RSA 加密及一些攻击方式

本文章转载自个人博客seandictionary.top同步更新可能不及时

原理

随机生成两个素数，p , q

令n = p*q

由欧拉公式计算出φ(n) = (p-1)(q-1)

规定e，使得e满足1<e<φ(n)，且gcd(e,φ(n)) = 1，一般e=65537或0x10001

此时就有了公钥=(e,n)

计算私钥

计算d，使得d满足ed≡1mod φ(n)，即称d是e在模φ(n)下的逆元

得到私钥=(d,n)

有关欧拉公式

• $n=p^{e_1}_1p^{e_2}_2\dots p^{e_k}_k$
• $\varphi(n)=n(1-\frac{1}{p_1})(1-\frac{1}{p_2})\dots (1-\frac{1}{p_k})$
• $\varphi(n)=p^{e_1-1}_1p^{e_2-1}_2\dots p^{e_k-1}_k(p_1-1)(p_2-1)\dots (p_k-1)$
• 当$gcd(n,x)=1$时，满足$x^{\varphi(n)}\equiv 1\ mod\ n$

RSA加密和解密

规定m表示明文，c表示密文

加密：c≡m^e mod n

解密：m≡c^d mod n

解密方式

正常解密

• 此方法适用于已知$p,q,e,c$的情况下

可以使用RSA Tool 2工具解密或者编写如下脚本

from Crypto.Util.number import *
p = ?
q = ?
e = ?
c = ?
n = p*q
d = inverse(e,(p-1) * (q-1))
m = pow(c,d,n)
print(long_to_bytes(m))

dp,dq泄露

• 此方法使用于已知$p,q,d_{p},d_{q},c$的情况下
• 然而使用此方法前可以先尝试常用e值，在发现无法确定e值的情况下再使用

实例BUUCTF-Crypto-21.RSA1

理论推导

• dp，dq泄露
• 理论推导 红色字 为重要结论

$$\because m \equiv c^d(mod\ n)$$

$$\therefore m = c^d+kn = c^d+k\cdot pq$$

$$m_{1} \equiv c^d(mod\ p)\ ①\ ,\ m_{2} \equiv c^d(mod\ q)\ ②$$

$$①\Rightarrow c^d=m_{1}+tp$$

$$代入②\Rightarrow m_{2}\equiv m_{1}+tp(mod\ q)$$

$$\therefore m_{2}=m_{1}+t\cdot p+r\cdot q\ \Rightarrow m_{2}-m_{1}=t\cdot p+r\cdot q$$

$$\therefore m_{2}-m_{1}=t\cdot p(mod\ q)$$

$$\therefore (m_{2}-m_{1})\cdot p^{-1}\equiv t(mod\ q)$$

$$\Rightarrow t=(m_{2}-m_{1})\cdot p^{-1}(mod\ q)$$

$$\therefore c^d=[(m_{2}-m_{1})\cdot p^{-1}(mod\ q)]p+m_{1}$$

$$\because m\equiv c^d(mod\ n)$$

$$\therefore {\color{Red} m\equiv [[(m_{2}-m_{1})\cdot p^{-1}(mod\ q)]p+m_{1}]mod\ n}$$

$$d\equiv d_{p}mod(p-1)\ , \ d\equiv d_{q}mod(q-1)$$

$$m_{1} \equiv c^dmod\ p\ ,\ m_{2} \equiv c^dmod\ q$$

$$\Rightarrow m_{1} \equiv c^\left .d_{p}\ +k_{1}\cdot (p-1)\right.\ mod\ p\ ,\ m_{2} \equiv c^\left .d_{q}\ +k_{2}\cdot (q-1)\right.\ mod\ p$$

$$\because c^\left.p-1\right.\equiv 1\ mod\ p$$

$$\Rightarrow {\color{Red} m_{1}\equiv c^\left .d_{p}\right . mod\ p\ ,\ m_{2}\equiv c^\left .d_{q}\right . mod \ q}$$

使用脚本解密

from Crypto.Util.number import *

p = ?
q = ?
dp = ?
dq = ?
c = ?

