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February 2002

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Cheating CHAP
Sebastian Krahmer krahmer@cs.uni-potsdam.de
February 2, 2002
Abstract
The Challenge Handshake Authentication Protocol (CHAP) is used to verify the
identity of a peer in a 3-way handshake and is usually embedded in other protocols,
commonly PPP. Several extensions (MS-CHAP) exist to allow the encryption of link
layer packets via CHAP authenticated connections. In this paper I will describe how
CHAP may be attacked, gaining unauthorized access to CHAP protected dialins or
VPN’s and show that CHAP is not the right protocol to authenticate clients in IP networks.

Disclaimer
This material 1 is for educational purposes only. It demonstrates weaknesses within authentication schemes often used in wireless networks for example. All described tests and
captures have been done in my own private LAN. It is strongly recommended that you do
not test these programs without prior written permission of the local authorities.

Introduction

CHAP is most frequently encountered in conjunction with PPP, the Point-to-Point Protocol
as used to connect to your ISP with a modem for instance. While I am not aware of any
non-PPP protocol that uses CHAP, there is no reason why it should not work on top of other
protocols despite the tight binding to PPP in the CHAP RFC 1994.
I am not going to explain the 100th insecurity of PPTP or that the hashing mechanism
some companies are using are weak. The weaknesses of PPTP have been described in
[PPTP ] and others papers.
The insecurities of CHAP do not depend on the underlying layer used, i.e. CHAP is
vulnerable regardless of wether PPP+CHAP is used for ISP dialins or with PPTP to protect
wireless networks. However, I will only focus on PPP+CHAP over IP (PPTP) because IP
network spoofing and sniffing techniques are widely known. As will be demonstrated later,
we will need these techniques.

How things work

RFC 1994 describes the CHAP authentication process. This process is summarized below
with a server authenticating a client, the authenticator and the peer:
• After the Link Establishment phase is complete, the authenticator sends a ”challenge” message to the peer.
• The peer responds with a value calculated using a ”one-way hash” function.
1 this paper and provided programs

HOW THINGS WORK

• The authenticator checks the response against its own calculation of the expected
hash value. If the values match, the authentication is acknowledged; otherwise the
connection SHOULD be terminated.
• At random intervals, the authenticator sends a new challenge to the peer, and repeats
steps 1 to 3.
The first three steps gave it the name challenge/response and are ok provided the following conditions are true.
1. The used hash function is cryptographicaly strong. Common algorithms are MD5 2
or SHA1.
2. The secret used to calculate the response is strong, i.e. it does not appear in a dictionary.
3. The challenge that is sent to the peer is truely random and not predictable by attackers. It is also assumed that the same challenge is never sent twice in a certain timeframe especially after reboots etc. That implies that the rand-pool is large enough.
The fourth RFC requirement for CHAP exists to protect channels from take-overs i.e.
once an attacker took over your communication-channel he may not know the answer to
the next challenge and is disconnected. This is meant to be a special protection within the
CHAP protocol but we will see that this fourth point will break CHAPs neck.
Challenge/response itself is a quite weak authentication method. It is similar to giving
the attacker your /etc/shadow file which should be avoided. This is because the second
point is often handled too lax, e.g. commonly students choose their password themself
when getting WLAN access.

2.1

CHAP packets found ”in the cage”

Figure 1 shows a captured challenge. The marked data part consists per RFC of:
• 1 Byte Code; 01 for Challenge 02 for Response
• 1 Byte Identifier
• 2 Byte Length of CHAP portion
• 1 Byte Value-Size to tell the peer the length of the Challenge or Response payload
• the Challenge/Response payload
• In challenge-case (code 01) the name of the server, in response-case (code 02) the
username
In this sample-session we see a challenge packet from host ’zehn’ to 10.0.0.2. The
servername is ’liane’ (last 5 byte of packet), the challenge is 16 byte long. The first packet
from 10.0.0.2 to zehn 3 initiated the talk and requested to use MD5 to hash the challenge.
Both sides now compute the hash as follows: They concatenate the identifier 4 with the
secret and the challenge and compute the MD5-hash. The secret is determined by looking
up in a certain file which holds the secrets (’passwords’).
Once the client computed the response he sends it to the server: A packet containing 16
bytes of response and the username the server should use to lookup the secret. If the client
sent the correct response the server now sends a success-packet. The captured responsepacket is seen in Figure 2.
2 CHAP is using MD5.
3 This packet was not captured because it did not contain useful information.
4 The second byte of the packet.

OPEN THE GATES!

Figure 1: A captured CHAP Challenge packet.

Open the gates!

