ECC_study

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用于汇总椭圆曲线加密算法的相关题目

[watevrCTF 2019]ECC-RSA

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from fastecdsa.curve import P521 as Curve
from fastecdsa.point import Point
from Crypto.Util.number import bytes_to_long, isPrime
from os import urandom
from random import getrandbits

def gen_rsa_primes(G):
	urand = bytes_to_long(urandom(521//8))
	while True:
		s = getrandbits(521) ^ urand

		Q = s*G
		if isPrime(Q.x) and isPrime(Q.y):
			print("ECC Private key:", hex(s))
			print("RSA primes:", hex(Q.x), hex(Q.y))
			print("Modulo:", hex(Q.x * Q.y))
			return (Q.x, Q.y)


flag = int.from_bytes(input(), byteorder="big")

ecc_p = Curve.p
a = Curve.a
b = Curve.b

Gx = Curve.gx
Gy = Curve.gy
G = Point(Gx, Gy, curve=Curve)


e = 0x10001
p, q = gen_rsa_primes(G)
n = p*q


file_out = open("downloads/ecc-rsa.txt", "w")

file_out.write("ECC Curve Prime: " + hex(ecc_p) + "\n")
file_out.write("Curve a: " + hex(a) + "\n")
file_out.write("Curve b: " + hex(b) + "\n")
file_out.write("Gx: " + hex(Gx) + "\n")
file_out.write("Gy: " + hex(Gy) + "\n")

file_out.write("e: " + hex(e) + "\n")
file_out.write("p * q: " + hex(n) + "\n")

c = pow(flag, e, n)
file_out.write("ciphertext: " + hex(c) + "\n")
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ECC Curve Prime: 0x1ffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff
Curve a: -0x3
Curve b: 0x51953eb9618e1c9a1f929a21a0b68540eea2da725b99b315f3b8b489918ef109e156193951ec7e937b1652c0bd3bb1bf073573df883d2c34f1ef451fd46b503f00
Gx: 0xc6858e06b70404e9cd9e3ecb662395b4429c648139053fb521f828af606b4d3dbaa14b5e77efe75928fe1dc127a2ffa8de3348b3c1856a429bf97e7e31c2e5bd66
Gy: 0x11839296a789a3bc0045c8a5fb42c7d1bd998f54449579b446817afbd17273e662c97ee72995ef42640c550b9013fad0761353c7086a272c24088be94769fd16650
e: 0x10001
p * q: 0x118aaa1add80bdd0a1788b375e6b04426c50bb3f9cae0b173b382e3723fc858ce7932fb499cd92f5f675d4a2b05d2c575fc685f6cf08a490d6c6a8a6741e8be4572adfcba233da791ccc0aee033677b72788d57004a776909f6d699a0164af514728431b5aed704b289719f09d591f5c1f9d2ed36a58448a9d57567bd232702e9b28f
ciphertext: 0x3862c872480bdd067c0c68cfee4527a063166620c97cca4c99baff6eb0cf5d42421b8f8d8300df5f8c7663adb5d21b47c8cb4ca5aab892006d7d44a1c5b5f5242d88c6e325064adf9b969c7dfc52a034495fe67b5424e1678ca4332d59225855b7a9cb42db2b1db95a90ab6834395397e305078c5baff78c4b7252d7966365afed9e

函数gen_rsa_primes()通过椭圆曲线的参数x,y来生成的RSA中的p,q，其中x,y满足

已知 故有

使用sage解这个方程

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a=-0x3
b=0x51953eb9618e1c9a1f929a21a0b68540eea2da725b99b315f3b8b489918ef109e156193951ec7e937b1652c0bd3bb1bf073573df883d2c34f1ef451fd46b503f00
n=0x118aaa1add80bdd0a1788b375e6b04426c50bb3f9cae0b173b382e3723fc858ce7932fb499cd92f5f675d4a2b05d2c575fc685f6cf08a490d6c6a8a6741e8be4572adfcba233da791ccc0aee033677b72788d57004a776909f6d699a0164af514728431b5aed704b289719f09d591f5c1f9d2ed36a58448a9d57567bd232702e9b28f
p0=0x1ffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff
R.<x>=Zmod(p0)[]
f=x**5+a*x**3+b*x**2-n**2
print( f.roots() )

得到

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[(6813140671672694477701511883397067876211159809088064490593325584756562268820329988116480298456252746748095410666300132267213094431909630229631434972416225885, 1), (4573744216059593260686660411936793507327994800883645562370166075007970317346237399760397301505506131100113886281839847419425482918932436139080837246914736557, 1), (1859314969084523636298100850823722544590555574470838518640063093117116629078281861281849586432508721074855657736668366212762253040197962779753163192386773060, 1)]

使用质数检测器容易知道，这三个数字里只有4573744216059593260686660411936793507327994800883645562370166075007970317346237399760397301505506131100113886281839847419425482918932436139080837246914736557是素数

得到 p 即可正常解RSA

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from Crypto.Util.number import long_to_bytes
import gmpy2

n = 0x118aaa1add80bdd0a1788b375e6b04426c50bb3f9cae0b173b382e3723fc858ce7932fb499cd92f5f675d4a2b05d2c575fc685f6cf08a490d6c6a8a6741e8be4572adfcba233da791ccc0aee033677b72788d57004a776909f6d699a0164af514728431b5aed704b289719f09d591f5c1f9d2ed36a58448a9d57567bd232702e9b28f
c = 0x3862c872480bdd067c0c68cfee4527a063166620c97cca4c99baff6eb0cf5d42421b8f8d8300df5f8c7663adb5d21b47c8cb4ca5aab892006d7d44a1c5b5f5242d88c6e325064adf9b969c7dfc52a034495fe67b5424e1678ca4332d59225855b7a9cb42db2b1db95a90ab6834395397e305078c5baff78c4b7252d7966365afed9e
p = 4573744216059593260686660411936793507327994800883645562370166075007970317346237399760397301505506131100113886281839847419425482918932436139080837246914736557
e = 0x10001
q  = n//p

phi = (p-1)*(q-1)
d = gmpy2.invert(e,phi)
m = pow(c,d,n)
print(long_to_bytes(m))

flag:watevr{factoring_polynomials_over_finite_fields_is_too_ez}