mp = pow(c,dp,p)
mq = pow(c,dq,q)

pi = inverse(p,q)

m = ((((mq-mp)pi)%q)p+mp)%(p*q)

print(long_to_bytes(m))

dp泄露

• 此方法使用于已知$e,n,d_{p},c$的情况下

实例BUUCTF-Crypto-27.RSA2

理论推导

• dp泄露
• 理论推导 红色字 为重要结论

$$d_{p} \equiv d mod (p-1)$$

$$\Rightarrow d\cdot e = k_{1}(p-1)+d_{p}\cdot e$$

$$\because d\cdot e = 1mod(p-1)(q-1) \Rightarrow d\cdot e=k_{2}(p-1)(q-1)+1$$

$$\therefore {\color{Red} (p-1)(k_{2}(q-1)-k_{1})+1 = d_{p}e}$$

$$\because d_{p}<d$$

$$\therefore (k_{2}(q-1)-k_{1})=x\in (1,e)$$

from Crypto.Util.number import *
e = 65537
n = ?
dp = ?
c = ?
a = dp*e-1
for x in range(2,e):
    if a%x == 0:
        p = a//x+1
        if n%p == 0:
            q = n//p
            break
d = inverse(e,(p-1)*(q-1))
m = pow(c,d,n)
print(long_to_bytes(m))

共模攻击

• 此方法使用于已知$n,e_{1},e_{2},c_{1},c_{2}$的情况下
• 且$gcd(e_{1},e_{2})=1$

实例BUUCTF-Crypto-28.RSA3

理论推导

• 共模攻击
• 理论推导 红色字 为重要结论
• 存在两种密钥对同一明文进行加密
• 下述为针对此题特殊情况下的特殊推导，需满足$gcd(e_{1},e_{2})=1$

$$c_{1}=m^\left. e_{1}\right. mod\ n\ \&\ m=c_{1}^\left.d_{1}\right.mod\ n$$

$$c_{2}=m^\left. e_{2}\right. mod\ n\ \&\ m=c_{2}^\left.d_{2}\right.mod\ n$$

$$构造一对(s_{1},s_{2})满足e_{1}s_{1}+e_{2}s_{2}=1其中s_{1},s_{2}\in Z,s_{1}>0,s_{2}<0$$

$$c_{1}=m^\left. e_{1}\right. mod\ n \Rightarrow c_{1}^\left. s_{1}\right. =m^\left. e_{1}s_{1}\right. mod\ n ①$$

$$c_{2}=m^\left. e_{2}\right. mod\ n \Rightarrow c_{2}^\left. s_{2}\right. =m^\left. e_{2}s_{2}\right. mod\ n ②$$

$$①\times ②\Rightarrow c_{1}^\left. s_{1}\right. c_{2}^\left. s_{2}\right. mod\ n = m^\left. e_{1}s_{1}+e_{2}s_{2}\right. mod\ n$$

$$\Rightarrow c_{1}^\left. s_{1}\right. c_{2}^\left. s_{2}\right. mod\ n=m\ mod\ n$$

$$\because m=c^d mod\ n \therefore m<n$$

$$\therefore {\color{red}m=c_{1}^\left. s_{1}\right. c_{2}^\left. s_{2}\right. mod\ n}$$

$$需要注意的是推论到上面就结束了，但是实际计算中上述式子计算过于复杂，所以应当使用下述式子$$

$${\color{red}m=(c_{1}^\left. s_{1}\right. mod\ n\cdot c_{2}^\left. s_{2}\right. mod\ n)mod\ n}$$

$$以及在计算s_{1},s_{2}过程中可以采用逆元的方式$$

$$(e_{1}-e{2})s_{1}+e{2}(s_{1}+s_{2})=1$$

$$(e_{1}-e{2})s_{1}\equiv 1\ mod\ e_{2}$$

$$s_{1}=(e_{1}-e{2})^\left. -1\right. $$

• 由此计算出一对$(s_{1},s_{2})$，进而算出明文m
• 编写脚本

from Crypto.Util.number import *
n = ?
c1 = ?
c2 = ?
e1 = ?
e2 = ?
e1_e2 = e1-e2
s1 = inverse(e1_e2,e2)
s2 = (1-e1*s1)//e2
m = pow(c1,s1,n)*pow(c2,s2,n)%n
print(long_to_bytes(m))

维纳攻击

• 此方法使用于$e$过大或过小的情况下
• 本质：满足$d < \frac{1}{3} N^{\frac{1}{4} }$则一定可以分解$N$
• 大致解密过程是用RSAwienerHacker库来攻击得到d
• 原理太复杂了要翻论文了-链接
• 板子

实例BUUCTF-Crypto-40.rsa2

from RSAwienerHacker import *
n = ?
e = ?
d = hack_RSA(e,n)

