  We prove that some natural "outside" property is equivalent (for a first
order class) to being stable.
  For a model, being resplendent is a strengthening of being kappa-saturated.
Restricting ourselves to the case kappa > |T| for transparency, a model M is
kappa-resplendent means: when we expand M by <kappa individual constants (c_i:i
< alpha), if (M, c_i)_ {i<alpha} has an elementary extension expandable to be a
model of T' where Th((M,c_i)_{i<alpha}) subseteq T', |T'| < kappa then already
(M,c_i)_{i<\alpha} can be expanded to a model of T'.
  Trivially any saturated model of cardinality lambda is lambda-resplendent. We
ask: how may kappa-resplendent models of a (first order complete) theory T of
cardinality lambda are there? Naturally we restrict ourselves to cardinals
lambda=lambda^ kappa + 2^ {|T|}. Then we get a complete and satisfying answer:
this depends just on T being stable or unstable. In this case proving that for
stable T we get few, is not hard; in fact every resplendent model of T is
saturated hence determined by its cardinality up to isomorphism. The inverse is
more problematic because naturally we have to use Skolem functions with any
alpha<kappa places. Normally we use relevant partition theorems (Ramsey theorem
or Erd\H{o}s-Rado theorem), but in our case the relevant partitions theorems
fails so we have to be careful.
