User harald k&#252;mmerle - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T21:47:03Z http://mathoverflow.net/feeds/user/11405 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/48000/direct-proof-of-special-case-of-hasses-theorem-for-elliptic-curves/48625#48625 Answer by Harald Kümmerle for Direct proof of special case of Hasse's theorem for elliptic curves Harald Kümmerle 2010-12-08T11:27:49Z 2010-12-08T11:27:49Z <p>I am looking for elementary proofs of (special cases) of Hasse's theorem. The above discussion was of great help for me in the stated case, but does somebody know where to find a proof for $\binom{\frac{p-1}{2}}{\frac{p-1}{3}}$ being small mod p?</p> <p>This should correspond with Hasse for the curve $y^2=x^3+1$.</p> http://mathoverflow.net/questions/48000/direct-proof-of-special-case-of-hasses-theorem-for-elliptic-curves/48625#48625 Comment by Harald Kümmerle Harald Kümmerle 2010-12-16T02:53:28Z 2010-12-16T02:53:28Z Lemmermeyer uses Jacobi sums. Looking for proofs of the other binomial coefficient statement that was given by me and failing to do so, I checked the general techniques involved. Chapter 8 in Ireland-Rosen's book develops a theory that suffices to prove the estimates for the first binomial coefficient (Exercise 8.26), the other one should be tackled similarly, I think. But that is not necessary, as one can directly prove Hasse's Theorem for all $y^2 = x^3 + Dx$, $y^2 = x^3 + D$ for all $F_p$ with $p&gt;3$ and almost arbitary nonzero integers D, using Jacobi sums only! See Chapter 18, &#167;3 and \$4. http://mathoverflow.net/questions/48000/direct-proof-of-special-case-of-hasses-theorem-for-elliptic-curves/48625#48625 Comment by Harald Kümmerle Harald Kümmerle 2010-12-09T00:02:34Z 2010-12-09T00:02:34Z Lemmermeyer indeed cites some of Gauss's works for many other congruences that are very similar to the first one discussed here. But at the moment, I cannot access his collected works on the server of the University of G&#246;ttingen. I'll have to try again later.