扩展维纳攻击

参考 CTF wiki

两个低解密指数

from Crypto.Util.number import *

e1 = ?
e2 = ?
N = ?
a = 5/14
D = diagonal_matrix(ZZ, [N, int(N^(1/2)), int(N^(1+a)), 1])
M = matrix(ZZ, [[1, -N, 0, N^2], [0, e1, -e1, -e1*N], [0, 0, e2, -e2*N], [0, 0, 0, e1*e2]])*D
L = M.LLL()
t = vector(ZZ, L[0])
x = t * M^(-1)
phi = int(x[1]/x[0]*e1)
d = inverse(e,phi)

三个低解密指数

e1 = ?
e2 = ?
e3 = ?
n  = ?
a = 2/5
D = diagonal_matrix(ZZ,[int(n^1.5), n, int(n^(a+1.5)), int(n^0.5), int(n^(a+1.5)), int(n^(a+1)), int(n^(a+1)), 1])
L = Matrix(ZZ,[[1, -n,   0,   n^2,   0,      0,      0,     -n^3],
               [0, e1, -e1, -n*e1, -e1,      0,   n*e1,   n^2*e1],
               [0,  0,  e2, -n*e2,   0,   n*e2,      0,   n^2*e2],
               [0,  0,   0, e1*e2,   0, -e1*e2, -e1*e2, -n*e1*e2],
               [0,  0,   0,     0,  e3,  -n*e3,  -n*e3,   n^2*e3],
               [0,  0,   0,     0,   0,  e1*e3,      0, -n*e1*e3],
               [0,  0,   0,     0,   0,      0,  e2*e3, -n*e2*e3],
               [0,  0,   0,     0,   0,      0,      0, e1*e2*e3]]) * D
Ge = L.LLL()[0]
x = vector(ZZ, Ge) / L
phi = int(x[1]/x[0]*e1)
d = inverse(e,phi)

通用情况

from Crypto.Util.number import *
isdigit = lambda x: ord('0') <= ord(x) <= ord('9')

def my_permutations(g, n):
    sub = []
    res = []
    def dfs(s, prev):
        if len(s) == n:
            res.append(s[::])
        for i in g:
            if i in s or i < prev:
                continue
            s.append(i)
            dfs(s, max(prev, i))
            s.remove(i)
    dfs(sub, 0)
    return res

class X3NNY(object):
    def __init__(self, exp1, exp2):
        self.exp1 = exp1
        self.exp2 = exp2

    def __mul__(self, b):
        return X3NNY(self.exp1 * b.exp1, self.exp2 * b.exp2)

    def __repr__(self):
        return '%s = %s' % (self.exp1.expand().collect_common_factors(), self.exp2)

class X_Complex(object):
    def __init__(self, exp):
        i = 0
        s = '%s' % exp
        while i < len(s):
            if isdigit(s[i]):
                num = 0
                while i < len(s) and isdigit(s[i]):
                    num = num*10 + int(s[i])
                    i += 1
                if i >= len(s):
                    self.b = num
                elif s[i] == '*':
                    self.a = num
                    i += 2
                elif s[i] == '/':
                    i += 1
                    r = 0
                    while i < len(s) and isdigit(s[i]):
                        r = r*10 + int(s[i])
                        i += 1
                    self.b = num/r
            else:
                i += 1
        if not hasattr(self, 'a'):
            self.a = 1
        if not hasattr(self, 'b'):
            self.b = 0

def WW(e, d, k, g, N, s):
    return X3NNY(e*d*g-k*N, g+k*s)
def GG(e1, e2, d1, d2, k1, k2):
    return X3NNY(e1*d1*k2- e2*d2*k1, k2 - k1)

def W(i):
    e = eval("e%d" % i)
    d = eval("d%d" % i)
    k = eval("k%d" % i)
    return WW(e, d, k, g, N, s)

def G(i, j):
    e1 = eval("e%d" % i)
    d1 = eval("d%d" % i)
    k1 = eval("k%d" % i)

    e2 = eval("e%d" % j)
    d2 = eval("d%d" % j)
    k2 = eval("k%d" % j)

    return GG(e1, e2, d1, d2, k1, k2)

def R(e, sn): # min u max v
    ret = X3NNY(1, 1)
    n = max(e)
    nn = len(e)
    l = set(i for i in range(1, n+1))
    debug = ''
    u, v = 0, 0
    for i in e:
        if i == 1:
            ret *= W(1)
            debug += 'W(%d)' % i
            nn -= 1
            l.remove(1)
            u += 1
        elif i > min(l) and len(l) >= 2*nn:
            ret *= G(min(l), i)
            nn -= 1
            debug += 'G(%d, %d)' % (min(l), i)
            l.remove(min(l))
            l.remove(i)
            v += 1
        else:
            ret *= W(i)
            l.remove(i)
            debug += 'W(%d)' % i
            nn -= 1
            u += 1
    # print(debug, end = ' ')
    return ret, u/2 + (sn - v) * a

