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共 15 个问题,第 1/1 页
椭圆曲线 MSE 2 票 2 回答 95 浏览 未读

Polynomial solutions to $A^4+B^4=C^4+D^2\,$ leading to numerical solutions to $w^4+x^4 = y^4+z^4$?

Tito Piezas III
An interesting MSE post was recently made by Koushik Pramanik. To give some background, there seems to be only one known polynomial solution to the equation in the first part of the title, namely, $$(17 p^2 - 12 p q - 13 q^2)^4 + (17 p^2 + 12 p q - 13 q^2)^4 = (17 p^2 - q^2)^4 + (289 p^4 + 14...
椭圆曲线 MSE 7 票 0 回答 115 浏览 未读

Can the new infinite family $a^4+b^4+c^4+d^4 = (a+27b+27c+27d)^4$ be split into two quadrics?

Tito Piezas III
In 2008, Jacobi-Madden found (essentially by data-mining the 25 smallest solutions) that $$a^4+b^4+c^4+d^4 = (a+b+c+d)^4 =e^4$$ was solvable and in fact a member of an infinite family. In early August 2026, Matej Veselovac and I were data-mining the first 45,000 smallest solutions of Eugene Go's...
椭圆曲线 MSE 1 票 0 回答 34 浏览 未读

Finding multigrade $(8,4,4)$ solutions satisfying $a^8+b^8+c^8+d^8=e^8+f^8+g^8+h^8$

jorisperrenet
A while ago, @Aleksandr posed this question, about finding new solutions to $$a^8+b^8+c^8+d^8=e^8+f^8+g^8+h^8$$ (where the solutions should be non-trivial and primitive) and he stated the known result that in 2006, Nuutti Kuosa discovered...
椭圆曲线 MSE 3 票 0 回答 59 浏览 未读

An Elliptic curve for solving $W^4+X^4=Y^4+Z^4$

Koushik Pramanik
It is well known that a parameterization for the equation $$W^2+X^2=Y^2+Z^2$$ is given by: $$(W,X,Y,Z)=(a+b,ab-1,ab+1,a-b).$$ I have found a similar type of parameterization for the fourth-power equation: $$W^4+X^4=Y^4+Z^4$$ where $W = d+e$, $X = cde-1$, $Y = d-e$, and $Z = cde+1$, provided that...
椭圆曲线 MSE 1 票 1 回答 64 浏览 未读

Elliptic curves for $a^4+b^4+c^4 = d^4+e^4$ with $d\neq \pm e$?

Tito Piezas III
(Moved from previous post since the answers were not elliptic curves.) I. Question We seek to find infinitely many primitive solutions to, $$a^4+b^4+c^4 = d^4+e^4$$ where $d \color{red}{\ne} e$ using polynomial solutions or elliptic curves. The most well-known case when $d = e$ is,...
椭圆曲线 MSE 3 票 3 回答 159 浏览 未读

Solutions to $a^4+b^4+c^4 = d^4+e^4$ with $d\neq e$?

Tito Piezas III
(Updated with a computer search.) I. Question We seek to find infinitely many primitive solutions to, $$a^4+b^4+c^4 = d^4+e^4$$ where $d \color{red}{\ne} e$. The most well-known case when $d = e$ is, $$a^4+b^4+(a+b)^4 = 2(a^2+ab+b^2)^2$$ where one then solves $a^2+ab+b^2 = z^k$ for $k=2$. (In...
椭圆曲线 MSE 0 票 0 回答 16 浏览 未读

Can Poncelet's invariant measure be generalized to pairs of quadrics in dimension 3?

user582761
Let $S\subset \mathbb R^3$ be a fixed sphere and let $E\subset \mathbb R^3$ be a fixed ellipsoid containing $S$. Consider tetrahedra $$A_1A_2A_3A_4$$ such that $$A_i\in E$$ and each face is tangent to $S$. Let $D_i\in S$ be the tangency point of the face opposite $A_i$. Poncelet’s closure...
椭圆曲线 MSE 3 票 1 回答 95 浏览 未读

Is the curve $y^2=x^4+1$ elliptic?

