 61, posted on June 22, 2019.

--------------------------------------------

I thank Issam Kaddoura for the answer 59.

--------------------------------------------

To answer Issam’s question about the mentioned chapter,

I see the title of the chapter in the table of contents

of the book posted on the following Web page:

---

Major book: Cubic Fields with Geometry

https://www.springer.com/us/book/9783030014025

---

Part or chapter? Cubic Pell Equations: Pages 205—245.

---

I do not have this book, and no hope to be able to buy it

before I finish my book of logic and start receiving

royalties from the future publisher of my future book.

--------------------------------------------

With best regards, Jean-Claude

A-62. Not all chains start with y =1

Answer 62, posted on June 22, 2019.

--------------------------------------------

The answer to my question posted in answer 56

is negative:

---

There exists a non-trivial solution of x^3 + Ny^3 = z^3

with N = 13:

---

The following solution was posted in answer 3:

(-2)^3 + (13)(3^3) = -8 + (13)(27) = - 8 +351 = 343 = 7^3.

---

But there are no solution with N = 13 and y = 1,

because 13 is neither a sum nor a difference

of two cubes.

---

Therefore, no chain of solutions with N = 13

starts with y = 1.

--------------------------------------------

With best regards, Jean-Claude

A-63. Start of a list of small solutions and

[A sum of 3 consecutive cubes] = [The next cube]

Answer 63, posted on June 22, 2019.

--------------------------------------------

Below is the start of a list of small non-trivial solutions

of x^3 + Ny^3 = z^3 in integers x, y, z, and N.

--------------------------------------------

“Trivial” means (x = 0) or (y = 0) or (z = 0).

--------------------------------------------

To avoid writing several times the “same” solutions,

we can add the following 4 conditions:

---

1. gcd(x, y, z) = 1.

2. N > 0.

3. N is not divisible by the cube of a prime.

4. y > 0.

--------------------------------------------

My current list is the following:

--------------------------------------------

N = 2: (-1)^3 + 2(1^3) = 1^3.

--------------------------------------------

N = 7: 1^3 + 7(1^3) = 2^3.

--------------------------------------------

N = 9: (-1)^3 + 9(1^3) = 2^3.

--------------------------------------------

N = 13: (-2)^3 + 13(3^3) = 7^3.

--------------------------------------------

N = 16: (-2)^3 + 16(1^3) = 2^3.

--------------------------------------------

N = 19: 2^3 + 19(1^3) = 3^3.

--------------------------------------------

N = 26: 1^3 + 26(1^3) = 3^3.

--------------------------------------------

N = 28: (-1)^3 + 28(1^3) = 3^3.

--------------------------------------------

N = 35: (-2)^3 + 35(1^3) = 3^3.

--------------------------------------------

N = 37: 3^3 + 37(1^3) = 4^3.

--------------------------------------------

N = 54: (-3)^3 + 54(1^3) = 3^3.

--------------------------------------------

N = 56: 2^3 + 56(1^3) = 4^3.

--------------------------------------------

N = 61: 4^3 + 61(1^3) = 5^3.

--------------------------------------------

N = 63: 1^3 + 63(1^3) = 4^3.

--------------------------------------------

N = 65: (-1)^3 + 65(1^3) = 4^3.

--------------------------------------------

N = 72: (-2)^3 + 72(1^3) = 4^3.

--------------------------------------------

N = 91: (-3)^3 + 91(1^3) = 4^3.

N = 91: 5^3 + 91(1^3) = 6^3.

--------------------------------------------

N = 98: 3^3 + 98(1^3) = 5^3.

--------------------------------------------

Remark 1: The two solutions

4^3 + 3^3 = 91 = 6^3 – 5^3

give the following sum of 3 consecutive cubes

that is equal to the next cube:

---

3^3 + 4^3 + 5^3 = 6^3.

--------------------------------------------

Remark 2: In answer 23, Sajal Mukherjee has given

a reference to a proof that for N = 3, there is only

the trivial solution x^3 + (3)(0^3) = x^3.

---

The next missing value of N in my above list is N = 4.

---

Is there a non-trivial solution for N = 4?

