 Prescribing a factor by elliptic-curve tripling

Let $N=pq$ with primes $p\equiv 1(\mathrm{mod}12)$ and $q\equiv 5(\mathrm{mod}12)$. quadratic reciprocity makes three a square modulo $p$; choose $r^2=3$ and impose $a=-3+2r(\mathrm{mod}p)$, $a=1(\mathrm{mod}q)$ by the Chinese remainder theorem. The curve has good reduction at both primes. Indeed its discriminant is $-16a^2(4a+27)$; the first condition could make a factor vanish only at $p=3$ or 11, while the second gives $a(4a+27)=31$ and the given residue class excludes $q=31$.

Doubling $P$ gives $x(2P)-1=(a^2+6a-3)/4$. This vanishes modulo $p$ and equals one modulo $q$. Modulo $p$ one also has $y(2P)=-1$, so $2P=-P$ and $3P$ is the identity. Modulo $q$ the addition is finite. Thus computing $3P$ modulo $N$ encounters a denominator whose greatest common divisor with $N$ is exactly $p$, revealing the factor. The numerator difference is minus two modulo $p$, so the failed inverse cannot be cancelled. This is a construction of successful elliptic curve method data from prescribed local group behavior.