Rational torsion on y squared equals x cubed plus positive k x ($E(\mathbb{Q}{)}_{\mathrm{tors}}\cong \mathbb{Z}/2\mathbb{Z}\text{ or }\mathbb{Z}/4\mathbb{Z}$)

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 $p\equiv 3(\mathrm{mod}4)$. The Dirichlet theorem on primes in arithmetic progressions excludes every odd torsion prime and bounds the two-primary part by four, by choosing $p\equiv 3(\mathrm{mod}8)$. 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 $k=4t^4$, $t\ge 1$. 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.