Timeline for Rank of $x (x^2 - 1) = c (c^2 - 1) y^2 $ over $\mathbb{Q}$ for given rational values of $c$
Current License: CC BY-SA 3.0
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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 |