Simpler square-root computation for Ed25519

Simpler square-root computation for Ed25519

Description:
Mark Wooden and Franck Rondepierre noted that the square-root-mod-p
operations used in the EdDSA RFC (RFC 8032) can be simplified.  For
Ed25519, instead of computing u*v^3 * (u * v^7)^((p-5)/8), we can
compute u * (u*v)^((p-5)/8).  This saves 3 multiplications and 2
squarings.  For more details (including a proof), see the following
message from the CFRG mailing list:

  https://mailarchive.ietf.org/arch/msg/cfrg/qlKpMBqxXZYmDpXXIx6LO3Oznv4/

Note that the Ed448 implementation (see
ossl_curve448_point_decode_like_eddsa_and_mul_by_ratio() in
./crypto/ec/curve448/curve448.c) appears to already use this simpler
method (i.e. it does not follow the method suggested in RFC 8032).

Testing:
Build and then run the test suite:

  ./Configure -Werror --strict-warnings
  make update
  make
  make test

Numerical testing of the square-root computation can be done using the
following sage script:

  def legendre(x,p):
      return kronecker(x,p)

  # Ed25519
  p = 2**255-19
  # -1 is a square
  if legendre(-1,p)==1:
      print("-1 is a square")

  # suppose u/v is a square.
  # to compute one of its square roots, find x such that
  #    x**4 == (u/v)**2 .
  # this implies
  #    x**2 ==  u/v, or
  #    x**2 == -(u/v) ,
  # which implies either x or i*x is a square-root of u/v (where i is a square root of -1).
  # we can take x equal to u * (u*v)**((p-5)/8).

  # 2 is a generator
  # this can be checked by factoring p-1
  # and then showing 2**((p-1)/q) != 1 (mod p)
  # for all primes q dividing p-1.
  g = 2
  s = p>>2  # s = (p-1)/4
  i = power_mod(g, s, p)

  t = p>>3  # t = (p-5)/8
  COUNT = 1<<18
  while COUNT > 0:
      COUNT -= 1

      r = randint(0,p-1)   # r = u/v
      v = randint(1,p-1)
      u = mod(r*v,p)

      # compute x = u * (u*v)**((p-5)/8)
      w = mod(u*v,p)
      x = mod(u*power_mod(w, t, p), p)

      # check that x**2 == r, or (i*x)**2 == r, or r is not a square
      rr = power_mod(x, 2, p)
      if rr==r:
          continue

      rr = power_mod(mod(i*x,p), 2, p)
      if rr==r:
          continue

      if legendre(r,p) != 1:
          continue

      print("failure!")
      exit()

  print("passed!")

Reviewed-by: Paul Dale <pauli@openssl.org>
Reviewed-by: Tomas Mraz <tomas@openssl.org>
(Merged from https://github.com/openssl/openssl/pull/17544)