2023 数据安全产业人才能力挑战赛 --- math_exam wp

题目

import os
from Crypto.Util.number import *
from secret import flag

bits = 512

def pad(msg, length):
    pad_length = length - len(msg) - 1
    pad_data = os.urandom(pad_length)
    return msg + b'\x00' + pad_data

def unpad(msg):
    return msg.split(b"\x00")[0]

def challenge1(m):
    p, q = [getPrime(bits) for i in range(2)]
    if p <= q:
        p, q = q, p
    e = 0x10001
    n = p * q
    c = pow(m, e, n)
    leak = (n + p) % (q-1)
    print('-------- challenge 1 --------')
    print(f'{e = }')
    print(f'{c = }')
    print(f'{n = }')
    print(f'{leak = }')

def challenge2(m):
    p, q = [getPrime(bits) for i in range(2)]
    e = 0x10001
    n = p * q
    d = inverse(e, (p-1)*(q-1))
    c = pow(m, e, n)
    leak = d + p + q
    print('-------- challenge 2 --------')
    print(f'{e = }')
    print(f'{c = }')
    print(f'{n = }')
    print(f'{leak = }')

def challenge3(m):
    p, q = [getPrime(bits) for i in range(2)]
    e = 0x10001
    n = p * q
    c = pow(m, e, n)
    leak = (pow(p, q, n) + pow(q, p, n)) % n
    print('-------- challenge 3 --------')
    print(f'{e = }')
    print(f'{c = }')
    print(f'{n = }')
    print(f'{leak = }')

assert len(flag) == 42
ms = []
for i in range(0, 42, 14):
    ms.append(bytes_to_long(pad(flag[i:i+14], bits//4-1)))

m1, m2, m3 = ms
challenge1(m1)
challenge2(m2)
challenge3(m3)

"""
-------- challenge 1 --------
e = 65537
c = 112742443814287255411540092433156061065388404049949520959549377871297566383025041892192679147481155020865118811016470498351633875090973546567374001852295013083192495299811476604405637385307524194793969533646755764136014187430115618114840780368311166911900457224593131166970676797547489278410997707815915932756
n = 121127425328043404860413278637978444801342472819958112597540188142689720655042880213001676368390521140660355813910726809125567752172921410143437643574528335234973793653775043030021036875866776532570781661875022102733555943967261003246543180935987772711036868216508554536086688819118597075508026787867088355603
leak = 216638862637129382765636503118049146067015523924032194492700294200289728064297722088882791754351329407138196573832392846467607399504585045028165699421278
-------- challenge 2 --------
e = 65537
c = 7964477910021153997178145480752641882728907630831216554750778499596527781702830885213467912351097301767341858663701574005489585561370961723264247818377063081744522471774208105250855114831033452448184392499682147532404562876275189577321587660597603848038824026981539659156304028998137796242331160312370913038
n = 140571013522095816880929287025269553867630639381779595547026503691829940612178900269986625350464874598461222087427155791855120339533208468121389480964471710028253589422629569889402475311387750348466199387760629889238062977271925350490110043385800605640905324122017637306715108727700910035925728362455954862209
leak = 58442382248753295429370894053397615609981110383986887405127350139482893508400422595729520437678203735054593866306478994471465948872565590901376309380029015549809468112086393107585011072503638322671608471684607214064187044372418770555236721845694224676090744181562673509234801011420696349507624867568099759003
-------- challenge 3 --------
e = 65537
c = 54161995127842474543974770981473422085334044100057089719350274921419091368361244533281599379235907845996678762379778310924192757650322930707785543132446159092950451255660204858292974657119337026589911330412367633761103944916751660957776230135927005700707688661350641600954072696774954805514477330339449799540
n = 88207747624007183083381863279444163105330473097729276113333026679597864128605555600000789783468271680476780366740448641311570797876037993255307716167149079618302706650018518487351604778857406170722209469765782625409279109832638886179654096975665134276856272488090272822541461702907181545730309689190333058151
leak = 19596671928335648228117128090384865424885102632642665068992144783391306491716530155291726644158221224616817878768426330717188403310818678195631582453246848
"""

过程分析

本题考点主要是关于RSA的基础数论分析，通过各种等式变换得到关于p和q的等式，从而计算出p和q。

challenge 1

$leak\equiv (n+p)mod(q-1)$
$\because n=p\ast q$
$\therefore leak\equiv p(q+1)mod(q-1)$
$\Rightarrow leak\equiv 2\ast pmod(q-1)$
$又\because p>q,\therefore pmod(q-1)=p-q+1$
$综上，leak=2(p-q+1)$
$此时联立n=p\ast q,可得如下方程组$
$\{\begin{array}{c}leak=2(p-q+1)\\ n=p\ast q\end{array}$
解方程即可得到p,q。

challenge 2

$ed\equiv 1modphi,leak=d+p+q$
$\because phi=(p-1)\ast (q-1)=n-(p+q)+1$
$\therefore p+q=n-phi+1$
$\Rightarrow e\ast (leak-p-q)=1+k\ast phi$
$\Rightarrow e\ast (leak-n+phi-1)=1+k\ast phi$
$\Rightarrow (e-k)\ast phi=e(n-leak+1)+1$
$\because 等式右边>0且phi>0，存在一个k使得上述等式成立，\therefore k\in [1,e)$

challenge 3

$leak\equiv p^q+q^{p}modn$
$leak\equiv p^{q}modq+q^{p}modp$
$根据费马小定理$
$a^{p-1}\equiv 1modp$
$\Rightarrow leak=p+q$
$此时联立n=p\ast q,可得如下方程组$
$\{\begin{array}{c}leak=p+q\\ n=p\ast q\end{array}$
解方程即可得到p,q.

