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Sep 19, 2014 at 1:49 comment added Greg Martin It's also worth noting that only finitely many primes (those dividing the discriminant of $f$) can simultaneously divide two of the $f_j(n)$. In fact, if one partitions the domain into suitable arithmetic progressions, one can avoid that possibility altogether - see Lemma 3.2 of math.ubc.ca/~gerg/index.shtml?abstract=AFNSVP
Aug 19, 2014 at 12:11 history edited Stanley Yao Xiao CC BY-SA 3.0
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Aug 19, 2014 at 11:28 history answered Stanley Yao Xiao CC BY-SA 3.0