---

If not, is there a proof that no such solution exists?

--------------------------------------------

With best regards, Jean-Claude

It is an important theorem ... Congratulations Prof. Issam Kaddoura .

A-66. Ny^3 = (-N)(-y)^3

Answer 66, posted on June 23, 2019.

--------------------------------------------

Thanks to the equality Ny^3 = (-N)(-y)^3,

it is sufficient to consider only the case N > 0.

--------------------------------------------

With best regards, Jean-Claude

The all primitive solutions (x, y, z) of the Diophanthine equation x3 – 9y3= z3 such that z > 0, |x| ≤ 1000 and |y| ≤ 1000 are obtained in Mathematica 9 as follows:

(-1, -1, 2), (-2, -1, 1), (17, -7, 20), (20, 7, 17), (271, -438, 919), (919, 438, 271).

The all primitive solutions (x, y, z) of the Diophanthine equation x3 – 91y3 = z3 such that z > 0, |x| ≤ 1000 and |y| ≤ 1000 are obtained in Mathematica 9 as folows:

(1, -2, 9), (-3, -1, 4), (-4, -1, 3), (5, -1, 6), (6, 1, 5) (9, 2, 1), (-23, -21, 94), (-94, -21, 23), (158, -57, 275), (204, -341, 1535), (275, 57, 158), (465, -37, 472), (472, 37, 465), (536, -111, 653), (653, 111, 536).

A-69. (a, b, c) is “the same” as (-c, b, -a)

Answer 69, posted on June 23, 2019.

--------------------------------------------

Because the equality [a^3 + Nb^3 = c^3]

is equivalent to the equality [(-c)^3 + Nb^3 = (-a)^3],

we have that the solution (x, y, z) = (a, b, c)

of the equation x^3 + Ny^3 = z^3

is “the same” as the solution (x, y, z) = (-c, b, -a).

--------------------------------------------

With best regards, Jean-Claude

A-70. Updated list of solutions

Answer 70, posted on June 23, 2019.

--------------------------------------------

I have added some of the new solutions

posted by Romeo Meštrović in answers 67 and 68

to our list of solutions started in answer 63.

I have not had time to add them all.

--------------------------------------------

I have posted our list in the following document:

Notes about the extended cubic Pell equation

https://www.researchgate.net/publication/333967839_Notes_about_the_extended_cubic_Pell_equation

--------------------------------------------

With best regards, Jean-Claude

A-72. Happy to save time to the contributors

Answer 72, posted on June 23, 2019.

--------------------------------------------

I thank Mercedes Orús-Lacort for the answer 71.

---

I am happy that my observations in answers 66 and 69

can save time to the contributors.

---

The problem of this thread posted by Issam Kaddoura

is terribly hard, and because of this, the contributions

are terribly time consuming.

--------------------------------------------

With best regards, Jean-Claude

A-74. Tables for [x^3 + Ny^3 = z^3] and [x^3 – dy^3 = 1]

Answer 74, posted on June 23, 2019.

--------------------------------------------

I thank a lot Mercedes for the answer 73.

---

I have added a table for x^3 – dy^3 = 1

after the table for x^3 + Ny^3 = z^3

in the document:

Notes about the extended cubic Pell equation

https://www.researchgate.net/publication/333967839_Notes_about_the_extended_cubic_Pell_equation

--------------------------------------------

With best regards, Jean-Claude

Nice project Jean-Claude Evard , We wish you success.

Dear Romeo Meštrović , Jean-Claude Evard

Thank you for your efforts with Matlab to show a family of

primitive solutions to the proposed Diophantine equation.

You know, the main goal of Diophantine equations to seek general solutions that may have a chance to apply in practical cryptographic research. Initial answers determine an initial clue, but closed formulas stay the master of all, few seconds is enough to generate solutions with a huge number of digits, where machines need hundred of years to discover them.

Best regards

Dear Jean-Claude Evard , Mercedes Orús-Lacort ,

Thank you for following up the problem and the nice contributions to the properties of the solution. We can sum up all in a useful Lemma that shows such properties that reflects the odd nature and the symmetric as well as homogeneous property of the equation. I wonder if we can study and extend the results to some Principal Ideal Domains.

