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/ Testing: Build and run the Ed25519 tests: mkdir build cd build cmake -GNinja .. ninja && ./crypto/crypto_test --gtest_filter="Ed25519Test*" 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") # 2 is a non-square if legendre(2,p)==-1: print("2 is a non-square") # 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. # 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). 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!") Change-Id: Iaa284d3365dd8c9fa18a4584121013f05a3f4cc6 Reviewed-on: https://boringssl-review.googlesource.com/c/boringssl/+/50965 Reviewed-by: David Benjamin <davidben@google.com> Reviewed-by: Adam Langley <agl@google.com> Commit-Queue: Adam Langley <agl@google.com>fips-20220613
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