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On why solutions to $x^4+y^4+z^4 = 1$ come in pairs

Solutions were found with $d<10^{26}.$ I. New u $(u,v)=(\dfrac{1744}{495}, \dfrac{135}{1208})$ with rank $3$. Then $\pm \sqrt{D^2}$ gives the pair, $$372623278887^4+435210480720^4+369168502640^4=...
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Elliptic curves - infinitely many $\mathbb{Q}$-rational points

As in comment, I'd assume $n \neq 0$. There are some rational points on the curve (for any $m$ and $n$): $(0, \pm 3n)$ and $(\pm (m^2 + 1), \pm 3n)$. Now, to show that the curve as infinitely many ...
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On counterexamples to Euler's conjecture using Bremner's and Durman's elliptic curves

I. Bremner's elliptic curve $$t^2 = 42856039590241 + 4301879366236v - 65877950554v^2 - 710638564v^3 - 109887359v^4$$ Quartic can be transformed to an elliptic curve as follows. $$E: Y^2= X^3 -X^2 + ...
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Example 3.5 Silverman's Elliptic Curves: problem with a point in the projective space

Be careful: the point is not that the equality $[-Y^2,Y(X-Z)]=[-Y,X-Z]$ holds at a point with $Y=0$ - rather, this equality holds on their common domain of definition and you can consider the map ...
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Points on Hyperelliptic Curve

The short answer: The "curved line" is not a straight line. The longer answer: It may be useful to have an explicit situation, where we add reduced divisors. Let us consider the Example from ...
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Why is $\wedge ^2 E[p] \cong \mu_p$?

Let $k$ be of characteristic coprime to $p$ and $E/k$. The first thing to note is that $E[p]$ is an $\mathbb{F}_p$-vector space of rank $2$. The exterior square of $E[p]$ is therefore an $\mathbb{F}_p$...
Mummy the turkey's user avatar
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Still more elliptic curves for $a^4+b^4+c^4=d^4$

Curve 1: $$4(110301312 + 10244932v - 1285119v^2 + 5299v^3 - 6260v^4) = t^2$$ Quartic can be transformed to an elliptic curve as follows. $$E: Y^2= X^3+ 2265722465761X -3154189403034549278$$ $E$ has ...
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Ramification of mod $\ell$ representation of elliptic curves

I might as well answer this. The first, and most important result here is the Néron-Ogg-Shafarevich criterion (see eg Serre-Tate, Good reduction of Abelian varieties – or just Silverman for elliptic ...
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$p$-adic numbers

Write the inequality as a congruence: it is asking you to solve $x^2 + 1 \equiv 0 \bmod 5^4$. It's very important to realize $p$-adic absolute value upper bounds are just congruences modulo a power of ...
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On why solutions to $x^4+y^4+z^4 = 1$ come in pairs

Here comes a table of new solutions for $u$, searched by the scheme proposed above, compare also with https://oeis.org/A003828. The last columns shows the obtained solution $x,y,z,t$, possibly ...
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On why solutions to $x^4+y^4+z^4 = 1$ come in pairs

Addendum Search results using group law with $d<10^{26}.$ u=233/60: (a,b,c,d) (4707813440, 7813353720, 11988496761,12558554489) (188195571677171463096, 335981923744570504065, 275897431444390465240,...
Tomita's user avatar
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Still more elliptic curves for $a^4+b^4+c^4=d^4$

(Updated with solutions from this table.) This is an addendum to Tomita's two answers and places the whole post in context. With new points on Curves 1,2,3,4 such that $d<10^{28}$, we have a better ...
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Sufficient condition for a quartic curve to be birationally equivalent to an elliptic curve

If the curve $X=V(y^2=Ax^4+Bx^3+Cx^2+Dx+E)\subset\Bbb A^2_k$ is nonsingular (equivalent to $Ax^4+Bx^3+Cx^2+Dx+E$ having no repeated roots), then it defines a genus-one curve: the morphism $X\to\Bbb A^...
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How to come up with the generating function of an elliptic curve?

Before starting, I need to make a confession: up to the very end, I hoped that someone could write this answer instead of me. That's because this story deserves to be explained and shared, but even ...
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Transforming $X^3 + Y^3 + Z^3 + 3XYZ = 0$ into an elliptic curve

I will follow Nagell's algorithm, as described for instance in the excellent Handbook of Ian Connell, §1.4, Cubic to Weierstraß, page 115 and following, a link found quickly is Ian Connell, Elliptic ...
dan_fulea's user avatar
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On why solutions to $a^4+b^4+c^4+d^4 = (a+b+c+d)^4$ also come in pairs

Thanks to a newer version of Tomita's table here, we find that Allan MacLeod found in 2017 four additional $u$ of small height, namely, \begin{array}{|c|c|c|c|c|} \hline \text{#} & \color{red}u &...
Tito Piezas III's user avatar
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Profinite completion of local Mordell-Weil group

You can easily check that $E(K)$ is a topological group that continuously embeds as a closed subspace $\mathbb{P}^2(K)$, hence it is a compact and totally disconnected topological group. That it is ...
Aphelli's user avatar
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Cubic curves and their lattices

So part a) is fine - the maps are $\sigma([x,y,z]) = [x,\sqrt{\lambda}y,z\lambda^{-1}]$, $\tau([x,y,z]) = [x-z, iy, -z]$ and $\nu$ is formed by firstly applying $\sigma$ and then $\tau$. It is ...
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1 vote

Is a twisted Edwards curve an elliptic curve?

Consider a twisted Edwards curve: $$a\ X^2 + Y^2 = 1 + d\ X^2 Y^2$$ Make the substitutions $X=\frac x y$ and $Y = \frac {x-1}{x+1}$ $$ \begin{align} a\,\left(\frac xy\right)^2 + \left(\frac {x-1}{x+1}\...
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