= Solution
For $p\nmid2k$, 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 $x$ gives
$$
\#\widetilde E(\mathbb F_p)=p+1+\sum_{x\in\mathbb F_p}\chi(x^3+kx).
$$
If $p\equiv3\pmod4$, 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 $p\equiv1\pmod4$, and put $h=(p-1)/2$. The <Euler criterion> and the sum of powers over a <finite field> show that the <Trace of Frobenius> satisfies
$$
a_p=-\sum_x\chi(x^3+kx)\equiv[x^{p-1}](x^3+kx)^h\pmod p.
$$
Indeed, 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> yields
$$
a_p\equiv\binom{h}{h/2}k^{h/2}\not\equiv0\pmod p.
$$
Neither $k$ nor the <binomial coefficient> vanishes modulo $p$. Hence $a_p\ne0$, proving the <point-count criterion for y squared equals x cubed plus k x>:
$$
\boxed{\#\widetilde E(\mathbb F_p)=p+1\iff p\equiv3\pmod4.}
$$
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 $j\geq2$ has <valuation> strictly greater than $v_p(t)$, since $jv_p(t)-v_p(j)>v_p(t)$. 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 $p\equiv3\pmod4$.
By the <Dirichlet theorem on primes in arithmetic progressions>, choose a good <prime> $p\equiv3\pmod8$. 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 $p\equiv3\pmod4$ and $p\equiv1\pmod\ell$, avoiding the finitely many <primes> dividing $2k$. But then $\ell\nmid p+1$, a contradiction. Therefore
$$
\boxed{\#E(\mathbb Q)_{\rm tors}\mid4.}
$$
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 is
$$
x(2P)=\frac{(x(P)^2-k)^2}{4x(P)(x(P)^2+k)}.
$$
Such 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 $t\geq1$. 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.
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