Best regards

Dear Adiya K. Hussein ,

Thank you for pointing out to the main results in

" Chapter The Cubic Analogue of Pell’s Equation" .

that are essential to study and compare the originality of any potential results.

Best wishes

More New results about the Cubic Pell's equation.

Theorem: If X³ - NY³ = Z³ has a primitive solution over the Gaussian integers,

then it has an infinite number of primitive roots.

In fact, this result can be extended for all principal ideal domains.

Example:

X3 -- 9Y3 = Z3

has the primitive solution over the Gaussian integers ( 1 - i, 1, 1 + 2i):

(1 - i)3 -- 9(1)3 = (1 + 2i)3

This solution generates the new primitive solution

( -190 + 82i, 83 + 54i, 53+106i) :

( -190 + 82i)3 -- 9(83 + 54i)3 = (53+106i)3

The work is in progress.

Best wishes

Further computational serach of primitive solutions (x, y, z) of the Diophanthine equations x3 – Ny3 = z3 with N = 91, 9 and 7, such that z > 0, |x| ≤ 5000 and |y| ≤ 5000 leads to two aditional solutions (1535, 341, 204) and (-1457, -1460, 6543) of the equation x3 – 91y3 = z3, while in this range the equation x3 – 9y3 = z^3 has no solution except six solutions given in my previous answer (1 day ago). Moreover, in this range the all primitive solutions (x, y, z) of the Diophanthine equation x3 –7y3 = z3 are: (1, -1, 2), (2, 1, 1), (-4, -3, 5), (-5, -3, 4), (17, -38, 73), (73, 38, 17), (1256, -183, 1265) are (1265, 183, 1256).

A-81. A conjecture

Answer 81, posted on June 24, 2019.

--------------------------------------------

By looking at the list of solutions posted by the contributors

to this thread, and that I have gathered into the following

document:

Notes about the extended cubic Pell equation

https://www.researchgate.net/publication/333967839_Notes_about_the_extended_cubic_Pell_equation

---

I see the following conjecture:

---

If N is a positive integer that is divisible by 3

and not divisible by 9, then the equation

x^3 + Ny^3 = z^3 in integers x, y, and z

has only the trivial solutions y = 0 and z = x.

--------------------------------------------

With best regards, Jean-Claude

23 - (3∙193)∙33 = (-25)3

43 + (3∙367)∙33 = 313

73 - (3∙103)∙33 = (-20)3,

etc. Hence, Conjecture proposed by Jean-Claude Evard (in the previous answer, 15 hours ago) is not true.

A-83. His counterexamples save me a lot of time

Answer 83, posted on June 25, 2019.

--------------------------------------------

I am extremely grateful to Romeo Meštrović

for his wonderful counterexamples

posted in answer 82.

---

His counterexamples will save me a lot of time,

as I will not lose time trying to prove that

my conjecture is true, given that he has proved

that it is false.

--------------------------------------------

Note that his counterexamples agree with the

theorem that I posted in Answer 36:

--------------------------------------------

Theorem. Let N be an integer

that is congruent to -4 or -3 or 3 or 4 modulo 9.

Let x, y, and z be integers such that x^3 + Ny^3 = z^3.

Then y is a multiple of 3.

--------------------------------------------

With best regards, Jean-Claude

A-84. Other conjectures

Answer 84, posted on June 25, 2019.

--------------------------------------------

For each of the missing values of N in our table

of solutions, we may conjecture that there are

no solutions with these missing values.

--------------------------------------------

Currently, the first missing values are:

N = 3, 4, 5, 6, 10, …

--------------------------------------------

Recall that the case N = 3 is solved:

---

In answer 23, Sajal Mukherjee said the following:

---

Trying to generate solutions for arbitrary N would be a next to impossible task, bcoz, there are values of N, such that the given equation does not have integral solutions. For example, a theorem of Legendre says that, for N=3, the equation is not solvable in integers. The proof uses very similar idea to that of the proof of Fermat's last theorem for cubics. See, for example, see Number Theory by Hardy and Wright. So, the first task should be to determine the values of N, for which the equation is solvable in integers. This should be an enormously challenging problem, of what I don't have any clue at present. It is truly a brilliant question due to Issam Kaddoura .

