The RSA algorithm is named after Ron Rivest, Adi Shamir and Len Adleman, who invented it in 1977 [RIVE78]. The basic technique was first discovered in 1973 by Clifford Cocks [COCK73] of CESG (part of the British GCHQ) but this was a secret until 1997. The patent taken out by RSA Labs has expired.
The RSA algorithm can be used for both public key encryption and digital signatures. Its security is based on the difficulty of factoring large integers.
Key Generation Algorithm
Generate two large random primes, p and q, of approximately equal size such that their product n = pq is of the required bit length, e.g. 1024 bits. [See note 1].
Compute n = pq and (φ) phi = (p-1)(q-1).
Choose an integer e, 1 < e < phi, such that gcd(e, phi) = 1. [See note 2].
Compute the secret exponent d, 1 < d < phi, such that ed ≡ 1 (mod phi). [See note 3].
The public key is (n, e) and the private key is (n, d). Keep all the values d, p, q and phi secret.
n is known as the modulus.
e is known as the public exponent or encryption exponent or just the exponent.
d is known as the secret exponent or decryption exponent.
Encryption
Sender A does the following:-
Obtains the recipient B's public key (n, e).
Represents the plaintext message as a positive integer m [see note 4].
Computes the ciphertext c = me mod n.
Sends the ciphertext c to B.
Decryption
Recipient B does the following:-
Uses his private key (n, d) to compute m = cd mod n.
Extracts the plaintext from the message representative m.
Digital signing
Sender A does the following:-
Creates a message digest of the information to be sent.
Represents this digest as an integer m between 0 and n-1. [See note 5].
Uses her private key (n, d) to compute the signature s = md mod n.
Sends this signature s to the recipient, B.
Signature verification
Recipient B does the following:-
Uses sender A's public key (n, e) to compute integer v = se mod n.
Extracts the message digest from this integer.
Independently computes the message digest of the information that has been signed.
If both message digests are identical, the signature is valid.
Notes on practical applications
To generate the primes p and q, generate a random number of bit length b/2 where b is the required bit length of n; set the low bit (this ensures the number is odd) and set the two highest bits (this ensures that the high bit of n is also set); check if prime (use the Rabin-Miller test); if not, increment the number by two and check again until you find a prime. This is p. Repeat for q starting with a random integer of length b-b/2. If p<q, swop p and q (this only matters if you intend using the CRT form of the private key). In the extremely unlikely event that p = q, check your random number generator. Alternatively, instead of incrementing by 2, just generate another random number each time.
There are stricter rules in ANSI X9.31 to produce strong primes and other restrictions on p and q to minimise the possibility of known techniques being used against the algorithm. There is much argument about this topic. It is probably better just to use a longer key length.
In practice, common choices for e are 3, 17 and 65537 (216+1). These are Fermat primes, sometimes referred to as F0, F2 and F4 respectively (Fx=2^(2^x)+1). They are chosen because they make the modular exponentiation operation faster. Also, having chosen e, it is simpler to test whether gcd(e, p-1)=1 and gcd(e, q-1)=1 while generating and testing the primes in step 1. Values of p or q that fail this test can be rejected there and then. (Even better: if e is prime and greater than 2 then you can do the less-expensive test (p mod e)!=1 instead of gcd(p-1,e)==1.)
To compute the value for d, use the Extended Euclidean Algorithm to calculate d = e-1 mod phi, also written d = (1/e) mod phi. This is known as modular inversion. Note that this is not integer division. The modular inverse d is defined as the integer value such that ed = 1 mod phi. It only exists if e and phi have no common factors.
2010-08-14: For more details of the extended Euclidean algorithm, see our new page The Euclidean Algorithm and the Extended Euclidean Algorithm which shows how to use the Euclidean algorithm, answer exam questions on it, and gives source code for an implementation.
When representing the plaintext octets as the representative integer m, it is usual to add random padding characters to make the size of the integer m large and less susceptible to certain types of attack. If m = 0 or 1 or n-1 there is no security as the ciphertext has the same value. For more details on how to represent the plaintext octets as a suitable representative integerm, see PKCS#1 Schemes below or the reference itself [PKCS1]. It is important to make sure that m < n otherwise the algorithm will fail. This is usually done by making sure the first octet of m is equal to 0x00.
Decryption and signing are identical as far as the mathematics is concerned as both use the private key. Similarly, encryption and verification both use the same mathematical operation with the public key. That is, mathematically, for m < n,
m = (me mod n)d mod n = (md mod n)e mod n
However, note these important differences in implementation:-
The signature is derived from a message digest of the original information. The recipient will need to follow exactly the same process to derive the message digest, using an identical set of data.
