ACEBEAR CTF 2019 - Cotan

Challenge

Challenge description, and script can be download here.

Solution

With some background math transform, @196 is able to convert the base to traditional DLP problem

p = 1361129467683753853853498429727072846149
g = 937857192022401732022326285294515252367
h = 71727917161216204087973385053390831556

Let’s factor order of the group p:

sage: factor(p-1)
2^2 \* 340282366920938463463374607431768211537

We call the sub-order of the group is ell, so we have:

• ell_small = 2
• ell_big = 340282366920938463463374607431768211537

Calculate DLP

Using CADO-NFS with two parameters like this:

$ ./cado-nfs.py -dlp -ell ell_big target=h p -t 6

Command explanation:

• -dlp: mean we calculate Discrete Log
• -ell: we input the subgroup order
• target=h p: we set target to value h, modulo p, which mean $2^x = h$ mod p
• -t 6: run on 6 cores

NOTE: We don’t specify the base here because we will have to calculate the base by ourselves

Calculate $log_2h$

$ ./cado-nfs.py -dlp -ell 340282366920938463463374607431768211537 target=71727917161216204087973385053390831556 1361129467683753853853498429727072846149 -t 6

Output $log_2h$

Info:root: p = 1361129467683753853853498429727072846149
Info:root: ell = 340282366920938463463374607431768211537
Info:root: log2 = 171268190177498693892391393563437542649
Info:root: log3 = 83622131975737922567870551344538854285
Info:root: Also check log(target) vs log(2) ...
Info:root: target = 71727917161216204087973385053390831556
Info:root: log(target) = 306425041562113865430846743034062879086
306425041562113865430846743034062879086

So we have log_h = 306425041562113865430846743034062879086

Calculate $log_2g$

$ ./cado-nfs.py -dlp -ell 340282366920938463463374607431768211537 target=937857192022401732022326285294515252367 1361129467683753853853498429727072846149 -t 6

Output $log_2g$

Info:root: p = 1361129467683753853853498429727072846149
Info:root: ell = 340282366920938463463374607431768211537
Info:root: log2 = 171268190177498693892391393563437542649
Info:root: log3 = 83622131975737922567870551344538854285
Info:root: Also check log(target) vs log(2) ...
Info:root: target = 937857192022401732022326285294515252367
Info:root: log(target) = 288756149835421404704013074339152764728
288756149835421404704013074339152764728

And we have log_g = 288756149835421404704013074339152764728

Like classical logarithm algorithm, to have to logarithm base g, which mean we are going to find $log_gh$ we do: $log(g)/log(h)$

sage: log_h \* inverse_mod(log_g, ell) % ell
17393774282928096980960357108851791532

NOTE: we only operate on x modulo ell, not x modulo (p-1) as we thought.

Now we have $x=log_g(h)$, next, we check if $g^x = h$ mod p or not, if it is then problem solved, otherwise we will do Chinese Reminder Theorem to figure out the full x modulo (p-1).

sage: p = 1361129467683753853853498429727072846149
....: g = 937857192022401732022326285294515252367
....: h = 71727917161216204087973385053390831556
....:
sage: log_h = 306425041562113865430846743034062879086
sage: log_g = 288756149835421404704013074339152764728
sage: x = log_h \* inverse_mod(log_g, ell) % ell
sage: power_mod(g, x, p)
71727917161216204087973385053390831556
sage: h
71727917161216204087973385053390831556
sage: assert power_mod(g, x, p) == h

Alright, seem like the solution is x = 17393774282928096980960357108851791532.

Now we are going to decrypt the flag

from pwn import \*
from Crypto.Cipher.AES import AESCipher

x = 17393774282928096980960357108851791532
x = hex(x).lstrip('0x')
key = unhex(x).decode('hex')
enc = '4e8f206f074f895bde336601f0c8a2e092f944d95b798b01449e9b155b4ce5a5ae93cc9c677ad942c32d374419d5512c'.decode('hex')
print(AESCipher(key).decrypt(enc))

And the flag is AceBear{_I_h0p3__y0u_3nj0y3d_1t_}