--------------------------------------------

With best regards, Jean-Claude

173 - 6∙213 = (-37)3

A-86. A new proof for x^3 – 3y^3 = 1

Answer 86, posted on June 26, 2019.

--------------------------------------------

I am extremely grateful to Romeo Meštrović

for the solution for N = 6 posted in answer 85.

--------------------------------------------

After many crashes I told about in answer 57,

I have finally found a new proof

that the cubic Pell equation x^3 – 3y^3 = 1

has only the trivial solution x = 1 and y = 0.

--------------------------------------------

I have posted my new proof after the first 6 lemmas,

in Theorem 7, at the beginning of the document:

---

Notes about the extended cubic Pell equation

https://www.researchgate.net/publication/333967839_Notes_about_the_extended_cubic_Pell_equation

---

As said by Sajal Mukherjee in answer 23,

a much more general theorem has been published in 1938

in the famous historical book:

---

An Introduction to the Theory of Numbers

by Godfrey Harold Hardy

https://en.wikipedia.org/wiki/G._H._Hardy

and Sir Edward Maitland Wright

https://en.wikipedia.org/wiki/E._M._Wright

--------------------------------------------

My theorem is only a special case of their theorem,

but mathematicians are interested in totally different

simple proofs of special cases of well-known theorems.

--------------------------------------------

With best regards, Jean-Claude

Dear's Sajal Mukherjee

Continued Fractions!!

Thank you for the nice answers and the fruitful examples generated by machines.

I announced before that I could generate solutions with n digits where n goes to infinity. I expect some theoretical results that support my previous mentioned theorems.

Let me provide a clue; this may help you to go in the right direction.

All that I want is to catch an initial primitive nontrivial solution; then I can generate infinitely many primitive solutions.

So as in the quadratic pell's equation, we can use the continued fractions to catch an initial solution.

I see many attempts to catch initial solutions for the cubic pell's equation using the same technique.

If one can make some original contribution in this direction, this encourages me to submit a joint article to an indexed journal.

Wish you great progress

A-89. The cubic FLT: Z versus Z[w] ?

Answer 89, posted on June 27, 2019.

--------------------------------------------

I thank Sajal Mukherjee very much for the answer 87.

--------------------------------------------

The proof of the cubic Fermat Last Theorem posted

on the free online encyclopedia Wikipedia

https://en.wikipedia.org/wiki/Proof_of_Fermat%27s_Last_Theorem_for_specific_exponents#n_=_3

uses the method of infinite descent and apparently

does not use the theory of Eisenstein integers.

---

I am not 100% sure about this, because it is said

---

… A crucial lemma shows that if s is odd

and if it satisfies an equation s^3 = u^2 + 3v^2,

then it can be written in terms of two coprime integers e and f

s = e^2 + 3t^2 …

---

I do not know whether the proof of this crucial lemma

uses the theory of Eisenstein integers or not.

--------------------------------------------

In the famous historical book

---

An Introduction to the Theory of Numbers

by Godfrey Harold Hardy

https://en.wikipedia.org/wiki/G._H._Hardy

and Sir Edward Maitland Wright

https://en.wikipedia.org/wiki/E._M._Wright

--------------------------------------------

the authors prove not only that the extended cubic Pell

equation x^3 + Ny^3 = z^3 in integers has only

the trivial solution x = 1 and y = 0,

but in addition, they prove that this is true

over the ring of Eisenstein integers.

--------------------------------------------

With best regards, Jean-Claude

A-90. A new conjecture

Answer 90, posted on June 27, 2019.

--------------------------------------------

By looking at the list of solutions posted by the contributors

to this thread, and that I have gathered into the following

document:

Notes about the extended cubic Pell equation

https://www.researchgate.net/publication/333967839_Notes_about_the_extended_cubic_Pell_equation

---

I see the following conjecture:

---

If N is a prime greater than 2 and congruent to -1 modulo 3,

then the extended cubic Pell equation x^3 + Ny^3 = z^3

in integers x, y, and z and in reduced form

has only the trivial solutions y = 0 and z = x.