The recommended methods for deriving the representative integers are different for encryption and signing (encryption involves random padding, but signing uses the same padding each time).
Summary of RSA
n = pq, where p and q are distinct primes.
phi, φ = (p-1)(q-1)
e < n such that gcd(e, phi)=1
d = e-1 mod phi.
c = me mod n, 1<m<n.
m = cd mod n.
Key length
When we talk about the key length of an RSA key, we are referring to the length of the modulus, n, in bits. The minimum recommended key length for a secure RSA transmission is currently 1024 bits. A key length of 512 bits is now no longer considered secure, although cracking it is still not a trivial task for the likes of you and me. The longer your information is needed to be kept secure, the longer the key you should use. Keep up to date with the latest recommendations in the security journals.
There is small one area of confusion in defining the key length. One convention is that the key length is the position of the most significant bit in n that has value '1', where the least significant bit is at position 1. Equivalently, key length = ceiling(log2(n+1)). The other convention, sometimes used, is that the key length is the number of bytes needed to store n multiplied by eight, i.e.ceiling(log256(n+1)).
The key used in the RSA Example paper [KALI93] is an example. In hex form the modulus is
0A 66 79 1D C6 98 81 68 DE 7A B7 74 19 BB 7F B0
C0 01 C6 27 10 27 00 75 14 29 42 E1 9A 8D 8C 51
D0 53 B3 E3 78 2A 1D E5 DC 5A F4 EB E9 94 68 17
01 14 A1 DF E6 7C DC 9A 9A F5 5D 65 56 20 BB AB
The most significant byte 0x0A in binary is 00001010'B. The most significant bit is at position 508, so its key length is 508 bits. On the other hand, this value needs 64 bytes to store it, so the key length could also be referred to by some as 64 x 8 = 512 bits. We prefer the former method. You can get into difficulties with the X9.31 method for signatures if you use the latter convention.
Minimum key lengths
The following table is taken from NIST's Recommendation for Key Management [NIST-80057]. It shows the recommended comparable key sizes for symmetrical block ciphers (AES and Triple DES) and the RSA algorithm. That is, the key length you would need to use to have comparable security.
Symmetric key algorithm
Comparable RSA key length
Comparable hash function
Bits of security
2TDEA*
1024
SHA-1
80
3TDEA
2048
SHA-224
112
AES-128
3072
SHA-256
128
AES-192
7680
SHA-384
192
AES-256
15360
SHA-512
256
* 2TDEA is 2-key triple DES - see What's two-key triple DES encryption.
Note just how huge (and impractical) an RSA key needs to be for comparable security with AES-192 or AES-256 (although these two algorithms have had some weaknesses exposed recently; AES-128 is unaffected).
The above table is a few years old now and may be out of date. Existing cryptographic algorithms only get weaker as attacks get better.
Computational Efficiency and the Chinese Remainder Theorem (CRT)
Key generation is only carried out occasionally and so computational efficiency is less of an issue.
The calculation a = be mod n is known as modular exponentiation and one efficient method to carry this out on a computer is thebinary left-to-right method. To solve y = x^e mod n let e be represented in base 2 as
e = e(k-1)e(k-2)...e(1)e(0)
where e(k-1) is the most significant non-zero bit and bit e(0) the least.
set y = x
for bit j = k - 2 downto 0
begin
y = y * y mod n /* square */
if e(j) == 1 then
y = y * x mod n /* multiply */
end
return y
The time to carry out modular exponentation increases with the number of bits set to one in the exponent e. For encryption, an appropriate choice of e can reduce the computational effort required to carry out the computation of c = me mod n. Popular choices like 3, 17 and 65537 are all primes with only two bits set: 3 = 0011'B, 17 = 0x11, 65537 = 0x10001.
The bits in the decryption exponent d, however, will not be so convenient and so decryption using the standard method of modular exponentiation will take longer than encryption. Don't make the mistake of trying to contrive a small value for d; it is not secure.
An alternative method of representing the private key uses the The Chinese Remainder Theorem (CRT).
2011-01-20: For an explanation of how the CRT is used with RSA, see Using the CRT with RSA.
