= Solution
<Bad reduction of an elliptic curve> must be tested on a <Minimal Weierstrass equation>, not inferred merely from primes dividing the displayed <elliptic-curve discriminant>. For an odd prime $p$, write $v_p(D)=2k+\epsilon$ with $\epsilon\in\{0,1\}$. The coordinate change $x=p^{2k}X$, $y=p^{3k}Y$ gives the same form with parameter $D/p^{2k}$. If $\epsilon=0$, that parameter is a unit at $p$ and its discriminant is a unit, so reduction is good.
If $\epsilon=1$, the new discriminant valuation is six. Discriminants of isomorphic Weierstrass models differ by a twelfth power under a coordinate change, so their valuations differ by a multiple of twelve. An integral model has nonnegative discriminant valuation; consequently a model of valuation six cannot be replaced by one of valuation zero. Reduction is bad. At $p=2$, oddness of $D$ makes the original discriminant valuation six, and the same argument gives <bad reduction of an elliptic curve> there as well. Thus
$$
\boxed{\operatorname{Bad}(E_D)=\{2\}\cup\{p\text{ odd}:v_p(D)\text{ is odd}\}.}
$$
These are the <bad primes of a congruent number curve with odd parameter>. If $D$ is squarefree this simplifies to the primes dividing $2D$. Squarefreeness is not stated in the PDF: for example, $D=9$ gives a curve isomorphic to $E_1$ by $x=9X,y=27Y$, so it has <good reduction> at three although the displayed discriminant is divisible by three.
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