--------------------------------------------

The first such primes are

5, 11, 17, 23, 29, 41, 47, 53, 59, 71, 83, 89, 101, …

---

All of them are not only primes in Z, but much more than this,

they are Eisenstein primes:

https://en.wikipedia.org/wiki/Eisenstein_prime

--------------------------------------------

What I call reduced form of x^3 + Ny^3 = z^3

is the form satisfying the following 7 conditions:

1. x ≠ 0.

2. y > 0.

3. z > 0.

4. z > x.

5. gcd(x, y, z) = 1.

6. N > 0.

7. N is not divisible by the cube of a prime.

---

Of course, the last condition is satisfied when N is a prime.

--------------------------------------------

With best regards, Jean-Claude

A-91. Cubic root for cubic Pell?

Answer 91, posted on June 28, 2019.

--------------------------------------------

In answer 88, Issam Kaddoura said:

---

… All that I want is to catch an initial primitive nontrivial solution;

then I can generate infinitely many primitive solutions.

So as in the quadratic pell's equation,

we can use the continued fractions to catch an initial solution.

I see many attempts to catch initial solutions

for the cubic pell's equation using the same technique.

If one can make some original contribution in this direction,

this encourages me to submit a joint article to an indexed journal. …

---

In the classical theory of the ordinary Pell’s equation x^2 – dy^2 = 1

over integers x, y, and d, that is taught in every first course

of Number Theory, it is proved that for every positive integer d,

one can obtain all the solutions

from a primitive fundamental solution,

and we can obtain this primitive solution

from the expression of the square root of d

in continued fraction.

--------------------------------------------

Will the theory be the same

for the cubic Pell’s equation x^3 – dy^3 = 1

with “cubic root” in place of “square root” ?

--------------------------------------------

Romeo Meštrović has posted a very good start

about this in answer 44.

--------------------------------------------

For the reader who have never worked or studied

the ordinary Pell’s equation and continued fractions,

I recall the main steps of the theory below:

--------------------------------------------

Pell's equation

https://en.wikipedia.org/wiki/Pell%27s_equation

---

Fundamental solution via continued fractions:

https://en.wikipedia.org/wiki/Pell%27s_equation#Fundamental_solution_via_continued_fractions

---

Continued fractions:

https://en.wikipedia.org/wiki/Continued_fraction

---

Infinite continued fractions and convergents

https://en.wikipedia.org/wiki/Continued_fraction#Infinite_continued_fractions_and_convergents

---

Continued fractions for square roots:

https://en.wikipedia.org/wiki/Continued_fraction#Square_roots

https://en.wikipedia.org/wiki/Methods_of_computing_square_roots#Continued_fraction_expansion

---

Periodic continued fractions:

https://en.wikipedia.org/wiki/Periodic_continued_fraction

… Euler was able to prove that

if x is a regular periodic continued fraction,

then x is a quadratic irrational number.

… Lagrange proved the converse of Euler's theorem:

if x is a quadratic irrational,

then the regular continued fraction expansion of x is periodic …

…

Quadratic irrational numbers:

https://en.wikipedia.org/wiki/Quadratic_irrational_number

…

Length of the repeating block:

https://en.wikipedia.org/wiki/Periodic_continued_fraction#Length_of_the_repeating_block

---

The length of the period of the simple continued fraction of d^(1/2)

J. H. E. Cohn

Pacific Journal of Mathematics

Volume 71, issue 1, pages 21—32, 1977

Open access:

https://projecteuclid.org/euclid.pjm/1102811631

---

Example: x^2 – 7y^2 = 1

https://en.wikipedia.org/wiki/Pell%27s_equation#Example

---

The smallest solution of Pell equation

x^2 – ny^2 = 1 for a given positive integer n

https://en.wikipedia.org/wiki/Pell%27s_equation#The_smallest_solution_of_Pell_equations

--------------------------------------------

With best regards, Jean-Claude

13+17∙73 = 183.

Hence, Conjecture proposed by Jean-Claude Evard (in the Answer 90, posted June 27, 2019) is not true.