The private key is represented as a quintuple (p, q, dP, dQ, and qInv), where p and q are prime factors of n, dP and dQ are known as the CRT exponents, and qInv is the CRT coefficient. The CRT method of decryption is about four times faster overall than calculating m = cd mod n. The extra values for the private key are:-
dP = (1/e) mod (p-1)
dQ = (1/e) mod (q-1)
qInv = (1/q) mod p where p > q
where the (1/e) notation means the modular inverse (see note 3 above). These values are pre-computed and saved along with pand q as the private key. To compute the message m given c do the following:-
m1 = c^dP mod p
m2 = c^dQ mod q
h = qInv(m1 - m2) mod p
m = m2 + hq
Even though there are more steps in this procedure, the modular exponentation to be carried out uses much shorter exponents and so it is less expensive overall.
[2008-09-02] Chris Becke has pointed out that most large integer packages will fail when computing h if m1 < m2. This can be easily fixed by computing
h = qInv(m1 + p - m2) mod p
or, alternatively, as we do it in our BigDigits implementation of RSA,
if (bdCompare(m1, m2) < 0)
bdAdd(m1, m1, p);
bdSubtract(m1, m1, m2);
/* Let h = qInv ( m_1 - m_2 ) mod p. */
bdModMult(h, qInv, m1, p);
Theory and proof of the RSA algorithm
Every man and his dog seems to have a proof of the RSA algorithm out there, all requiring varying degrees of understanding of number theory. This is our version of a proof of the RSA algorithm, re-written October 2006 (PDF version) and some hints on understanding it.
A very simple example of RSA encryption
This is an extremely simple example using numbers you can work out on a pocket calculator (those of you over the age of 35 45 can probably even do it by hand).
Select primes p=11, q=3.
n = pq = 11.3 = 33
phi = (p-1)(q-1) = 10.2 = 20
Choose e=3
Check gcd(e, p-1) = gcd(3, 10) = 1 (i.e. 3 and 10 have no common factors except 1),
and check gcd(e, q-1) = gcd(3, 2) = 1
therefore gcd(e, phi) = gcd(e, (p-1)(q-1)) = gcd(3, 20) = 1
Compute d such that ed ≡ 1 (mod phi)
i.e. compute d = e-1 mod phi = 3-1 mod 20
i.e. find a value for d such that phi divides (ed-1)
i.e. find d such that 20 divides 3d-1.
Simple testing (d = 1, 2, ...) gives d = 7
Check: ed-1 = 3.7 - 1 = 20, which is divisible by phi.
Public key = (n, e) = (33, 3)
Private key = (n, d) = (33, 7).
This is actually the smallest possible value for the modulus n for which the RSA algorithm works.
Now say we want to encrypt the message m = 7,
c = me mod n = 73 mod 33 = 343 mod 33 = 13.
Hence the ciphertext c = 13.
To check decryption we compute
m' = cd mod n = 137 mod 33 = 7.
Note that we don't have to calculate the full value of 13 to the power 7 here. We can make use of the fact that
a = bc mod n = (b mod n).(c mod n) mod n
so we can break down a potentially large number into its components and combine the results of easier, smaller calculations to calculate the final value.
One way of calculating m' is as follows:-
m' = 137 mod 33 = 13(3+3+1) mod 33 = 133.133.13 mod 33
= (133 mod 33).(133 mod 33).(13 mod 33) mod 33
= (2197 mod 33).(2197 mod 33).(13 mod 33) mod 33
= 19.19.13 mod 33 = 4693 mod 33
= 7.
Now if we calculate the ciphertext c for all the possible values of m (0 to 32), we get
m 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
c 0 1 8 27 31 26 18 13 17 3 10 11 12 19 5 9 4
m 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32
c 29 24 28 14 21 22 23 30 16 20 15 7 2 6 25 32
Note that all 33 values of m (0 to 32) map to a unique code c in the same range in a sort of random manner. In this case we have nine values of m that map to the same value of c - these are known as unconcealed messages. m = 0, 1 and n-1 will always do this for any n, no matter how large. But in practice, higher values shouldn't be a problem when we use large values for n in the order of several hundred bits.
If we wanted to use this system to keep secrets, we could let A=2, B=3, ..., Z=27. (We specifically avoid 0 and 1 here for the reason given above). Thus the plaintext message "HELLOWORLD" would be represented by the set of integers m1, m2, ...
(9,6,13,13,16,24,16,19,13,5)
Using our table above, we obtain ciphertext integers c1, c2, ...
(3,18,19,19,4,30,4,28,19,26)
Note that this example is no more secure than using a simple Caesar substitution cipher, but it serves to illustrate a simple example of the mechanics of RSA encryption.
Remember that calculating me mod n is easy, but calculating the inverse c-e mod n is very difficult, well, for large n's anyway. However, if we can factor n into its prime factors p and q, the solution becomes easy again, even for large n's. Obviously, if we can get hold of the secret exponent d, the solution is easy, too.