bxhlywzzcr
The curve $y^2=P(x)$ over the field of complex numbers, where $P(x)$ is a polynomial of degree $4$ without repeating roots, can be transformed with a birational transformation into $Y^2=Q(X)$ with $Q(x)$ of degree $3$ without repeating roots. That is, an elliptic curve. However, if $P(x)$ does...
椭圆曲线 MSE 0 票 0 回答 52 浏览 未读

New primitives $3(a^3+b^3+c^3+a+b+c)+5(a^2+b^2+c^2)+2(a^2b+b^2c+ac^2)+4(a^2c+bc^2+ab^2+ab+bc+ac)=0$

Aleksandr
$$3(a^3+b^3+c^3+a+b+c)+5(a^2+b^2+c^2)+2(a^2b+b^2c+ac^2)+4(a^2c+bc^2+ab^2+ab+bc+ac)=0$$ A table of primitives known to me $(a, b, c)$, $a \in{Z}$, $b\in \mathbb{Z}$, $c\in \mathbb{Z}^+$ $$ \boxed{\begin{array} {|r|r|r|r|}\hline №(a,b,c)& a_n & b_n & c_n \\ \hline S_1 & 0 & 0 & 0 \\ \hline S_2 & 0...
椭圆曲线 MSE 1 票 0 回答 18 浏览 未读

$\Pi$-orbits of elliptic curve covers

J. Zimmerman
Let $$ \mathcal S=(\mathcal I,\Gamma,\Pi) $$ be a seam marked seed built from four compact oriented $2$-dimensional complex orbifold sheets. Each sheet is assumed to be a football type orbifold: its coarse underlying Riemann surface is $$ |\mathcal O_i|\cong \mathbb P^1 $$ and it has two...
椭圆曲线 MSE 5 票 2 回答 101 浏览 未读

An elliptic curve for $x_1^5+x_2^5+x_3^5=y_1^5+2y_2^5$?

Tito Piezas III
In a prior post, the equation, $$x_1^5+2x_2^5 = y_1^5+2y_2^5$$ was considered. It has only one known primitive solution. This present post considers the similar, $$x_1^k+x_2^k+x_3^k = y_1^k+2y_2^k$$ valid for both $k = (1,5)$. Duncan Moore found only one primitive solution, namely, $$85333^k +...
椭圆曲线 MSE 3 票 1 回答 57 浏览 未读

Question about Theorem 4.2 from Rational Points on Elliptic Curves

KnobbyWan
I have a question about a certain part of the proof of Theorem 4.2 from Rational Points on Elliptic Curves. Let $p$ be a prime so that $p=1\;\mathrm{mod}\;3$. Let $R=\{x^{3}|x\in\mathbb{F}_{p},x\neq 0\}$. Notation: $[XYZ]$ is the number of triples $(x,y,z)$ so that $x+y+z=0,x\in X,y\in Y,z\in...
椭圆曲线 MSE 1 票 0 回答 96 浏览 未读

Is there a completely elementary way to prove that $Y^2=X^3-32X$ has rank 1?

Kieren MacMillan
I’m working on a paper in which I end up considering the biquadratic rational curve $$u^2v^2 - u^2 - v^2 - 6uv + 8 = 0. \tag{$1$}$$ To complete the remainder of my proof/method, I need to prove that it has rank 1. I believe it can be transformed to the Weierstrass form $$Y^2=X^3-32X,$$ and then...
椭圆曲线 MSE 1 票 1 回答 128 浏览 未读

Near to Euler’s 4th power taxicab equation solution using $W^{4}+X^{4}=Y^{2}+Z^{4}$?

Pure Mathematics lover
The above given equation solution is very easy just make it to an elliptic curve. For $$ W^{4}+X^{4}=Y^{2}+Z^{4} $$ Divide both sides $Z^{4}$, we wil get $$ \left(\frac {W}{Z}\right)^4 + \left(\frac{X}{Z}\right)^4 = \left(\frac{Y}{Z^2}\right)^2 +1 $$ If we substitute $\frac{W}{Z} = (u+v)$,...
椭圆曲线 MSE 2 票 0 回答 50 浏览 未读

Cubic Diophantine equation

Odail Gouttai
Problem:I am looking for help with the following Diophantine equation: $$y^2 = x^3 - x^2 + 16$$ By working through the equation, I have successfully found 8 distinct non- negative integer solutions. The largest value of $x$ among all the solutions I found is $x = 112$ (which gives $y = 1180$)....