Schaefer:  Can Sha(E)[5] be arb. large?
Yes, but... can't prove it.

  0 -> E(Q)/pE(Q) -> H^1(Qbar/Q,E[p]) -> H^1(Qbar/Q,E)[p] -> 0.

  0 -> E(Q)/pE(Q) -> S^p(Q,E) -> Sha(E)[p] -> 0.
                     finite       finite
                     computable   

Obstruction:  Sha(Q,E) 

Facts: Because of (alternating) CT pairing, if Sha is finite, then 
order must be a square.  Bolling, Cassels ==> Sha[2], Sha[3] arb. large.
Cathy: Rohlrich "Sha[3] arbitrarily large."

Look at BSD: 

(X(5) x X(5)) / Diagonal.



  0 -> E(Q)/5E(Q) -> S^5(Q,E) -> Sha(E)[5] -> 0.

  0 -> F(Q)/5F(Q) -> S^5(Q,F) -> Sha(F)[5] -> 0.


Theorem (Schaefer):

#S^{phi'}(Q,E)
--------------  = computable, easily.
#S^{phi}(Q,E')