A slightly less simple example of the RSA algorithm
This time, to make life slightly less easy for those who can crack simple Caesar substitution codes, we will group the characters into blocks of three and compute a message representative integer for each block.
ATTACKxATxSEVEN = ATT ACK XAT XSE VEN
In the same way that a decimal number can be represented as the sum of powers of ten, e.g.
135 = 1 x 102 + 3 x 101 + 5,
we could represent our blocks of three characters in base 26 using A=0, B=1, C=2, ..., Z=25
ATT = 0 x 262 + 19 x 261 + 19 = 513
ACK = 0 x 262 + 2 x 261 + 10 = 62
XAT = 23 x 262 + 0 x 261 + 19 = 15567
XSE = 23 x 262 + 18 x 261 + 4 = 16020
VEN = 21 x 262 + 4 x 261 + 13 = 14313
For this example, to keep things simple, we'll not worry about numbers and punctuation characters, or what happens with groups AAA or AAB.
In this system of encoding, the maximum value of a group (ZZZ) would be 263-1 = 17575, so we require a modulus n greater than this value.
We "generate" primes p=137 and q=131 (we cheat by looking for suitable primes around √n)
n = pq = 137.131 = 17947
phi = (p-1)(q-1) = 136.130 = 17680
Select e = 3
check gcd(e, p-1) = gcd(3, 136) = 1, OK and
check gcd(e, q-1) = gcd(3, 130) = 1, OK.
Compute d = e-1 mod phi = 3-1 mod 17680 = 11787.
Hence public key, (n, e) = (17947, 3) and private key (n, d) = (17947, 11787).
Question: Why couldn't we use e=17 here?
To encrypt the first integer that represents "ATT", we have
c = me mod n = 5133 mod 17947 = 8363.
We can verify that our private key is valid by decrypting
m' = cd mod n = 836311787 mod 17947 = 513.
Overall, our plaintext is represented by the set of integers m
(513, 62, 15567, 16020, 14313)
We compute corresponding ciphertext integers c = me mod n, (which is still possible by using a calculator)
(8363, 5017, 11884, 9546, 13366)
You are welcome to compute the inverse of these ciphertext integers using m = cd mod n to verify that the RSA algorithm still holds. However, this is now outside the realms of hand calculations unless you are very patient.
To help you carry out these modular arithmetic calculations, download our free modular arithmetic command line programs (last updated 18 June 2009).
Note that this is still a very insecure example. Starting with the knowledge that the modulus 17947 is probably derived from two prime numbers close to its square root, a little testing of suitable candidates from a table of prime numbers would get you the answer pretty quickly. Or just work methodically through the table of prime numbers dividing n by each value until you get no remainder. You could also write a simple computer program to factor n that just divides by every odd number starting from 3 until it reaches a number greater than the square root of n.
A real example
In practice, we don't have a series of small integers to encrypt like we had in the above examples, we just have one big one. We put our message into an octet string, most significant byte first, and then convert to a large integer. Also, rather than trying to represent the plaintext as an integer directly, we generate a random session key and use that to encrypt the plaintext with a conventional, much faster symmetrical algorithm like Triple DES or AES-128. We then use the much slower public key encryption algorithm to encrypt just the session key.
The sender A then transmits a message to the recipient B in a format something like this:-
Session key encrypted with RSA = xxxx
Plaintext encrypted with session key = xxxxxxxxxxxxxxxxx
The recipient B would extract the encrypted session key and use his private key (n,d) to decrypt it. He would then use this session key with a conventional symmetrical decryption algorithm to decrypt the actual message. Typically the transmission would include in plaintext details of the encryption algorithms used, padding and encoding methods, initialisation vectors and other details required by the recipient. The only secret required to be kept, as always, should be the private key.
If Mallory intercepts the transmission, he can either try and crack the conventionally-encrypted plaintext directly, or he can try and decrypt the encryped session key and then use that in turn. Obviously, this system is as strong as its weakest link.
When signing, it is usual to use RSA to sign the message digest of the message rather than the message itself. A one-way hash function like SHA-1 or SHA-256 is used. The sender A then sends the signed message to B in a format like this
Hash algorithm = hh
Message content = xxxxxxxxx...xxx
Signature = digest signed with RSA = xxxx
The recipient will decrypt the signature to extract the signed message digest, m; independently compute the message digest, m', of the actual message content; and check that m and m' are equal. Putting the message digest algorithm at the beginning of the message enables the recipient to compute the message digest on the fly while reading the message.