def H(n):
    if n == 0:
        return [0]
    if n == 2:
        return [(), (1,), (2,), (1, 2)]
    ret = []
    for i in range(3, n+1):
        ret.append((i,))
        for j in range(1, i):
            for k in my_permutations(range(1, i), j):
                ret.append(tuple(k + [i]))
    return H(2) + ret

def CC(exp, n):
    cols = [0 for i in range(1<<n)]

    # split exp
    texps = ('%s' % exp.exp1.expand()).strip().split(' - ')
    ops = []
    exps = []
    for i in range(len(texps)):
        if texps[i].find(' + ') != -1:
            tmp = texps[i].split(' + ')
            ops.append(0)
            exps.append(tmp[0])
            for i in range(1, len(tmp)):
                ops.append(1)
                exps.append(tmp[i])
        else:
            ops.append(0)
            exps.append(texps[i])
    if exps[0][0] == '-':
        for i in range(len(exps)):
            ops[i] = 1-ops[i]
        exps[0] = exps[0][1:]
    else:
        ops[0] = 1
    # find e and N
    l = []
    for i in range(len(exps)):
        tmp = 1 if ops[i] else -1
        en = []
        j = 0
        while j < len(exps[i]):
            if exps[i][j] == 'e':
                num = 0
                j += 1
                while isdigit(exps[i][j]):
                    num = num*10 + int(exps[i][j])
                    j += 1
                tmp *= eval('e%d' % num)
                en.append(num)
            elif exps[i][j] == 'N':
                j += 1
                num = 0
                if exps[i][j] == '^':
                    j += 1
                    while isdigit(exps[i][j]):
                        num = num*10 + int(exps[i][j])
                        j += 1
                if num == 0:
                    num = 1
                tmp *= eval('N**%d' % num)
            else:
                j += 1
        if tmp == 1 or tmp == -1:
            l.append((0, ()))
        else:
            l.append((tmp, tuple(sorted(en))))

    # construct h
    mp = H(n)
    for val, en in l:
        cols[mp.index(en)] = val
    # print(cols)
    return cols

def EWA(n, elist, NN, alpha):
    mp = H(n)
    var('a')
    S = [X_Complex(n*a)]
    cols = [[1 if i == 0 else 0 for i in range(2^n)]]
    for i in mp[1:]:
        eL, s = R(i, n)
        cols.append(CC(eL, n))
        S.append(X_Complex(s))

    alphaA,alphaB = 0, 0
    for i in S:
        alphaA = max(i.a, alphaA)
        alphaB = max(i.b, alphaB)
    # print(alphaA, alphaB)
    D = []
    for i in range(len(S)):
        # print((alphaA-S[i].a), (alphaB - S[i].b))
        D.append(
            int(NN^((alphaA-S[i].a)*alpha + (alphaB - S[i].b)))
        )
    kw = {'N': NN}
    for i in range(len(elist)):
        kw['e%d' % (i+1)] = elist[i]

    B = Matrix(ZZ, Matrix(cols).T(**kw)) * diagonal_matrix(ZZ, D)
    L = B.LLL(0.5)
    v = Matrix(ZZ, L[0])
    x = v * B**(-1)
    phi = int(x[0,1]/x[0,0]*elist[0])
    return phi

def attack(NN, elist, alpha):
    phi = EWA(len(elist), elist, NN, alpha)
    print(phi)
    return phi

n = ?   # n是d的比特位数
NN = ?
elist = [e1,e2,e3, ...,en]

alpha = n / int(NN).bit_length()
for i in range(1, len(elist)+1):
    var("e%d" % i)
    var("d%d" % i)
    var("k%d" % i)
g, N, s = var('g'), var('N'), var('s')

for i in range(len(elist)):
    elist[i] = Integer(elist[i])
phi = attack(NN, elist, alpha)
d = inverse(e, phi)

Boneh Durfee攻击

• 这是维纳攻击的一种延伸，拓展了可以攻击的d的范围
• $d<n^{0.292}$

GitHub仓库，在此基础上修改

import time
from Crypto.Util.number import *

"""
Setting debug to true will display more informations
about the lattice, the bounds, the vectors...
"""
debug = True

