Nuclear Sale

2 minutes to read

We are given a PCAP file (challenge.pcap). If we analyze it using Wireshark, we will see some SMTP emails:

Nuclear Sale 1

The challenge is based on a series of emails sent within a company. There is some information encrypted there. These are the emails:

Hello everyone,

A potential Buyer approached us asking for a HUGE amount of plutonium. Are we even allowed to  
sell this much?

Best Regards,
Sales Dept

We are very XORry but the management does not approve such a sale. It may damage our business.  
Who is the buyer?

Best Regards,
Management

He is a high profile individual. His information is encrypted below:

6b65813f4fe991efe2042f79988a3b2f2559d358e55f2fa373e53b1965b5bb2b175cf039  

You know what you have to do.

Best Regards,
Sales Dept

Here is the ciphertext encrypted with our key.

fd034c32294bfa6ab44a28892e75c4f24d8e71b41cfb9a81a634b90e6238443a813a3d34  

Best Regards,
Management

Encrypting again with our key...

de328f76159108f7653a5883decb8dec06b0fd9bc8d0dd7dade1f04836b8a07da20bfe70  

Best Regards,
Sales Dept

Oh my... This changes everything. We cannot refuse selling to this guy. He can literally destroy us.  
Move the process.

Best Regards,
Management

Alright, we will process the order. Thanks!  

Best Regards,
Sales Dept

There is a hint in the second email telling that XOR cipher will be used (“We are very XORry”). Then, we have three hexadecimal strings that might be the result of XOR ciphers.

The above ciphertexts (in order, $c_0$, $c_1$, $c_2$) are related as follows (where $p$ is the plaintext, $k_x$ is the client’s key and $k$ is “our key”):

$$ \begin{cases} c_0 = p \oplus k_x \newline c_1 = c_0 \oplus k \newline c_2 = p \oplus k \newline \end{cases} $$

Because of XOR properties:

$$ \begin{align} c_1 \oplus c_2 & = (c_0 \oplus k) \oplus (p \oplus k) \newline & = c_0 \oplus p \newline & = (p \oplus k_x) \oplus p \newline & = k_x \end{align} $$

And then $p = c_0 \oplus k_x$. So, we can get the plaintext using the following operation:

$$ p = c_0 \oplus k_x = c_0 \oplus c_1 \oplus c_2 $$

This can be easily done with CyberChef:

Nuclear Sale 2

And also in Python:

$ python3 -q
>>> from pwn import unhex, xor
>>> c0 = unhex('6b65813f4fe991efe2042f79988a3b2f2559d358e55f2fa373e53b1965b5bb2b175cf039')  
>>> c1 = unhex('fd034c32294bfa6ab44a28892e75c4f24d8e71b41cfb9a81a634b90e6238443a813a3d34')
>>> c2 = unhex('de328f76159108f7653a5883decb8dec06b0fd9bc8d0dd7dade1f04836b8a07da20bfe70')
>>> xor(c0, c1, c2)
b'HTB{s3cr3t_sh4r1ng_w1th_x0r_15_l4m3}'