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Mar 13, 2015 at 18:09 vote accept R.P.
Mar 13, 2015 at 22:41
Oct 12, 2014 at 19:51 comment added Michael Stoll Stating the obvious: the mere existence of a function $f$ is trivial, just take $f(n)$ to be the (finite) number of solutions...
Jan 28, 2014 at 14:57 comment added GH from MO @joro: I see, then René has answered your question.
Jan 28, 2014 at 14:48 comment added joro @GHfromMO Thanks. I asked about the largest number of solutions known.
Jan 28, 2014 at 14:46 comment added GH from MO @René: I think joro asked what is known with proof.
Jan 28, 2014 at 14:46 comment added GH from MO @joro: I don't know what is the best current bound on $f(n)$.
Jan 28, 2014 at 11:17 comment added R.P. Euler noticed that $59^4+158^4=133^4+134^4$. This already gives $16$ solutions to $x^4+y^4=n$, where $n=635318657$ (apply the symmetries $(x,y)\mapsto(\pm x,\pm y)$ and $(x,y)\mapsto(y,x)$). According to Hardy and Wright then, this must be conjectured as being the record number of solutions for any $n$.
Jan 28, 2014 at 5:49 comment added joro @GHfromMO what is the current record for number of solutions?
Jan 27, 2014 at 23:12 history edited R.P. CC BY-SA 3.0
deleted 2 characters in body
Jan 27, 2014 at 23:03 history edited R.P. CC BY-SA 3.0
added 101 characters in body
Jan 27, 2014 at 19:09 comment added GH from MO We certainly have $f(n)\ll_\epsilon n^\epsilon$ for any $\epsilon>0$, and we should have $f(n)\leq 16$ but this is widely open. See my comments to ACL's response.
Jan 27, 2014 at 19:05 comment added Qiaochu Yuan Your question is stated slightly confusingly - your friend asks for a bound depending on $n$ but you ask for a bound independent of $n$, and perhaps this is what confused ACL.
Jan 27, 2014 at 15:50 vote accept R.P.
Jan 27, 2014 at 16:12
Jan 27, 2014 at 15:46 answer added ACL timeline score: 6
Jan 27, 2014 at 15:45 history edited R.P. CC BY-SA 3.0
in description of the main question, x,y should be integers
Jan 27, 2014 at 15:35 history asked R.P. CC BY-SA 3.0