# PicoCTF 2014 Write-ups

## SSH Backdoor - 100 (Forensics)

#### Writeup by Oksisane

Created: 2014-11-07 20:25:54

### Problem

Some hackers have broken into my server backdoor.picoctf.com and locked my user out (my username is jon). I need to retrieve the flag.txt file from my home directory. The last thing we noticed in out network logs show is the attacker downloading this. Can you figure out a way to get back into my account?

### Hint

The attackers left a backdoor in the openssh server. Maybe you could compare it to the real code?

### Overview

Compare the real openssh-6.7p1 with the one supplied, identify/decrypt a one char xor cipher, and use the resulting plaintext to login to a server with ssh.

### Details

The file left for us is a archive containing a directory which claims it is openssh-6.7p1. The hint us to real oppenssh-6.7p1 which we were able to find hosted here. To compare the two, we used Beyond Compare, which allows us to easily see the differences in files between the two versions of the code. Here is an example of the comparison:

a Right away we can see that the auth.c auth.h and auth-passwd.c files are the only ones that have been changed, and a little further investigation reveals the functions added by the hackers (reproduced below).

static int frobcmp(const char *chk, const char *str) {
int rc = 0;
size_t len = strlen(str);
char *s = xstrdup(str);
memfrob(s, len);

if (strcmp(chk, s) == 0) {
rc = 1;
}

free(s);
return rc;
}

}




The frobcmp() function serves as an additional authentication mode meaning that if it returns 1, we are allowed into the server! Looking at the code we can see the string CGCDSE_XGKIBCDOY^OKFCDMSE_XLFKMY is being passed in to be compared to our password after memfrob is run on it. A little googling reveals that memfrob is a rudimentary encryption technique that encrypts the input data by preforming an XOR with it an the number 42. XOR is the term for a Binary Exclusive OR. XOR has the unique property of being the inverse of itself, that is to say if you XOR something twice with the same key, you get your original text back. Since we have the text (CGCDSE_XGKIBCDOY^OKFCDMSE_XLFKMY) which the XORed version of our password is being compared with, we can simply XOR it with 42 to get what our password should be before it is XORed. Here is an example of xoring every character with 42 in python:

''.join(chr(i) for i in [ord(a) ^ 42 for a in "CGCDSE_XGKIBCDOY^OKFCDMSE_XLFKMY"])


Which gives us the password iminyourmachinestealingyourflags. However, this is not quite the solution. Going back to the problem text, we realize we have to login to backdoor.picoctf.com with the username jon and the password, we just figured out, iminyourmachinestealingyourflags. In the shell server, this command will do that for us:

ssh [email protected]


Once we are in, we simply cat flag.txt which prints the flag to our console!

### Flag

ssshhhhh_theres_a_backdoor
`