= Rational torsion on y squared equals x cubed plus positive k x
{title2=$E(\mathbb Q)_{\rm tors}\cong\mathbb Z/2\mathbb Z\text{ or }\mathbb Z/4\mathbb Z$}
For positive integral $k$, the rational <torsion subgroup> of $y^2=x^3+kx$ has order dividing four. Indeed, <torsion-freeness of the formal group over Qp for odd p>, <reduction of torsion points on an elliptic curve>, and the <point-count criterion for y squared equals x cubed plus k x> bound its order by $p+1$ at all good <primes> $p\equiv3\pmod4$. The <Dirichlet theorem on primes in arithmetic progressions> excludes every odd torsion <prime> and bounds the two-primary part by four, by choosing $p\equiv3\pmod8$. There is exactly one nonzero rational <2-torsion> point, namely $(0,0)$. A rational half of it must have $x^2=k$ by the <elliptic-curve addition formula>. Writing $k=s^2$ with $s>0$, its equation is $y^2=2s^3$, so $s=2t^2$ and $k=4t^4$. Conversely $(2t^2,4t^3)$ has order four. Thus the cyclic group of order four occurs exactly for $k=4t^4$, $t\geq1$. The order bound itself holds for every nonzero integral $k$ of either sign: positivity was used only to classify the rational points of orders two and four.
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