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For $p\nmid 2k$, the displayed integral equation has nonzero elliptic-curve discriminant modulo $p$, so it has good reduction. Let $\chi$ be the Legendre symbol, with $\chi (0)=0$. Counting the two, one, or zero possible ordinates over each $x$ givesIf $p\equiv 3(\mathrm{mod}4)$, then $\chi (-1)=-1$ and the terms for $x$ and $-x$ cancel. The term at zero vanishes. Thus the count is $p+1$.

For the converse, suppose $p\equiv 1(\mathrm{mod}4)$, and put $h=(p-1)/2$. The Euler criterion and the sum of powers over a finite field show that the Trace of Frobenius satisfiesIndeed, among the positive exponents in this polynomial, of degree $3(p-1)/2<2(p-1)$, only $p-1$ has a nonzero sum over ${\mathbb{F}}_p$, and that sum is $-1$. Since $(x^3+kx{)}^h=x^h(x^2+k{)}^h$, the coefficient formula for trace of Frobenius modulo p yieldsNeither $k$ nor the binomial coefficient vanishes modulo $p$. Hence $a_p\ne 0$, proving the point-count criterion for y squared equals x cubed plus k x:

We use the following formal logarithm fact, also useful for the next part. For a formal group law over ${\mathbb{Z}}_p$, its logarithm has coefficients $c_j/j$ with $c_j\in {\mathbb{Z}}_p$, obtained by integrating its integral invariant differential. If $p$ is odd and $t\in p{\mathbb{Z}}_p\setminus \{0\}$, every term of degree $j\ge 2$ has valuation strictly greater than $v_p(t)$, since $j{v}_p(t)-v_p(j)>v_p(t)$. Thus the logarithm converges and is nonzero at $t$. Its homomorphism into the additive characteristic-zero group proves torsion-freeness of the formal group over Qp for odd p. Consequently reduction injects the entire rational torsion subgroup at any odd prime of good reduction, including its $p$-primary torsion. In particular the torsion subgroup is finite, and its order divides $p+1$ at every good $p\equiv 3(\mathrm{mod}4)$.

By the Dirichlet theorem on primes in arithmetic progressions, choose a good prime $p\equiv 3(\mathrm{mod}8)$. Then $v_2(p+1)=2$, bounding the two-primary part by four. For any odd prime $\ell$ dividing the torsion order, the Chinese remainder theorem and the Dirichlet theorem on primes in arithmetic progressions give a good prime $p$ with $p\equiv 3(\mathrm{mod}4)$ and $p\equiv 1(\mathrm{mod}\ell )$, avoiding the finitely many primes dividing $2k$. But then $\ell \nmid p+1$, a contradiction. Therefore

The order four does occur. For $k=4$, the elliptic-curve addition formula gives $2(2,4)=(0,0)$, so $(2,4)$ has order four. More precisely, the only nonzero rational 2-torsion point when $k>0$ is $(0,0)$, so order four means a cyclic group with a rational half of this point. The duplication formula isSuch a half must have $x(P{)}^2=k$. Writing $k=s^2$, positivity of $y(P{)}^2=2kx(P)$ forces $x(P)=s>0$. Then $y(P{)}^2=2s^3$, so $2s$ is a rational square, equivalently $s=2t^2$ for an integer $t\ge 1$. Conversely $(2t^2,4t^3)$ is a half of $(0,0)$ when $k=4t^4$. The rational torsion on y squared equals x cubed plus positive k x is therefore cyclic of order four exactly for these $k$, and otherwise cyclic of order two.

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.