# It's Not My Fault 1

What do you mean RSA with CRT has an attack that's not a fault attack? Connect with nc mercury.picoctf.net 26695. not_my_fault.py

- 1.The first step of this problem is to get past the MD5 proof-of-work. We need to find a string that starts with a random 5 character long string and creates an MD5 hash that ends with a string of 6 random hex characters. This can be bruteforced with a simple script.
- 2.After we complete the md5 proof-of-work section, we are shown the public modulus and a clue, which is the public exponent. There is a function called
`get_flag`

that is called and will print the flag if we pass in`p+q`

in less than 15 minutes. The public exponent,`e`

, was generated using the two Chinese Remainder Theorem (CRT) exponents,`d_p`

and`d_q`

.`d_p`

is at most 20 bits (a number between 1 and 1048576). - 3.We can try to bruteforce
`d_p`

and thus find`p`

and`q`

using the approach discussed at Attacking RSA for fun and CTF points – part 4, which I found by searching "rsa crt small dp bruteforce". However, searching for "rsa crt attack -fault" finds this answer on MathOverflow, which discusses a much more complicated bruteforce that will execute much faster. This method is described on page 506 of Galbraith's book "Mathematics of Public Key Cryptography", where it is attributed to Pinch. - 4.>>> n=95580702933509662811256129990158655210667121276245053843875590334281563078868202152845967187641817281520364662600110239110410372520340630639373679599982371620736610194814723749147422221945978800055101110346161945811520158431287139909125886966214800526831490560384144156085296004816333892025839072729987354233>>> e=1817084480271067137841898198122075168542117135135738925285694555698012943264936112861815937200507849960517390660821911331068907250788900674614345400567411>>> m = 7516789928765 # random number>>> for dp in range(1000000):... f = gmpy2.gcd(m-pow(m, e*dp, n), n)... if f > 1:... print(dp, f)... break...187261 8275629468590614667884614599278593237258686111405345888268221129814081809682982742676180514534238891248302334619164139839173447495925780801832743975865311>>> p = 8275629468590614667884614599278593237258686111405345888268221129814081809682982742676180514534238891248302334619164139839173447495925780801832743975865311>>> q = n // p>>> p*q == nTrue
- 5.In the solution script, I implement the above bruteforce using Python's
`multiprocessing`

module to run it on multiple threads because we only have 15 minutes. Without multiprocessing, it would take about 1h20m to run completely on my computer. - 6.[+] Opening connection to mercury.picoctf.net on port 26695: Done[*] MD5 PoW string must start with 31950 and hash must end with bdf98c[+] Bruteforcing MD5 PoW: MD5 PoW String Found: 319507851758[+] Getting public modulus and clue: Success[+] Bruteforcing RSA-CRT d_p: p=6755585209747918163868616886189762801438723814293780076178965401337533787532793973718733625630017793302088239336261323736906089254764788312712004326117117, q=13014663099444252202371035556694916836922288463424074950271603159273262042649683760419705683944766998631464971606746251919923961539027359455369007457245187[*] Switching to interactive modepicoCTF{1_c4n'7_b3l13v3_17'5_n07_f4ul7_4774ck!!!}[*] Got EOF while reading in interactive

`picoCTF{1_c4n'7_b3l13v3_17'5_n07_f4ul7_4774ck!!!}`

Last modified 1yr ago