解题脚本

import gmpy2
from Crypto.Util.number import *
from sympy import *

def decode1():
    #leak = (n + p) % (q-1)
    #leak = (p*q+p) % (q-1)
    #leak = p(q+1) % (q-1),(q+1) %(q-1) = 2
    #leak = 2*p % (q-1),因为p>q 所以 p %(q-1) = (p-q+1)
    #leak = 2*(p-q+1),构建方程组求p,q
    e = 65537
    c = 112742443814287255411540092433156061065388404049949520959549377871297566383025041892192679147481155020865118811016470498351633875090973546567374001852295013083192495299811476604405637385307524194793969533646755764136014187430115618114840780368311166911900457224593131166970676797547489278410997707815915932756
    n = 121127425328043404860413278637978444801342472819958112597540188142689720655042880213001676368390521140660355813910726809125567752172921410143437643574528335234973793653775043030021036875866776532570781661875022102733555943967261003246543180935987772711036868216508554536086688819118597075508026787867088355603
    leak = 216638862637129382765636503118049146067015523924032194492700294200289728064297722088882791754351329407138196573832392846467607399504585045028165699421278
    p,q = symbols("p q")
    eq = [p*q-n,2*(p-q+1)-leak]
    result = list(nonlinsolve(eq, [p, q]))
    p,q =int(result[1][0]),int(result[1][1])
    phi =(p-1)*(q-1)
    d = gmpy2.invert(e,phi)
    m = pow(c,d,n)
    flag1 = long_to_bytes(m)
    return flag1[:14]

def decode2():
    #e*d = 1 +k*phi,leak = d+p+q
    #e*(leak-p-q) = 1 + k*phi
    #n-(p+q)+1 = phi   ---->>>  p+q = n-phi+1
    #e*(leak-n-1+phi) = 1 +k*phi
    #e*(leak-n-1)+e*phi = 1+k*phi
    #(e-k)*phi = e*(n-leak+1)+1,爆破k找到符合条件的k即可求出phi，接下来就是RSA解密
    e = 65537
    c = 7964477910021153997178145480752641882728907630831216554750778499596527781702830885213467912351097301767341858663701574005489585561370961723264247818377063081744522471774208105250855114831033452448184392499682147532404562876275189577321587660597603848038824026981539659156304028998137796242331160312370913038
    n = 140571013522095816880929287025269553867630639381779595547026503691829940612178900269986625350464874598461222087427155791855120339533208468121389480964471710028253589422629569889402475311387750348466199387760629889238062977271925350490110043385800605640905324122017637306715108727700910035925728362455954862209
    leak = 58442382248753295429370894053397615609981110383986887405127350139482893508400422595729520437678203735054593866306478994471465948872565590901376309380029015549809468112086393107585011072503638322671608471684607214064187044372418770555236721845694224676090744181562673509234801011420696349507624867568099759003
    for k in range(1,65537):
        if (e*(n-leak+1)+1)%(e-k)==0:
            phi = (e*(n-leak+1)+1)//(e-k)
            d = gmpy2.invert(e, phi)
            try:
                m = pow(c,d,n)
                flag2 = long_to_bytes(m)
                if flag2[4:5]==b'-' and flag2[9:10]==b'-':
                    return flag2[:14]
            except:
                pass


def decode3():
    #leak = p**q %n + q**p %n
    #leak = p**q %q+q**p%p,费马小定理：a**(p-1) = 1 %p
    #leak = p+q，构建方程组求p,q
    e = 65537
    c = 54161995127842474543974770981473422085334044100057089719350274921419091368361244533281599379235907845996678762379778310924192757650322930707785543132446159092950451255660204858292974657119337026589911330412367633761103944916751660957776230135927005700707688661350641600954072696774954805514477330339449799540
    n = 88207747624007183083381863279444163105330473097729276113333026679597864128605555600000789783468271680476780366740448641311570797876037993255307716167149079618302706650018518487351604778857406170722209469765782625409279109832638886179654096975665134276856272488090272822541461702907181545730309689190333058151
    leak = 19596671928335648228117128090384865424885102632642665068992144783391306491716530155291726644158221224616817878768426330717188403310818678195631582453246848
    p,q = symbols("p q")
    eq = [p*q-n,p+q-leak]
    result = list(nonlinsolve(eq, [p, q]))
    p,q =int(result[1][0]),int(result[1][1])
    phi =(p-1)*(q-1)
    d = gmpy2.invert(e,phi)
    m = pow(c,d,n)
    flag3 = long_to_bytes(m)
    return flag3[:14]

if __name__ == '__main__':
    print(decode1()+decode2()+decode3())

flag:

flag{9dc4b6dd-b162-479c-b087-01d351073d14}

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