An observation related to the sum of cubes:

If (x1, y1, z1) and (x2 ,y2 ,z2) are two primitive solutions for

x3 - Ny3 = z3,

then (x1y2)3 + (y1z2)3 = (x2y1)3 + (y2z1)3

And using the previous Theorem, we have proved the following:

Corollary:

There exists an infinite number of primitive cubic chains such that

A31+ B31 = A32 + B32 = . .. = A3n + B3n = ...

This is quite fascinating in light of what I was able to show regarding x^3+y^3+z^3=a^3 provided x is a perfect square. It seems that there are a lot of unusual relationships between the variables in these cubic equations but it’s not so easy to figure them out. I wish I could somehow determine a general solution for the equation I’ve played with that encompasses those equalities where x is not a perfect square - but I don’t think this is possible. And yet I find it very difficult to comprehend that there exist such solutions where x is not a perfect square. Can anyone explain this? (But remember - I’m a Doctor, not a mathematician, so my knowledge and insight is limited).

Dear Issam Kaddoura,

Do Equalities (x1y2)3 + (y1z2)3 = (x2y1)3 + (y2z1)3

and (x1y3)3 + (y1z3)3 = (x3y1)3 + (y3z1)3 imply the equality (x2y1)3 + (y2z1)3 = (x1y3)3 + (y1z3)3?

A-96. An example for answer 94

Answer 96, posted on June 29, 2019.

--------------------------------------------

Here is an example as a first answer to the question

posted by Louis John Scheinman in answer 94:

---

By applying the wonderful method found

by Issam Kaddoura and posted in answer 93

to the following equalities:

---

1^3 + (7)(1^3) = 2^3,

(-4)^3 + (7)(3^3) = 5^3,

---

I obtain:

---

[(-4)^3][38^3] – [5^3][38^3] = [3^3][17^3] – [3^3][73^3]

---

and none of these 4 products is a square.

--------------------------------------------

I took the above first two equalities from

the first table posted in the following document,

choosing two solutions with the same N = 7:

---

Notes about the extended cubic Pell equation

https://www.researchgate.net/publication/333967839_Notes_about_the_extended_cubic_Pell_equation

--------------------------------------------

With best regards, Jean-Claude

A-97. A big question

Answer 97, posted on June 29, 2019.

--------------------------------------------

In answer 88, Issam Kaddoura explained that

he is interested in any contribution to find an initial

solution of a cubic Pell equation with a given N,

using continued fractions.

--------------------------------------------

The main corresponding theorem for the ordinary

Pell equation that is in all books for first courses

of Number Theory is the following:

---

Let x, y, d, and n be integers such that x^2 – dy^2 = n,

d > 0, d is not a square, and |n| < d.

Then x/y is a convergent of the continued fraction

for the square root of d.

---

The big question is whether a similar theorem

has been or can be established for the cubic Pell

equation and the cubic root of d.

--------------------------------------------

With best regards, Jean-Claude

Dear Romeo Meštrović ,

If (x₁,y₁,z₁) , (x₂,y₂,z₂) and (x₃,y₃,z₃) are primitive roots,

then all the following three equations are true

(x₁y₂)³+(y₁z₂)³ = (x₂y₁)³+(y₂z₁)³

(x₁y₃)³+(y₁z₃)³ = (x₃y₁)³+(y₃z₁)³

(x₂y₁)³+(y₂z₁)³ = (x₁y₃)³+(y₁z₃)³

I think the converse is not true.

That is, if all equations are true,

then (x₁,y₁,z₁) , (x₂,y₂,z₂) and (x₃,y₃,z₃) are primitive roots.

One need time to construct a counterexample.

I will do when time permits.

Best regards

Dear Issam Kaddoura,

I agree! Hence, is it correct for your assertion in Corollary (in the answer, 4 hours ago): “There exists an infinite number of primitive cubic chains such that A13 + B13 = A23 + B23 = … = An3 + Bn3 =…”???. For example, please, find the part of this infinite chain induced by the following three solutions of the equation x3–7y3 = z3: (1, -1, 2), (-4, -3, 5) and (17, -38, 73)!!!

Best regards