Despite the fact that CHAP was designed to send as little information as possible across
the link we still gather:
• username
• servername
• client and server IP
• the ID used to compute response
• challenge and associated response
The most important thing is missing: the secret. Yes, challenge response was designed
to keep the secret secret.
The easiest thing we may now do is try a dictionary attack because we have all the necessary information to calculate the response. Only the secret is unknown which allows us
to try a dictionary-attack. We assume the secret is strong 5 . Further, the challenge is probably unique and the randomness is good enough, so we won’t succeed with precomputing
or replaying attacks either.
5 this is probably not the case, such an attack is always worth a try, but theres a better way, so lets continue

OPEN THE GATES!

Figure 2: A captured CHAP response packet.

CONCLUSION

However, looking again over the data we already gathered and thinking about the protocol, immediately the 4th requirement in the RFC comes in mind: The client MUST answer
any challenge it receives. Aha! What if we send a request to the server, await the challenge
and let an already authenticated client where the handshake has just been captured compute
the response?
Analyzing the sourcecode of pppd 6 shows that this could work. The server does not
send repeated challenges, but the client will answer any challenge it receives.
Since PPP (the protocol itself) does not know about IP addresses there is no way for the
pppd to determine who was asking and it happily sends a response. Thus, in theory the pppd
is answering any question. In practise things are a bit more hairy since the PPP packets are
wrapped in PPTP frames because we use PPTP to tunnel PPP over IP. PPTP however uses
sequence numbers 7 and source/destination addresses to match incoming packets to the
daemons that should handle them. That means the 10.0.0.2 client in the example will not
respond to challenges from 10.0.73.50 but only to challenges from 10.0.0.1 which is the
legitimate PPTP server in this configuration. Now you know why we use IP. It is trivial to
send a spoofed challenge to the client which appears to originate from the original server.
The client computes the response and sends it to the server. Two things happen then. First
we sniff the response and use it to authenticate ourself, second the server also obtains the
challenge from the legit client which he just answers with a success packet as required by
the RFC.
As already stated the PPTP tunnel also uses 32 Bit sequencenumbers to keep track
of already sent packets. This is not a big obstacle since we may sniff the last valid sequencenumber. But things are even simpler because the pptp program accepts any sequencenumber which is larger than the last one, i.e. we may safely send a PPTP packet with
sequencenumber of 0x00ffffff. Using this sequencenumber we can be pretty sure that it is
greater than the last valid sequencenumber the server has seen from this client.

3.1 Finetuning
Figure 3 shows successfull exploitation of the CHAP weakness inside a test-LAN (hubbed).
The first step is to gather valid client and server IP addresses and the login-name/servername pair. The special pppd will use these informations to log into the VPN without
knowing the secret accociated with the login-name it is using. Additionally the challenge
and the response have been saved which allows dictionary attacks when apropriate.
When using the patched pppd from team TESO [7350pppd ] to use other clients to
authenticate ourself, as a sideeffect this victim-client will lose its connection. The pptpd
running on the server drops all packets from the legit client because the sequencenumber
the client is using are too small to be accepted by the server. This might be avoided by
keeping track of sent sequence numbers and only choosing last seen+1 to ask the victimclient for the response. That way, the client will only miss one packet from the server. A
second way is to let the sequence-number wrap-around, i.e. sending MAXSEQ-1 for the
fake-request and after obtaining the response, sending a MAXSEQ packet so the sequence
number wraps around. Any following packets from the client will have bigger sequence
numbers and are passed through.

Conclusion

The take home message of this paper should be that CHAP is not the right protocol for
authentication to be used within IP based networks. I am not sure how the authentication6 the program responsible for handling CHAP on both ends
7 The sequencenumbers in its own header, do not mix this with TCP seqence-numbers. TCP never plays any
role in our discussion.

REFERENCES

bypassing could be done at dialins because it lacks sniffing/spoofing capabilities. However,
this does not mean its secure.
Never ever use CHAP for authentication in IP networks, wireless LANs in particular.
Think about strong authentication schemes such as RSA authentication or Kerberos.

Acknowledgments
I would like to thank grugq for proof-reading this paper. It was hard to understand all
the grammatics but I think I got it right at the end. Same to scut for kicking my ass to
make new screenshots. The old ones really sucked. And finally again thanks to segfault.net
consortium for giving valuable discussion about crypto and why NOC people should watch
their cables :-).

References
[PPTP ] Bruce Schneier and Mudge. Cryptoanalysis of Microsoft’s Point-to-Point Tunneling Protocol (PPTP) .
[7350pppd ] TESO
An implementation of the attack described in this paper.
http://stealth.7350.org/7350pppd.tgz .

REFERENCES

Figure 3: A successfull login without knowing the secret.