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Let $m$ and $n$ be relatively prime integers and let$$A = |(m^3 + 32n^3)m|,\text{ }B = |4(m^3 - 4n^3)n|.$$Denote by $D$ the greatest common divisor of $A$ and $B$. In order to show the inequality in question, it suffices to show that $D$ is a divisor of $144$. For, if that is the case and if the $x$-coordinate of $P$ is given by $m/n$ ($n \neq 0$) in lowest ...

3

Let $C$ be the curve defined by the equation $$3(a^2 + 1)(b^2 + 1) = 10( a(b^2 + 1) + b(a^2 + 1)),$$ so rational points on $C$ correspond to solutions of your original equation with $c = 3$. Let $E$ be the elliptic curve with equation $$y^2 = x^3 + x^2 + 65x + 3458.$$ The point is that $E$ and $C$ are "birationally equivalent" over the rational numbers ...

3

Because it doesn't induce an automorphism of the function field of the curve. Automorphisms of an elliptic curve $E$ are endomorphisms $\varphi \colon E\to E$ which have degree $1$. Recall that the degree is defined as $[k(E)\colon \varphi^*(k(E)) ]$, where $k(E)$ is the function field of the curve and $\varphi^*$ is the induced map on $k(E)$. The Frobenius ...

3

Here’s how to do it for an elliptic curve over $\Bbb Q$: Start with $y^2=x^3+ax+b$, with the neutral point $\Bbb O$ up at infinity. So you homogenize and then set $y=1$ to get $z=x^3+axz^2+bz^3$. I like to think of this as a recursive schema for expanding $z$ as a series in $x$, but you can do it in various ways, to get  z = x^3 + ax^7 + bx^9 + ...

3

What makes Fermat's equation really hard is the presence of one of the unknown variables, n, as an exponent. This makes it an example of an "exponential Diophantine equation". If one has a "polynomial Diophantine equation", (polynomial function in some number of variables) = 0, then finding rational solutions is still rather difficult, but not usually quite ...

3

Elliptic curves are very famous examples of algebraic varieties. You seem to have forgotten that the function can have more than one variable; take the function $f(x,y)=y^2-x(x^2+5)$, then $E$ is just the variety $V(f)$.

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If $p \neq 1 \pmod 3$ then $x \mapsto x^3$ is a permutation of $\Bbb F_p$, and so for each $y \in \Bbb F_p$ there is a unique $x \in \Bbb F_p$ such that $x^3 = y^2-17$

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Since $+$ is a rational map $E \times E \to E$, and this map is the restriction of $+$ to $E \times \{P\}$, it is a rational map.

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Let $E_1, E_2$ be elliptic curves. Fix a point $Q \in E_2$. Then the map $\phi : E_1 \rightarrow E_2$ defined by $P \mapsto Q$ is a constant morphism. This is because it is a rational map (it is just effectively a constant polynomial) and is defined everywhere trivially.

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