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Feb 1, 2014 at 14:54 comment added Lev Borisov In particular, all such $c$ would have to be algebraic.
Feb 1, 2014 at 13:32 comment added Joe Silverman @Mike Thank you, I'll fix that typo.
Feb 1, 2014 at 13:32 history edited Joe Silverman CC BY-SA 3.0
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Feb 1, 2014 at 13:32 comment added Joe Silverman @DavidLoeffler Well, each finite set will corrrespond to the finitely many $c$'s that satisfy $\sum a_n\sqrt{n^2+c^2}=0$.
Feb 1, 2014 at 13:06 comment added David Loeffler It it obvious that the same finite set $\{a_n\}$ can't work for more than one real $c$?
Feb 1, 2014 at 12:31 comment added mme Minor nitpick: your definition of $S_c$ doesn't have $c$ in it.
Feb 1, 2014 at 12:12 history answered Joe Silverman CC BY-SA 3.0