= Solution
{c}
Since neither $n$ nor $-n$ is a square, both cubics have exactly one rational root, zero. Thus each <torsion point of an elliptic curve> <group> has exactly one nonidentity <2-torsion>, namely $T=(0,0)$.
Neither curve has a <rational point> of order four. Indeed on $E_m:y^2=x^3+mx$ the duplication formula gives
$$
x(2P)=\frac{(x(P)^2-m)^2}{4y(P)^2}.
$$
If $2P=T$, then $P$ is neither the identity nor a <2-torsion>, so the denominator is nonzero and $x(P)^2=m$. Thus $m$ must be a rational square; an <integer> that is a rational square is an <integer> square. Applying this with $m=n$ or $m=-n$ contradicts the hypotheses.
If $3\nmid n$, both curves have <good reduction of an elliptic curve> at three and exactly four points over $\mathbb F_3$. To see this, $x^3=x$ there, so the equation is $y^2=(1+m)x$. For $m=1$ the right side runs through all <residues>, and for $m=-1$ it is always zero; either way there are three affine points. Injection of <torsion point of an elliptic curve> at this odd good <prime> bounds each <torsion element> order by four. An order-four <torsion subgroup> with just one nonzero <2-torsion> would contain an order-four point, which was excluded. Therefore in this case both <torsion subgroups> equal $\{o,T\}$.
The remaining case is $3\mid n$. The literal hypothesis $15\nmid n$ then ensures $5\nmid n$, so both curves have <good reduction of an elliptic curve> at five. Direct square counting gives
$$
\begin{array}{c|rrrr}
m\bmod5&1&2&3&4\\\hline
\#E_m(\mathbb F_5)&4&2&10&8.
\end{array}
$$
For example the square <residues> are $0,1,4$; substituting $x=0,\ldots,4$ gives this table, including the point at infinity. The pair $m=n,-n$ has <residues> $(1,4)$ or $(2,3)$ in some order. In either pair at least one reduction <group> has order two or four. The same injection and absence of order-four points then force that curve's <torsion point of an elliptic curve> to have order two. Thus
$$
\boxed{\text{at least one of }E(\mathbb Q)_{\rm tors},
F(\mathbb Q)_{\rm tors}\text{ is }\{o,(0,0)\}.}
$$
This <torsion alternative for opposite quartic-twist coefficients> uses only $15\nmid n$; it does not strengthen the source hypothesis to coprimality with fifteen.
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