PKCS#1 Schemes
The most common scheme using RSA is PKCS#1 version 1.5 [PKCS1]. This standard describes schemes for both encryption and signing. The encryption scheme PKCS#1v1.5 has some known weaknesses, but these can easily be avoided. See Weaknesses in RSA below.
There is an excellent paper by Burt Kalinski of RSA Laboratories written in the early 1990s [KALI93] that describes in great detail everything you need to know about encoding and signing using RSA. There are full examples right down to listing out the bytes. OK, it uses MD2 and a small 508-bit modulus and obviously doesn't deal with refinements built up over the last decade to deal with more subtle security threats, but it's an excellent introduction.
The conventions we use here are explained below in Notation and Conventions.
Encryption using PKCS#1v1.5
Algorithm: Encryption using PKCS#1v1.5
INPUT: Recipient's RSA public key, (n, e) of length k = |n| bytes; data D (typically a session key) of length |D| bytes with |D|<=k-11.
OUTPUT: Encrypted data block of length k bytes
Form the k-byte encoded message block, EB,
EB = 00 || 02 || PS || 00 || D
where || denotes concatenation and PS is a string of k-|D|-3 non-zero randomly-generated bytes (i.e. at least eight random bytes).
Convert the byte string, EB, to an integer, m, most significant byte first,
m = StringToInteger(EB, k)
Encrypt with the RSA algorithm
c = m^e mod n
Convert the resulting ciphertext, c, to a k-byte output block, OB
OB = IntegerToString(m, k)
Output OB.
The conversions in steps (2) and (4) from byte string to large integer representative and back again may not be immediately obvious. Large integers and byte (bit) strings are conceptually different even though they may both be stored as arrays of bytes in your computer.
Worked Example
Bob's 1024-bit RSA encryption key in hex format:
n=
A9E167983F39D55FF2A093415EA6798985C8355D9A915BFB1D01DA197026170F
BDA522D035856D7A986614415CCFB7B7083B09C991B81969376DF9651E7BD9A9
3324A37F3BBBAF460186363432CB07035952FC858B3104B8CC18081448E64F1C
FB5D60C4E05C1F53D37F53D86901F105F87A70D1BE83C65F38CF1C2CAA6AA7EB
e=010001
d=
67CD484C9A0D8F98C21B65FF22839C6DF0A6061DBCEDA7038894F21C6B0F8B35
DE0E827830CBE7BA6A56AD77C6EB517970790AA0F4FE45E0A9B2F419DA8798D6
308474E4FC596CC1C677DCA991D07C30A0A2C5085E217143FC0D073DF0FA6D14
9E4E63F01758791C4B981C3D3DB01BDFFA253BA3C02C9805F61009D887DB0319
A randomly-generated one-off session key for AES-128 might be
D=4E636AF98E40F3ADCFCCB698F4E80B9F
The encoded message block, EB, after encoding but before encryption, with random padding bytes shown in green,
0002257F48FD1F1793B7E5E02306F2D3228F5C95ADF5F31566729F132AA12009
E3FC9B2B475CD6944EF191E3F59545E671E474B555799FE3756099F044964038
B16B2148E9A2F9C6F44BB5C52E3C6C8061CF694145FAFDB24402AD1819EACEDF
4A36C6E4D2CD8FC1D62E5A1268F496004E636AF98E40F3ADCFCCB698F4E80B9F
After RSA encryption, the output is
3D2AB25B1EB667A40F504CC4D778EC399A899C8790EDECEF062CD739492C9CE5
8B92B9ECF32AF4AAC7A61EAEC346449891F49A722378E008EFF0B0A8DBC6E621
EDC90CEC64CF34C640F5B36C48EE9322808AF8F4A0212B28715C76F3CB99AC7E
609787ADCE055839829E0142C44B676D218111FFE69F9D41424E177CBA3A435B
Note that the output for encryption will be different each time (or should be!) because of the random padding used.
Encrypting a message
For a plaintext message, say,
PT="Hello world!"