"""
Setting strict to true will stop the algorithm (and
return (-1, -1)) if we don't have a correct
upperbound on the determinant. Note that this
doesn't necesseraly mean that no solutions
will be found since the theoretical upperbound is
usualy far away from actual results. That is why
you should probably use `strict = False`
"""
strict = False

"""
This is experimental, but has provided remarkable results
so far. It tries to reduce the lattice as much as it can
while keeping its efficiency. I see no reason not to use
this option, but if things don't work, you should try
disabling it
"""
helpful_only = True
dimension_min = 7 # stop removing if lattice reaches that dimension

############################################
# Functions
##########################################

# display stats on helpful vectors
def helpful_vectors(BB, modulus):
    nothelpful = 0
    for ii in range(BB.dimensions()[0]):
        if BB[ii,ii] >= modulus:
            nothelpful += 1

    print (nothelpful, "/", BB.dimensions()[0], " vectors are not helpful")

# display matrix picture with 0 and X
def matrix_overview(BB, bound):
    for ii in range(BB.dimensions()[0]):
        a = ('%02d ' % ii)
        for jj in range(BB.dimensions()[1]):
            a += '0' if BB[ii,jj] == 0 else 'X'
            if BB.dimensions()[0] < 60:
                a += ' '
        if BB[ii, ii] >= bound:
            a += '~'
        print (a)

# tries to remove unhelpful vectors
# we start at current = n-1 (last vector)
def remove_unhelpful(BB, monomials, bound, current):
    # end of our recursive function
    if current == -1 or BB.dimensions()[0] <= dimension_min:
        return BB

    # we start by checking from the end
    for ii in range(current, -1, -1):
        # if it is unhelpful:
        if BB[ii, ii] >= bound:
            affected_vectors = 0
            affected_vector_index = 0
            # let's check if it affects other vectors
            for jj in range(ii + 1, BB.dimensions()[0]):
                # if another vector is affected:
                # we increase the count
                if BB[jj, ii] != 0:
                    affected_vectors += 1
                    affected_vector_index = jj

            # level:0
            # if no other vectors end up affected
            # we remove it
            if affected_vectors == 0:
                print ("* removing unhelpful vector", ii)
                BB = BB.delete_columns([ii])
                BB = BB.delete_rows([ii])
                monomials.pop(ii)
                BB = remove_unhelpful(BB, monomials, bound, ii-1)
                return BB

            # level:1
            # if just one was affected we check
            # if it is affecting someone else
            elif affected_vectors == 1:
                affected_deeper = True
                for kk in range(affected_vector_index + 1, BB.dimensions()[0]):
                    # if it is affecting even one vector
                    # we give up on this one
                    if BB[kk, affected_vector_index] != 0:
                        affected_deeper = False
                # remove both it if no other vector was affected and
                # this helpful vector is not helpful enough
                # compared to our unhelpful one
                if affected_deeper and abs(bound - BB[affected_vector_index, affected_vector_index]) < abs(bound - BB[ii, ii]):
                    print ("* removing unhelpful vectors", ii, "and", affected_vector_index)
                    BB = BB.delete_columns([affected_vector_index, ii])
                    BB = BB.delete_rows([affected_vector_index, ii])
                    monomials.pop(affected_vector_index)
                    monomials.pop(ii)
                    BB = remove_unhelpful(BB, monomials, bound, ii-1)
                    return BB
    # nothing happened
    return BB

"""
Returns:
* 0,0   if it fails
* -1,-1 if `strict=true`, and determinant doesn't bound
* x0,y0 the solutions of `pol`
"""
def boneh_durfee(pol, modulus, mm, tt, XX, YY):
    """
    Boneh and Durfee revisited by Herrmann and May

    finds a solution if:
    * d < N^delta
    * |x| < e^delta
    * |y| < e^0.5
    whenever delta < 1 - sqrt(2)/2 ~ 0.292
    """

    # substitution (Herrman and May)
    PR.<u, x, y> = PolynomialRing(ZZ)
    Q = PR.quotient(x*y + 1 - u) # u = xy + 1
    polZ = Q(pol).lift()

    UU = XX*YY + 1

    # x-shifts
    gg = []
    for kk in range(mm + 1):
        for ii in range(mm - kk + 1):
            xshift = x^ii * modulus^(mm - kk) * polZ(u, x, y)^kk
            gg.append(xshift)
    gg.sort()

