All Questions
Tagged with cubics elementary-number-theory
39
questions
-2
votes
2
answers
118
views
Prime numbers $p$ such that $7p+1$ is a cube [closed]
I am stuck on this assignment. I have to find every prime number $p$ such that $7p+1$ is a cube number. After exploring enough I must say there is no prime $p$ satisfying this condition. I have tried ...
- 79
2
votes
1
answer
103
views
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
3
votes
3
answers
381
views
Find $a, b, c$ such that $a^3+b^3+c^3-3abc=2017$.
Find all natural numbers $a, b, c$ such that $a\leq b\leq c$ and $a^3+b^3+c^3-3abc=2017$.
My Attempt
$$a^3+b^3+c^3-3abc=2017$$
$$(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=2017*1$$
Now, $a+b+c$ can't be equal to $...
user1017704
4
votes
2
answers
194
views
Pairs of integers $ (x,m)$ for which $\sqrt[3]{\sqrt[3]{x-2}+m}+\sqrt[3]{-\sqrt[3]{x-2}+m}=2$ hold?
Find all pairs of integers $(x,m)$ for which $$\sqrt[3]{\sqrt[3]{x-2}+m}+\sqrt[3]{-\sqrt[3]{x-2}+m}=2$$ hold.
I have used this property :
Property:
if $$a+b+c=0 \implies a^3+b^3+c^3=3abc, $$ I come ...
- 7,195
0
votes
1
answer
298
views
Find all triples of prime numbers $(p,q,r)$ that satisfy both $p+q+r\mid pq+qr+rp$ and $p+q+r\mid p^3+q^3+r^3-2pqr$
Find all triples of prime numbers $(p,q,r)$ that satisfy:
$p+q+r\mid pq+qr+rp$ and $p+q+r\mid p^3+q^3+r^3-2pqr$. I would like you to only check my reasoning. This is an outline. From the problem ...
- 521
12
votes
8
answers
595
views
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
0
votes
3
answers
272
views
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
182
views
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
1
vote
3
answers
204
views
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
5
votes
5
answers
1k
views
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
0
votes
1
answer
49
views
$x^3-bx^2+4hx-ph=0$
$$x^3-bx^2+4hx-ph=0$$
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
vote
2
answers
379
views
Find all positive integer $a,b,c$ that $a^3+b^3+c^3$ can be divided by $a^2b,b^2c,c^2a$ [closed]
Find all triplets of positive integers $(a,b,c)$ for which
$$a^3+b^3+c^3$$
is divisible by $a^2b$, $b^2c$ and $c^2a$.
I just found that $a=b=c$ satisfies the problem. Are there any other ...
- 168
-1
votes
3
answers
1k
views
How do you solve a quadratic equation containing a negative power?
I was trying to factor a polynomial with Wolfram and I noticed a quadratic form I've never considered.
$$n^2-4+\frac{6}{n}=0$$
The purpose of it is not important but it made me wonder. How do you ...
- 6,416
8
votes
3
answers
178
views
Primitive instances where $c^3|(a^3+b^3)$ with $a,b,c\in\mathbb{N}$
It is known that by Fermat's Last Theorem there are no solutions to $a^3+b^3=c^3$ for $a,b,c\in\mathbb{N}$. I wondered about how multiplying the $c^3$ by a constant would change this fact.
...
- 3,054
2
votes
3
answers
74
views
Simplify this expansion : $\omega=(28+(\frac{5290}{3})^{\frac{3}{2}})^{\frac{1}{3}}+(28-(\frac{5290}{3})^{\frac{3}{2}})^{\frac{1}{3}}$
Find a simple closed form of :
$\omega=(28+(\frac{5290}{3})^{\frac{3}{2}})^{\frac{1}{3}}+(28-(\frac{5290}{3})^{\frac{3}{2}})^{\frac{1}{3}}$
My try :
Let :
$A=(28+(\frac{5290}{3})^{\frac{3}{2}})^{...
- 2,323