that is, the 12 bytes in hex format,
PT=48656C6C6F20776F726C6421
Then, using the 128-bit session key from above,
KY=4E636AF98E40F3ADCFCCB698F4E80B9F
and the uniquely-generated 128-bit initialization vector (IV)
IV=5732164B3ABB6C4969ABA381C1CA75BA
the ciphertext using AES-128 in CBC mode with PKCS padding is,
CT=67290EF00818827C777929A56BC3305B
The sender would then send a transmission to the recipient (in this case, Bob) including the following information in some agreed format
Recipient: Bob
Key Encryption Algorithm: rsaEncryption
Encrypted Key:
3D2AB25B1EB667A40F504CC4D778EC399A899C8790EDECEF062CD739492C9CE5
8B92B9ECF32AF4AAC7A61EAEC346449891F49A722378E008EFF0B0A8DBC6E621
EDC90CEC64CF34C640F5B36C48EE9322808AF8F4A0212B28715C76F3CB99AC7E
609787ADCE055839829E0142C44B676D218111FFE69F9D41424E177CBA3A435B
Content Encryption Algorithm: aes128-cbc
IV: 5732164B3ABB6C4969ABA381C1CA75BA
Encrypted Content:
67290EF00818827C777929A56BC3305B
The usual formats used for such a message are either a CMS enveloped-data object or XML, but the above summary includes all the necessary info (well, perhaps "Bob" might be defined a bit more accurately).
Cryptographic Message Syntax (CMS) [CMS] is a less-ambiguous version of the earlier PKCS#7 standard (also of the same name) and is designed to be used in S/MIME messages. CMS enveloped-data objects (yes, that's enveloped not encrypted) use ASN.1 and are encoded using either DER- or BER-encoding. (DER-encoding is a stricter subset of BER).
The terminology for CMS and ASN.1 may sound messy, but the end results are well-defined and universally-accepted. On the other hand, the XML cryptographic standards are, to be honest, a complete mess: see XML is xhite. Pretty Good Privacy (PGP) also has a format for RSA messages, although PGP stopped using RSA because of patent issues back in the 1990s.
Nothing, of course, stops you and your recipient from agreeing on your own format and using that. But be careful, even the experts get these things wrong and accidentally give away more than they realise.
Signing using PKCS#1v1.5
Algorithm: Signing using PKCS#1v1.5
INPUT: Sender's RSA private key, (n, d) of length k = |n| bytes; message, M, to be signed; message digest algorithm, Hash.
OUTPUT: Signed data block of length k bytes
Compute the message digest H of the message,
H = Hash(M)
Form the byte string, T, from the message digest, H, according to the message digest algorithm, Hash, as follows
Hash
T
MD5
30 20 30 0c 06 08 2a 86 48 86 f7 0d 02 05 05 00 04 10 || H
SHA-1
30 21 30 09 06 05 2b 0e 03 02 1a 05 00 04 14 || H
SHA-256
30 31 30 0d 06 09 60 86 48 01 65 03 04 02 01 05 00 04 20 || H
SHA-384
30 41 30 0d 06 09 60 86 48 01 65 03 04 02 02 05 00 04 30 || H
SHA-512
30 51 30 0d 06 09 60 86 48 01 65 03 04 02 03 05 00 04 40 || H
where T is an ASN.1 value of type DigestInfo encoded using the Distinguished Encoding Rules (DER).
Form the k-byte encoded message block, EB,
EB = 00 || 01 || PS || 00 || T
where || denotes concatenation and PS is a string of bytes all of value 0xFF of such length so that |EB|=k.
Convert the byte string, EB, to an integer m, most significant byte first,
m = StringToInteger(EB, k)
Sign with the RSA algorithm
s = m^d mod n
Convert the resulting signature value, s, to a k-byte output block, OB
OB = IntegerToString(s, k)
Output OB.
Worked Example
Alice's 1024-bit RSA signing key in hex format:
n=
E08973398DD8F5F5E88776397F4EB005BB5383DE0FB7ABDC7DC775290D052E6D
12DFA68626D4D26FAA5829FC97ECFA82510F3080BEB1509E4644F12CBBD832CF
C6686F07D9B060ACBEEE34096A13F5F7050593DF5EBA3556D961FF197FC981E6
F86CEA874070EFAC6D2C749F2DFA553AB9997702A648528C4EF357385774575F
e=010001
d=
00A403C327477634346CA686B57949014B2E8AD2C862B2C7D748096A8B91F736
F275D6E8CD15906027314735644D95CD6763CEB49F56AC2F376E1CEE0EBF282D
F439906F34D86E085BD5656AD841F313D72D395EFE33CBFF29E4030B3D05A28F
B7F18EA27637B07957D32F2BDE8706227D04665EC91BAF8B1AC3EC9144AB7F21
The message to be signed is, of course,
M="abc"
that is, the 3 bytes in hex format,
PT=616263
The message digest algorithm is SHA-1, so
H = Hash("abc") = A9993E364706816ABA3E25717850C26C9CD0D89D
The DigestInfo value for SHA-1 is
T=
3021300906052B0E03021A05000414A9993E364706816ABA3E25717850C26C9C
D0D89D
The encoded message block, EB, after encoding but before signing is
0001FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF
FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF
FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF00302130
0906052B0E03021A05000414A9993E364706816ABA3E25717850C26C9CD0D89D
After RSA signing, the output is
60AD5A78FB4A4030EC542C8974CD15F55384E836554CEDD9A322D5F4135C6267
A9D20970C54E6651070B0144D43844C899320DD8FA7819F7EBC6A7715287332E
C8675C136183B3F8A1F81EF969418267130A756FDBB2C71D9A667446E34E0EAD
9CF31BFB66F816F319D0B7E430A5F2891553986E003720261C7E9022C0D9F11F
Weaknesses in RSA
Small encryption exponent
If you use a small exponent like e=3 and send the same message to different recipients and just use the RSA algorithm without adding random padding to the message, then an eavesdropper could recover the plaintext.
