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>
James Muir
0fc57bef18