All Questions
Tagged with cubics number-theory
45
questions
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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
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1
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Rational solutions for $x^3+y^3=1$ where both x and y are non-negative
How can I find rational solutions for $x^3+y^3=1$ where both x and y are non-negative?
Edit: One of the answer in this post for general form of solutions
$$(a,b) \mapsto \left( \frac{a(a^3 + 2b^3)}{a^...
- 59
1
vote
0
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185
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Solving Vieta's jumping problem (1988 IMO problem 6) with a cubic polynomial
I have recently watched this Numberphile video explaining how Zvezdelina Stankova solved the notorious 1988 IMO problem 6 as a student.
(Most of you will know this problem)
Let $x$ and $y$ be ...
- 181
1
vote
1
answer
479
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Solving Cubic Systems of Diophantine Equations
What techniques are there for solving systems of Cubic Diophantine equations? I know there is no general purpose technique and looking at some papers it can quickly go over my head even for just a ...
- 13
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2
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173
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Integer solutions of the cubic equation $x^3-a^2bx-1-2ab=0$
Given the equation $x^3-a^2bx-1-2ab=0$. Is there a way to know if any integer solutions exist for $a,b$ integers greater than 1.
I've plotted graphs and tried to brute force it but found no solutions.
...
- 691
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1
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Is this identity $ (1+6x^3+9x^4)^3+(1-6x^3+3x-9x^4)^3+(1-9x^3-6x^2)^3=9x^2+9x+3$ known with $x$ is an integer?
It is known that, let $x$ arbitrary integer $$(9x^4)^3+(3x-9x^4)^3+(1-9x^3)^3=1\tag{1}$$ discovred by Kurt Mahler in 1936, and $$(1+6x^3)^3+(1-6x^3)^3+(-6x^2)^3=2\tag{2}$$ discovred by A. S. ...
- 7,195
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Counterexample to the Hasse principle of the form $x^3 + y^3 + z^3 + nt^3$
Selmer's cubic is a counterexample to the Hasse principle for ternary cubic forms.
We also know that the Hasse principle does not hold for quaternary cubic forms, as
$$
5x^3 + 12y^3 + 9z^3 + 10t^3
$$
...
- 2,907
4
votes
2
answers
102
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Do there exist $n,b\in \mathbb{N}, b<n$ such that $(n-b)(n^2-b)$ divides $n^3$
My friend invented this problem in his spare time and asked me if I could solve it:
Do there exist $n,b\in \mathbb{N}, b<n$ such that $(n-b)(n^2-b)$ divides $n^3$ ?
I tried my best and all I can ...
- 127
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0
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126
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Cubic root in a quadratic number field
Suppose we are given an element $\alpha = a+b\sqrt{d}$ in the quadratic number field $\mathbb{Q}(\sqrt{d})$, where $a$, $b$ and $d$ are all rationals and $d$ does not have a rational square root. I ...
- 325
2
votes
1
answer
182
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Proving the sums of three cubes conjecture by the Hasse principle
In his Cours d'arithmétique Serre applies the Hasse-Minkowski theorem to quadratic forms of the form:
$$
x^2 + y^2 + z^2 = n
$$
for $n \in \mathbb{N}$ to prove that a natural number $n$ is a square if ...
- 2,907
4
votes
2
answers
200
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Sum of cube roots of two conjugate quadratic integers makes an integer.
Consider the following expression:
$$(20-\sqrt{392})^{1/3}+(20+\sqrt{392})^{1/3}.$$
This equals $4$, but how can I show this?
Note that I do not want to make use of the following line of reasoning: 4 ...
- 53
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5
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Given that $f(x)$ is a polynomial of degree $3$, and its remainders are $2x - 5$ and $-3x + 4$ when divided by $x^2 - 1$ and $x^2 - 4$ respectively.
So here is the Question :-
Given that $f(x)$ is a polynomial of degree $3$, and its remainders are $2x - 5$ and $-3x + 4$ when divided by $x^2 - 1$ and $x^2 - 4$ respectively. Find $f(-3)$ .
What I ...
- 4,254
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49
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$x^3-bx^2+4hx-ph=0$
$$x^3-bx^2+4hx-ph=0$$
Taking the equation above, is there away to prove that for every positive prime p:
There are positive integer values for b and h so that the cubic has 3 positive integer roots
...
- 107
1
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1
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Show that $x^3+3y^3+9z^3-9xyz=1$ has infinitely many integer solutions. [duplicate]
Show that $x^3+3y^3+9z^3-9xyz=1$ has infinitely many integer solutions.
I have found that (1,0,0) and (1,-18,12) are two solutions and tried (1,-18+n,12-n).
There is a hint saying that I should try ...
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