2010-10-23: For an example of this, see Cracking RSA on our page on the The Chinese Remainder Theorem.
Using the same key for encryption and signing
Given that the underlying mathematics is the same for encryption and signing, only in reverse, if an attacker can convince a key holder to sign an unformatted encrypted message using the same key then she gets the original.
Acting as an oracle
There are techniques to recover the plaintext if a user just blindly returns the RSA transformation of the input. So don't do that.
Solutions
Don't use the same RSA key for encryption and signing.
If using PKCS#v1.5 encoding, use e=0x10001 for your public exponent.
Always format your input before encrypting or signing.
Always add fresh random padding - at least 8 bytes - to your message before encrypting.
When decrypting, check the format of the decrypted block. If it is not as expected, return an error, not the decrypted string.
Similarly, when verifying a signature, if there is any error whatsoever, just respond with "Invalid Signature".
More Advanced Schemes
The underlying RSA computations
c = me mod n, m' = cd mod n; s = md mod n, m' = se mod n
are always the same, but there are many variants of how these can be used inside an encryption or digital signature scheme. Here are some of them.
RSAES-OAEP
The OAEP encoding technique for encryption is described in PKCS#1 version 2 and in IEEE P136. It is more secure than the PKCS#1v1.5 encoding method described above, perhaps provably secure. The encoding technique involves a mask generation function (MGF) based on a hash function and there is no obvious structure in the encoded block, unlike the PKCS#1v1.5 encoding method. Despite being the recommended method for the last decade, OAEP is not used in practice as much as you'd expect.
RSASSA-PSS
The PSS encoding method is used to encode before creating a signature. The technique is described in PKCS#1v2.1 and is similar in design to the OAEP encoding used for encryption involving an MGF based on a hash function. However, there are active patents associated with this method, so sensible implementors avoid it like the plague. Since there are currently no known weaknesses with the PKCS#1v1.5 signature scheme and the nasty smell of patents still lingers, PSS is also not used in practice very much.
X9.31 Signature Scheme
ANSI standard X9.31 [AX931] requires using strong primes derived in a way to avoid particular attacks that are probably no longer relevant. X9.31 uses a method of encoding the message digest specific to the hash algorithm. It expects a key with length an exact multiple of 256 bits. The same algorithm is also specified in P1363 [P1363] where it is called IFSP-RSA2. The scheme allows for the public exponent to be an even value, but we do not consider that case here; all our values of e are assumed to be odd. The message digest hash, H, is encapsulated to form a byte string as follows
EB = 06 || PS || 0xBA || H || 0x33 || 0xCC
where PS is a string of bytes all of value 0xBB of length such that |EB|=|n|, and 0x33 is the ISO/IEC 10118 part number for SHA-1. The byte string, EB, is converted to an integer value, the message representative, f.
Algorithm: Forming an X9.31/RSA2 signature value from the message representative (for odd e).
INPUT: Signer's RSA private key, (n, d); integer, f, where 0 <= f < n and f ≡ 12 (mod 16).
OUTPUT: Signature, an integer s, 0 <= s < n/2, i.e. a value at least one bit shorter than n.
t = fd mod n
s = min{t, n-t}
Output s.
The integer, s, is converted to a byte string of length |n| bytes.
Algorithm: Extracting the message representative from an X9.31/RSA2 signature value (for odd e).
INPUT: Signer's RSA public key, (n, e); signature, s.
OUTPUT: Message representative, f, such that t ≡ 12 (mod 16), or "invalid signature".
If s is not in [0,(n-1)/2], output "invalid signature" and stop.
Compute t = se mod n
If t ≡ 12 (mod 16) then let f = t.
Else let f = n-t. If NOT f ≡ 12 (mod 16), output "invalid signature" and stop.
Output f.