    # x-shifts list of monomials
    monomials = []
    for polynomial in gg:
        for monomial in polynomial.monomials():
            if monomial not in monomials:
                monomials.append(monomial)
    monomials.sort()

    # y-shifts (selected by Herrman and May)
    for jj in range(1, tt + 1):
        for kk in range(floor(mm/tt) * jj, mm + 1):
            yshift = y^jj * polZ(u, x, y)^kk * modulus^(mm - kk)
            yshift = Q(yshift).lift()
            gg.append(yshift) # substitution

    # y-shifts list of monomials
    for jj in range(1, tt + 1):
        for kk in range(floor(mm/tt) * jj, mm + 1):
            monomials.append(u^kk * y^jj)

    # construct lattice B
    nn = len(monomials)
    BB = Matrix(ZZ, nn)
    for ii in range(nn):
        BB[ii, 0] = gg[ii](0, 0, 0)
        for jj in range(1, ii + 1):
            if monomials[jj] in gg[ii].monomials():
                BB[ii, jj] = gg[ii].monomial_coefficient(monomials[jj]) * monomials[jj](UU,XX,YY)

    # Prototype to reduce the lattice
    if helpful_only:
        # automatically remove
        BB = remove_unhelpful(BB, monomials, modulus^mm, nn-1)
        # reset dimension
        nn = BB.dimensions()[0]
        if nn == 0:
            print ("failure")
            return 0,0

    # check if vectors are helpful
    if debug:
        helpful_vectors(BB, modulus^mm)

    # check if determinant is correctly bounded
    det = BB.det()
    bound = modulus^(mm*nn)
    if det >= bound:
        print ("We do not have det < bound. Solutions might not be found.")
        print ("Try with highers m and t.")
        if debug:
            diff = (log(det) - log(bound)) / log(2)
            print ("size det(L) - size e^(m*n) = ", floor(diff))
        if strict:
            return -1, -1
    else:
        print ("det(L) < e^(m*n) (good! If a solution exists < N^delta, it will be found)")

    # display the lattice basis
    if debug:
        matrix_overview(BB, modulus^mm)

    # LLL
    if debug:
        print ("optimizing basis of the lattice via LLL, this can take a long time")

    BB = BB.LLL()

    if debug:
        print ("LLL is done!")

    # transform vector i & j -> polynomials 1 & 2
    if debug:
        print ("looking for independent vectors in the lattice")
    found_polynomials = False

    for pol1_idx in range(nn - 1):
        for pol2_idx in range(pol1_idx + 1, nn):
            # for i and j, create the two polynomials
            PR.<w,z> = PolynomialRing(ZZ)
            pol1 = pol2 = 0
            for jj in range(nn):
                pol1 += monomials[jj](w*z+1,w,z) * BB[pol1_idx, jj] / monomials[jj](UU,XX,YY)
                pol2 += monomials[jj](w*z+1,w,z) * BB[pol2_idx, jj] / monomials[jj](UU,XX,YY)

            # resultant
            PR.<q> = PolynomialRing(ZZ)
            rr = pol1.resultant(pol2)

            # are these good polynomials?
            if rr.is_zero() or rr.monomials() == [1]:
                continue
            else:
                print ("found them, using vectors", pol1_idx, "and", pol2_idx)
                found_polynomials = True
                break
        if found_polynomials:
            break

    if not found_polynomials:
        print ("no independant vectors could be found. This should very rarely happen...")
        return 0, 0

    rr = rr(q, q)

    # solutions
    soly = rr.roots()

    if len(soly) == 0:
        print ("Your prediction (delta) is too small")
        return 0, 0

    soly = soly[0][0]
    ss = pol1(q, soly)
    solx = ss.roots()[0][0]

    #
    return solx, soly

def example(N,e,delta):
    ############################################
    # How To Use This Script
    ##########################################

    #
    # Lattice (tweak those values)
    #

    # you should tweak this (after a first run), (e.g. increment it until a solution is found)
    m = 4 # size of the lattice (bigger the better/slower)

    # you need to be a lattice master to tweak these
    t = int((1-2*delta) * m)  # optimization from Herrmann and May
    X = 2*floor(N^delta)  # this _might_ be too much
    Y = floor(N^(1/2))    # correct if p, q are ~ same size

    #
    # Don't touch anything below
    #

    # Problem put in equation
    P.<x,y> = PolynomialRing(ZZ)
    A = int((N+1)/2)
    pol = 1 + x * (A + y)

    #
    # Find the solutions!
    #

    # Checking bounds
    if debug:
        print ("=== checking values ===")
        print ("* delta:", delta)
        print ("* delta < 0.292", delta < 0.292)
        print ("* size of e:", int(log(e)/log(2)))
        print ("* size of N:", int(log(N)/log(2)))
        print ("* m:", m, ", t:", t)

