All Questions
Tagged with cubics diophantine-equations
56
questions
0
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The Diophantine equation $P_1^3 + P_2^3 + P_3^3 = P_4^3$
Consider the Diophantine equation
$$P_1^3 + P_2^3 + P_3^3 = P_4^3$$
Where $P_n$ are distinct odd primes.
What are the smallest solutions ?
Is there even a solution ?
Or is there a reason no such ...
- 16.4k
7
votes
1
answer
230
views
More $a^3+b^3+c^3=(c+1)^3,$ and $\sqrt[3]{\cos\tfrac{2\pi}7}+\sqrt[3]{\cos\tfrac{4\pi}7}+\sqrt[3]{\cos\tfrac{8\pi}7}=\sqrt[3]{\tfrac{5-3\sqrt[3]7}2}$
I. Solutions
In a previous post, On sums of three cubes of form $a^3+b^3+c^3=(c+1)^3$, an example of which is the well-known,
$$3^3+4^3+5^3=6^3$$
we asked if there were polynomial parameterizations ...
- 54.9k
1
vote
1
answer
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Solving Cubic Systems of Diophantine Equations
What techniques are there for solving systems of Cubic Diophantine equations? I know there is no general purpose technique and looking at some papers it can quickly go over my head even for just a ...
- 13
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votes
2
answers
173
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Integer solutions of the cubic equation $x^3-a^2bx-1-2ab=0$
Given the equation $x^3-a^2bx-1-2ab=0$. Is there a way to know if any integer solutions exist for $a,b$ integers greater than 1.
I've plotted graphs and tried to brute force it but found no solutions.
...
- 691
1
vote
1
answer
106
views
Finding Lattice points on a Cubic
I want to study the rational points on a cubic. Eventually I found Nagell's algorithm from http://webs.ucm.es/BUCM/mat/doc8354.pdf, but I cannot immediately apply it because I don't know a rational ...
- 779
0
votes
3
answers
96
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Finding all possible roots of the equation
Find all possible solutions to the equation $$(x^3-x)+(y^3-y)=z^3-z$$ where $(x,y,z)\gt1$ and $\in\mathbb{Z}$ and not all three of them are equal.
The original question didn't have the last condition ...
- 860
2
votes
1
answer
103
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On Réalis’s solution of the “cubic Markov equation”
I am interested in the Diophantine equation
$$X^3+Y^3+Z^3=3XYZ$$
and its solutions. In Nouv. Corr. Math. 1879 (page 7), Réalis claims that, other than the trivial solution $x=y=z$, the complete ...
- 7,929
0
votes
0
answers
135
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The Diophantine equation $x^2\pm iy^2=z^3$ in Gaussian integers
I am trying to find any reference about the solutions of the Diophantine equation $x^2\pm iy^2=z^3$. It seems to me that I read about this somewhere before, but I can't remember where.
Any ...
- 119
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votes
8
answers
595
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Solving $X^2-6Y^2=Z^3$ in positive integers
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 ...
- 7,929
3
votes
3
answers
251
views
Solve for integers $x, y$ and $z$: $x^2 + y^2 = z^3.$
Solve for integers $x, y$ and $z$:
$x^2 + y^2 = z^3.$
I tried manipulating by adding and subtracting $2xy$ , but it didn't give me any other information, except the fact that $z^3 - 2xy$ and $z^3+2xy$...
user907745
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votes
3
answers
272
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Find all triples of natural numbers $x$, $y$, $z$ such that $x+2y+2z=xyz$.
Find all triples of natural numbers $x$, $y$, $z$ such that $x+2y+2z=xyz$.
I found a solution, but it involves a lot of case work. Can someone help me find a solution which doesn't involve a lot of ...
- 255
2
votes
1
answer
459
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Cubic Diophantine equations
We know how to solve Diophantine equations of first degree (linear) and of the second degree. What is available for solving Diophantine equations of the third degree? Let us say that we are given a ...
- 287
3
votes
2
answers
189
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Integral solutions to the diophantine equation $y^2=x^3+2017$.
Case 1: $y^2=0 \mod{3}$
\begin{align*}
& y^2 = 3b \quad (b=3k^2) \implies x^3=3a +2 \\
& 3b = 3a + 2 + 2017 \implies b-a = 673 \\
& k^2 = \frac{a-2}{3} + 225 \\
\end{align*}
Then I just ...
- 860
1
vote
3
answers
204
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Cubic diophantine equation in two variables
I am trying to solve the Diophantine equation $xy^2 + 2xy + x - 243y = 0$.
I simplified it to $x(y^2 +2y +1) = 243y$ but I am stuck on what to do now. Any help would be appreciated.
- 59
1
vote
2
answers
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$6(m + 1)(n − 1), 6 + (m − 1)(n + 1)$ and $(m − 2)(n + 2)$ are simultaneously perfect cubes.
Find all pairs of integers (m, n) such that the integers $6(m + 1)(n − 1), 6 +
(m − 1)(n + 1)$ and $(m − 2)(n + 2)$ are simultaneously perfect cubes.
I assumed $6(m + 1)(n − 1), 6 +
(m − 1)(n + 1)$ ...
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