The integer f is converted to a byte string of length |n| bytes and then parsed to confirm that all bytes match the required format
EB = 06 || PS || 0xBA || H || 0x33 || 0xCC
If not, output "invalid signature" and stop; otherwise output the extracted message digest hash, H.
ISO/IEC 9796
IOS/IEC 9796 is an old standard devised before there was a recognised message digest function like MD5 or SHA-1. It allows the entire message to be recovered. Unfortunately, it is considered broken for signing plain text messages, but is still OK for signing message digest values. It is used in the AUTACK scheme described in [EDIFACT].
The encapsulation mechanism weaves the input bytes into a format exactly one bit shorter than the RSA key. The signing mechanism is similar to that in ANSI X9.31 described above, but the message representative, f, is required to be f ≡ 6 (mod 16), instead of modulo 12. In other words, make sure the last 4 bits are equal to 0x6 instead of 0xC.
RSA-KEM
The RSA-KEM Key Transport Algorithm encrypts a random integer with the recipient's public key, and then uses a symmetric key-wrapping scheme to encrypt the keying data. KEM stands for Key Encapsulation Mechanism. The general algorithm is as follows
Generate a random integer z between 0 and n-1.
Encrypt the integer z with the recipient's RSA public key: c = ze mod n.
Derive a key-encrypting key KEK from the integer z.
Wrap the keying data using KEK to obtain wrapped keying data WK.
Output c and WK as the encrypted keying data.
This method has a higher security assurance than PKCS#1v1.5 because the input to the underlying RSA operation is random and independent of the message, and the key-encrypting key KEK is derived from it in a strong way. The downside is that you need to implement a key derivation method (of which there are many varieties - see How many KDFs are there?) and a key wrapping algorithm. The encoding of the final data into the recommended ASN.1 format is messy, too. For more details, see [CMSRSAKEM].
We published our own set of test vectors for RSA-KEM in January 2008 (now out of date).
Ferguson-Schneier Encryption
In their book [FERG03], Niels Ferguson and Bruce Schneier suggest a much simpler method of encryption. They suggest using the same modulus n for both encryption and signatures but to use e=3 for signatures and e=5 for encryption. They use RSA to encrypt a random integer and use a hash function to derive the actual content encryption key, thus removing any structural similarities between the actual CEK and the data encrypted by the RSA. F&S recommend using the function, Hash(x):=SHA256(SHA256(x)), for hashing data.
Algorithm: Ferguson-Schneier Encrypt Random Key with RSA.
INPUT: Recipient's RSA public key, (n, e).
OUTPUT: Content encryption key, CEK; RSA-encrypted CEK, c.
Compute the exact bit length of the RSA key, k = ceiling(log2(n+1)).
Choose a random r in the interval [0, 2k-1].
Compute the content encryption key by hashing r, CEK=Hash(r).
c = re mod n.
Output CEK and c.
For a plaintext message, m, the transmission sent to the recipient is IntegerToString(c) || ECEK(m), where ECEK(m) is the result of encrypting m with a symmetrical encryption algorithm using key, CEK. Given that the recipient knows the size of the RSA key and hence the exact number of bytes needed to encode c, it is a simple matter to parse the input received from the sender.
For example code of this algorithm in Visual Basic (both VB6 and VB.NET) using our CryptoSys PKI Toolkit, see Ferguson-Schneier RSA Encryption.
Notation and Conventions
We use the following notation and conventions in this page.
A || B denotes concatenation of byte (or bit) strings A and B.
|B| denotes the length of the byte (or bit) string B in bytes.
|n| denotes the length of the non-negative integer n in bytes, |n| = ceiling(log256(n+1)).
IntegerToString(i, n) is an n-byte encoding of the integer i with the most significant byte first (i.e. in "big-endian" order). So, for example,
IntegerToString(1, 4)="00000001",
IntegerToString(7658, 3)="001DEA"
StringToInteger(S, n) is the integer represented by the byte string S of length n bytes with the most significant byte first.
ceiling(x) is the smallest integer, n, such that n ≥ x.
Implementation in C and VB
We show an example implementation of the RSA algorithm in C in our BigDigits library. It's not necessarily the most efficient way, and could be improved in its security, but it shows the maths involved. Look in the BigDigits Test Functions.
There is an example in VB6/VBA code at RSA and Diffie-Hellman in Visual Basic.
For a professional implementation, see our commercial CryptoSys PKI Toolkit which can be used with Visual Basic, VB6, VBA, VB2005+, C/C++ and C# applications. There are examples using the `raw' RSA functions to carry out RSA Encryption and RSA Signing.