    # boneh_durfee
    if debug:
        print ("=== running algorithm ===")
        start_time = time.time()

    solx, soly = boneh_durfee(pol, e, m, t, X, Y)

    # found a solution?
    if solx > 0:
        print ("=== solution found ===")
        if False:
            print ("x:", solx)
            print ("y:", soly)

        d = int(pol(solx, soly) / e)
        print ("private key found:", d)
    else:
        print ("=== no solution was found ===")

    if debug:
        print("=== %s seconds ===" % (time.time() - start_time))
    return d

if __name__ == "__main__":
    n = ?
    e = ?
    c = ?
    # the hypothesis on the private exponent (the theoretical maximum is 0.292)
    delta = 0.28 # this means that d < N^delta
    d = example(n,e,delta)
    print(long_to_bytes(int(pow(c,d,n))))

e和φ不互质

1.二次剩余+CRT

• 此方法适用于$e$非质，即$gcd(e,\varphi )=2$
• 要求解任意模数二次剩余先要求解奇素数模数二次剩余–具体理论查看此链
• 据此推导过程编写脚本

实例[0xGame 2024] Number-Theory-CRT

from Crypto.Util.number import *
import sympy
from sympy.ntheory.modular import crt
c = ?
e = ?
n = ?
p = ?
q = ?
phi = (p-1) * (q-1)
print(GCD(e,phi)) # 发现公约数2,分析得到二次剩余

def find_quadratic_residues(a, p):
    # 首先检查 a 是否是模 p 下的二次剩余
    if not sympy.is_quad_residue(a, p):
        return None  # 如果 a 不是二次剩余，返回 None

    # 使用 sympy 的 nthroot_mod 找到一个解
    x = sympy.nthroot_mod(a, 2, p, all_roots=False)

    # 计算另一个解
    second_solution = p - x

    return (x, second_solution)

x1 = find_quadratic_residues(c,p)       # 求解模p下的二次剩余
x2 = find_quadratic_residues(c,q)       # 求解模q下的二次剩余

for i in x1:
    for j in x2:
        remainders = [i,j]
        mods = [p,q]
        m_ = crt(mods, remainders)[0]   # CRT合并得到模n的二次剩余解
        c_ = m_%n

        e_ = e//2
        d = inverse(e_,phi)
        m = pow(c_,d,n)
        print(long_to_bytes(m))

2.用Sage求解有限域上开方

• 理论来说应该可以解决任意公约数的不互质问题，但是运行时间较长
• 自认为Method 1适用于gcd小、n大的情况，Method 2适用于gcd大、n小的情况

from Crypto.Util.number import *
from sympy.ntheory.modular import crt

p = ?
q = ?
n = ?
c = ?
e = ?

phi = (p-1)*(q-1)
gcd = GCD(e,phi)
d = inverse(e//gcd,phi)

# Method 1 求解同余方程
R.<x> = PolynomialRing(Zmod(p))
f = x^gcd - c
res1 = f.roots(multiplicities=False)

R.<x> = PolynomialRing(Zmod(q))
f = x^gcd - c
res2 = f.roots(multiplicities=False)

# Method 2 开根号
res1 = Zmod(p)(c).nth_root(gcd, all=True)
res2 = Zmod(q)(c).nth_root(gcd, all=True)

for i in res1:
    for j in res2:
        m = crt([p,q],[int(i),int(j)])
        if m is not None:
            try:
                print(long_to_bytes(int(pow(m[0],d,n))).decode())
            except Exception as e:
                continue

Rabin加密

• 可以看作是RSA中$e=2$的特殊情况，且$p\equiv q\equiv 3\ mod\ 4$
• 解法原理同上e和φ不互质

给定一些关于$p,q$计算后的值

• 已知$n,x_1p+y_1q,x_2p+y_2q$
• 已知$n,(xp+yq)^{n-p-q}mod\ n$

[长城杯 2024]Crypto-rasnd

• 已知$p^2+q^2$

[国城杯 2024]EZ_sign

# SageMath
N = ?
f = ZZ[I](N)
divisors_f = divisors(f)
for d in divisors_f:
    a,b = d.real(), d.imag()
    if a**2 + b**2 == N:
        p = abs(int(a))
        q = abs(int(b))
        if is_prime(p) and is_prime(q):
            print("p =",p)
            print("q =",q)
            break

• 已知$p^q\ mod\ n$
• 易得$p^q\ mod\ n = p$

[HGAME 2024]ezRSA

部分泄露

已知p高位

Coppersmith 攻击

对于512位素数，需要未知至多227位

对于1024位素数，需要未知至多454\455位，可能性大致对半开

如果无法满足上述条件，可以爆破几位数

from Crypto.Util.number import *

n = ?
c = ?
e = 65537
p_high = ?

