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Let $E$ be an elliptic curve over $\mathbb{Q}$ given by a Weierstrass equation: $$y^2 +a_1 xy + a_3 y = x^3 + a_2 x^2 + a_4 x + a_6 \;\;\;(*)$$

In Silverman's book, The Arithmetic of Elliptic Curves Chapter VIII, Lemma 4.1, it is brought up that every affine $\mathbb{Q}$-rational point to the equation $y^2 = x^3 + A x + B$ can be written of the form $\displaystyle (x, y) = \left(\frac{a}{d^2}, \frac{b}{d^3}\right),$ where each coordinate is a fraction written in lowest terms. I then read somewhere else that this also holds for a general Weierstrass equation $(*)$, with neither source providing a proof to back these statements up.

First, why are these statements true? I have tried proving this as if it were an elementary number theory exercise, but I haven't been fruitful (or perhaps creative enough...).

Second, I just wanted to express my shock that this fact isn't emphasized more. This gives an immediate sanity check to rule out any $\mathbb{Q}$-rational solutions to a general Weierstrass equation. It also eliminates many fractions such as $\frac{1}{7}$ or more generally, any reduced fraction that has a squarefree denominator as either an $x$ or $y$-coordinate to the solution set.

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    Did you try applying the same argument as for the 1st case ? Let $x = \frac{u}{v}, y = \frac{w}{z}, gcd(u,v)=gcd(w,z) =1$, then $y^2 = x^3+ax+b$ means $\frac{w^2}{z^2} = \frac{u^3}{v^3}+a\frac{u}{v}+b=\frac{u^3+a uv^2+bv^3}{v^3}$ where $gcd(w^2,z^2) = gcd(u^3+auv^2+bv^3,v^3)= 1$ so that $z^2=v^3 = d^6,w^2 = u^3+auv^2+bv^3$ and $x= \frac{u}{d^2}, y = \frac{w}{d^3}$ – 2017-01-29
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    How did you get that second GCD condition? I can't seem to resolve the $auv^2$ term. – 2017-01-29
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    $gcd(u^3+auv^2+bv^3,v)=gcd(u^3,v)= 1$ – 2017-01-29
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    Ah, thank you very much! – 2017-01-29

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