= Solution
The <Lutz–Nagell theorem> says that for a nonsingular short equation $y^2=x^3+ax+b$ with $a,b\in\mathbb Z$, every nonidentity rational <torsion point of an elliptic curve> $(x,y)$ has $x,y\in\mathbb Z$, and either $y=0$ or
$$
\boxed{y^2\mid 4a^3+27b^2.}
$$
This is a necessary condition for torsion, not a converse.
To prove integrality, fix a prime $q$. If $v_q(x)<0$, the $x^3$ term dominates the right side of the equation. Hence $2v_q(y)=3v_q(x)$, so $v_q(x)=-2s$, $v_q(y)=-3s$ for some $s\ge1$. The parameter $t=-x/y$ has valuation $s$ and identifies the point with the formal neighbourhood of the identity, namely the <formal group of an elliptic curve> evaluated on $q\mathbb Z_q$.
For odd $q$, this group is a <torsion-free group> by the logarithm argument of Question 2. For $q=2$, use the special short-model structure. Negation sends $t$ exactly to $-t$, so the formal multiplication series $[2](t)$ is odd and has integral coefficients:
$$
[2](t)=2t+t^3g(t),\qquad g(t)\in\mathbb Z_2[[t]].
$$
If $s=v_2(t)\ge1$, the first term has valuation $s+1$ and all the others have valuation at least $3s>s+1$. Thus $v_2([2](t))=s+1$, and no iteration of doubling kills a nonzero point. Multiplication by an odd integer has unit linear coefficient and also cannot kill it. This proves the <torsion-free formal subgroup for a short Weierstrass equation>, including at two. Consequently a rational <torsion point of an elliptic curve> cannot have $v_q(x)<0$ at any prime. Its $x$ is integral, and the equation then makes its rational $y$ integral too.
For the divisibility conclusion suppose $y\ne0$. Then $2P\ne O$ and is also a rational <torsion point of an elliptic curve>, so $x(2P)$ is integral. Its tangent slope $\lambda=f'(x)/(2y)$ satisfies $x(2P)=\lambda^2-2x$. Thus $\lambda^2\in\mathbb Z$; a rational number with integral square is itself integral. In particular $y^2\mid f'(x)^2$. Reduce the supplied polynomial identity modulo $y^2=f(x)$: both terms on the left are divisible by $y^2$, so the right side $4a^3+27b^2$ is too. This is the <divisibility proof in the Nagell–Lutz theorem>.
For the specific curve, $4a^3+27b^2=3^7 5^4$. The computed equality $2P_1=-P_1$ makes $P_1$ a point of exact order three. The coordinate $y(P_3)=17$ has square not dividing $3^7 5^4$, so $P_3$ has infinite order. To decide $P_2$, use the already computed point $P_1+P_2=(15,-60)$. Its $y$-coordinate has even square, which cannot divide the odd number $3^7 5^4$, so that sum has infinite order. Since $P_1$ is torsion, $P_2$ must have infinite order as well. \b[$P_1$ has order three; $P_2$ and $P_3$ both have infinite order.]
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