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Jun 18, 2012 at 5:17 comment added Noam D. Elkies $c$ is rational but not necessarily integral, so it's not just primes dividing $c(c^2-1$). Instead write $c = m/n$ and $|mn(m^2-n^2)| = r^2 d$ with $d \in \bf Z$ squarefree, and ask for the number of prime factors in $2d$. If I remember right the actual bound is something like the number of prime factors plus the number of prime factors congruent to $1 \bmod 4$.
Jun 17, 2012 at 20:11 history answered Remke Kloosterman CC BY-SA 3.0