安全-RSA（i春秋）

Posted 2021-09-09 小狐狸FM

文章目录

• 一、题目
• 二、WriteUp

一、题目

This time John managed to use RSA " correctly "&ellipsis; I think he still made some mistakes though。

N=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

e=0x10001

phi=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

d=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

c=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

二、WriteUp

RSA公式如下

n = p*q

φ(n) = (p-1)*(q-1)

c = ( m^e ) mod n

d = 1 mod φ(n) / e

m = ( c^d ) mod n

目前已知的参数是n、e、p、d、c，需要求的是明文m
0x开头的数字是十六进制，需要先将其转换成十进制，使用python代码进行进制转换int(字符串, 进制类型)

N=43210326036290106251088810298667915702007744567467854988107937233622821671752930583378558808381197036542903396534337305723496107154050324953536530993744267386008120658910242143992656773360792182698142273761182047011119261779988377408208858109503177413194541524543447827777831409903664749520527375514748678462101261445039900408999429969510786960471887158957679337190091527913340383310225813322177657695256994315048270952315001100548949554323965404643005206732777361386623591544461533797263652780595639288158892122219626720481472616420118039436680647298714768071797941290307081010086729667552569828323354008441335674251

e=65537

phi=43210326036290106251088810298667915702007744567467854988107937233622821671752930583378558808381197036542903396534337305723496107154050324953536530993744267386008120658910242143992656773360792182698142273761182047011119261779988377408208858109503177413194541524543447827777831409903664749520527375514748678461684001179681025836243272641520046001108394411608315423967170123709569701951917513847951924936347574671212789694222728087028519090194131250999358518159691864169187723815719657688598576820104123456547706995794001020837756695923476582226622677087727699446323459489485731800382017980557710365098279949382831681952

d=36745622940545343875759976434157197886755506358760982257308567037006074391719705315666781200065625116203199598633622026070492775743618607966652393996419663853350085914252797887742782661594880295499227386075929594641556658033207382848075073319786244161193801795105901007640174254798831863226544268004149920625395925259715625248417078457317167458592444532825039828013768181860349705192246622240475711133444054985367520564542488697167606480838294747710396401026533820191299116891186474240520253953309474166963956334430798860084072450328931700881244717326902973061667440442315315084591766881188371668945135391290476758145

c=37209877026138824302301697292319328185824059912506644719041273895972747926698556612194455367172368277268883774102719113194850674012999971337550561061697826458468363574428378679179721672530515881761949297613967812683948622068020382771205609975220407119022966600675469044732891210789248284410739482876937857726026431593283960162298707892143970513804892248370427455318472402011536095802255121284911517882268092785246985981687913310965628793178703498961617628661113277249071490584774764571170891391355080033791745662119139170834529306548644716069025161989103880671580075279656495472299747401794635781847901588156099495017

因为使用下方的公式就能得到m的值，且c、d和n的值是已知的，所以可以直接求解
使用python计算出m的十进制后，再将其转换成十六进制

m = ( c^d ) mod n

m=3843655260524402023604596518050334491485822435243281383499136834535067384556161639265107050668678281151778547364113350618891028501821403003350717660361853

转换成十六进制数

m=0x4963654354467b7273615f69735f617765736f6d655f7768656e5f757365645f636f72726563746c795f6275745f686f727269626c655f7768656e5f6e6f747d

使用在线工具将hex转换为str

IceCTF{rsa_is_awesome_when_used_correctly_but_horrible_when_not}