R.<x> = PolynomialRing(Zmod(n))
f = p_high + x
res = f.small_roots(X = 2^256,beta = 0.4)
if res != []:
    p = p_high + int(res[0])
    q = n // p
    d = inverse(e,(p-1)*(q-1))
    m = pow(c,d,n)
    print(long_to_bytes(int(m)))

已知m高位

[HGAME 2024]MidRSA

似乎此题的e通常为3，实测0x10001运行几分钟都没结果

from Crypto.Util.number import long_to_bytes

n = ?
c = ?
e = 3
m1 = ?
bits = ?

R.<x> = PolynomialRing(Zmod(n))
f = (m1 + x)^e - c
res = f.small_roots(X = 2^bits,beta = 1)

已知d高位

d高位低位泄露的主要推导基于此论文-An Attack on RSA Given a Small Fraction of the Private Key Bits

d高位泄露

满足上述条件时可以计算出$\tilde{k}$然后通过恒定附加误差爆破即可得到正确$k$

再依据上述结论计算出$x= p\ mod\ e$

剪枝

p^(q>>nbits)

p = getPrime(lenth)
q = getPrime(lenth)
n = p*q
gift = p ^ (q >> bits)

主要思路是计算p和q，基于以下两个已知事实，去遍历确定每个比特位

• 将 p、q 未确定的比特位全填 0，乘积应小于 n
• 将 p、q 未确定的比特位全填 1，乘积应大于 n

n = ?
gift = ?
bits = ?
lenth = ?

p = bin(gift)[2:2+bits]
q = ""

for i in range(lenth-bits):
    tmp = int(bin(gift)[2:][bits+i])
    for a in range(2):
        b = tmp^a
        if (int(p+str(a),2)<<(lenth-bits-1-i)) * (int(q+str(b),2)<<(lenth-1-i)) < n and ((int(p+str(a),2)<<(lenth-bits-1-i)) + 2**(lenth-bits-1-i) - 1) * ((int(q+str(b),2)<<(lenth-1-i)) + 2**(lenth-1-i) - 1) > n:
            p = p+str(a)
            q = q+str(b)
            break

p = int(p,2)
assert n%p == 0
q = n//p

RSA Leak Oracle

RSA Byte Oracle

如果有一个预言机可以提供RSA解密操作，但是只会返回最后一个字节。可以在$log_256n$的时间内计算出明文$m$

已知$c=m^e\ mod\ n$和$n,e$

显然由于RSA是乘法同态的加密算法，于是可以构造一个已知的明文，加密后和密文相乘，使得解密后的明文乘上构造的明文，即

$$a = (256^i)^e\ mod\ n$$

$$ac^d \equiv 256^im\ mod\ n=256^im-kn$$

由于预言机只能给出最后一个字节所以我们实际上得到的是$output = -kn\ mod\ 256$

易得$k\equiv -n^{-1}output\ mod\ 256$

很容易注意到这里$k$是对256取模的，也就是说当$i>1$时显然$k<256^i$并不能准确计算出$k$

但是能发现当$i=1$的时候很容易计算得到此时的$k_1=-n^{-1}output_1$有$k_1n\le 256c\le (k_1+1)n$

那么当$i=2$时，有$k_2n\le 256^2c\le (k_2+1)n$，这两个范围肯定有并集因此可以有

$$ \left\{\begin{matrix} 256k_1n\le 256^2c\le 256(k+1)n\\ k_2n\le 256^2c\le (k_2+1)n \end{matrix}\right. \Rightarrow \left\{\begin{matrix} 256k_1<(k_2+1)\\ k_2<256(k_1+1) \end{matrix}\right. $$

$$\Rightarrow k_2=256k_1+t$$

很amazing啊，推广后能发现k除以256的商正好就是上一轮的k而且余数也是能通过上述推导得来的，因此就能计算出k从而确定明文了。

MatrixRSA

咕咕ing

*论文题

关键字：RSA、相同私钥、格基规约、格、同一明文多次加密、LLL

论文链接

实例