RSA Algorithm
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 cryptosystem is the most widelyused public key cryptography algorithm in the world. It can be used to encrypt a message without the need to exchange a secret key separately.
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.
Party A can send an encrypted message to party B without any prior exchange of secret keys. A just uses B's public key to encrypt the message and B decrypts it using the private key, which only he knows. RSA can also be used to sign a message, so A can sign a message using their private key and B can verify it using A's public key.
We look into the mathematics behind the algorithm on our RSA Theory page.
Contents
Recommended reading
 Cryptography Engineering by Niels Ferguson, Bruce Schneier and Tadayoshi Kohno. Look Inside
 Handbook of Applied Cryptography by Alfred J. Menezes, Paul C. van Oorschot and Scott A. Vanstone
 Cryptanalysis of RSA and Its Variants by M. Jason Hinek
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Affiliate disclosure: we get a small commission for purchases made through the above links
 Key generation algorithm
 A practical key generation algorithm
 Encryption
 Decryption
 Digital signing
 Signature verification
 Notes on practical application
 Summary of RSA
 Theory and proof of the RSA algorithm
 Key length
 Computational efficiency and the Chinese Remainder Theorem
 A very simple example
 A slightly less simple example
 A real example
 PKCS#1 Schemes
 Weaknesses in RSA
 More Advanced Schemes
 Notation and Conventions
 What is the difference between a bit string and an integer?
 Implementation in C and VB
 References
 Author
 Contact
 Comments
Key Generation Algorithm
This is the original 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 $\phi = (p1)(q1)$. [See note 6].
 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 $ed \equiv 1 \bmod \phi$. [See note 3].
 The public key is $(n, e)$ and the private key $(d, p, q)$. Keep all the values d, p, q and $\phi$ secret. [Sometimes the private key is written as $(n, d)$ because you need the value of n when using d. Other times we might write the key pair as $((N, e), d)$.]
 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.
A practical key generation algorithm
Incorporating the advice given in the notes below, a practical algorithm to generate an RSA key pair is given below. Typical bit lengths are $k = 1024, 2048, 3072, 4096,...$, with increasing computational expense for larger values. You will not go far wrong if you choose $e$ as 65537 (=0x10001) in step (1).
Algorithm: Generate an RSA key pair.
INPUT: Required modulus bit length, $k$.
OUTPUT: An RSA key pair $((N,e), d)$ where N is the modulus, the product of two primes ($N=pq$) not exceeding $k$ bits in length; $e$ is the public exponent, a number less than and coprime to $(p1)(q1)$; and $d$ is the private exponent such that $ed \equiv 1 \bmod {(p1)(q1)}$.
 Select a value of $e$ from ${3, 5, 17, 257, 65537}$
 repeat
 p ← genprime(k/2)
 until $(p \bmod e) \ne 1$
 repeat
 q ← genprime(k  k/2)
 until $(q \bmod e) \ne 1$
 N ← pq
 L ← (p1)(q1)
 d ← modinv(e, L)
 return $(N, e, d)$
The function genprime(b)
returns a prime of exactly $b$ bits, with the $b$th bit set to 1.
Note that the operation $k/2$ is integer division giving the integer quotient with no fraction.
If you've chosen $e = 65537$ then the chances are that the first prime returned in steps (3) and (6) will pass the tests in steps (4) and (7), so each repeatuntil loop will most likely just take one iteration. The final value of $N$ may have a bit length slightly short of the target $k$. This actually does not matter too much (providing the message m is always < N), but some schemes require a modulus of exact length. If this is the case, then just repeat the entire algorithm until you get one. It should not take too many goes. Alternatively, use the trick setting the two highest bits in the prime candidates described in note 1.
Encryption
Sender A does the following:
 Obtains the recipient B's public key $(n, e)$.
 Represents the plaintext message as a positive integer $m$ with $1 < m < n$ [see note 4].
 Computes the ciphertext $c = m^e \bmod n$.
 Sends the ciphertext $c$ to B.
Decryption
Recipient B does the following:
 Uses his private key $(n, d)$ to compute $m = c^{d} \bmod n$.
 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 1 and $n1$ [See note 5].
 Uses her private key $(n, d)$ to compute the signature $s = m^{d} \bmod n$.
 Sends this signature $s$ to the recipient, B.
Signature verification
 Uses sender A's public key $(n, e)$ to compute integer $v = s^{e} \bmod n$.
 Extracts the message digest $H$ from this integer.
 Independently computes the message digest $H'$ of the information that has been signed.
 If both message digests are identical, i.e. $H=H'$, the signature is valid.
 Uses sender A's public key $(n, e)$ to compute integer $v = s^{e} \bmod n$.
 Independently computes the message digest $H'$ of the information that has been signed.
 Computes the expected representative integer $v'$ by encoding the expected message digest $H'$.
 If $v=v'$, the signature is valid.
Notes on practical applications
 To generate the primes $p$ and $q$,
generate a random number of bit length $k/2$ where
$k$ is the required bit length of the modulus $n$; 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 RabinMiller 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 $kk/2$.
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 certain 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, 5, 17, 257 and 65537 $(2^{16}+1)$.
These particular values are chosen because they are primes and make the modular exponentiation operation faster, having only two bits of value 1.
> Aside: These five numbers are the first five Fermat numbers, referred to as $F_{0}$ to $F_{4}$, where $F_x=2^{2^x} + 1$. Just be careful, these first five Fermat numbers are prime ("Fermat primes"), but the numbers $F_{5}$ and above are not prime. For example, $F_{5} = 4294967297 = 641 \times 6700417$.The usual choice for $e$ is $F_{4} = 65537$ = 0x10001. Also, having chosen $e$, it is simpler to test whether $\gcd(e, p1)=1$ and $\gcd(e, q1)=1$ 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 an odd prime then you can do the lessexpensive test $(p \bmod e) \ne 1$ instead of $\gcd(p1,e) = 1$.Why is that? If $e$ is an odd prime then $\gcd(p1, e) \ne 1$ if and only if $p1$ is a multiple of $e$. If $p1$ is a multiple of $e$ then $p  1 \equiv 0 \pmod e$ or $p \equiv 1 \pmod e$. Conversely, if $p\not\equiv 1 \pmod e$ then $p1\not\equiv 0 \pmod e$ and $p1$ is not a multiple of $e$. So $\gcd(p1, e)=1$.
 To compute the value for $d$,
use the Extended Euclidean Algorithm to calculate
$d = e^{1} \bmod \phi$, also written $d = (1/e) \bmod \phi$.
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 $ed = 1 \bmod \phi$.
It only exists if $e$ and $\phi$ have no common factors.
For more details of the extended Euclidean algorithm, see our 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 important 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 n1 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 integer $m$, 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 $m < n$,
$m = (m^{e}\bmod n)^{d} \bmod n = (m^{d} \bmod n)^{e} \bmod n$
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).
 The original definition of RSA uses the Euler totient function
$\phi(n) = (p1)(q1)$. More recent standards use the Charmichael function $\lambda(n) = \text{lcm}(p1, q1)$ instead.
$\lambda(n)$ is smaller than $\phi(n)$ and divides it. The value of $d'$ computed by $d' = e^{1} \bmod \lambda(n)$
is usually different from that derived by $d = e^{1} \bmod \phi(n)$, but the end result is the same.
Both $d$ and $d'$ will decrypt a message $m^{e} \bmod n$ and both will give the same signature value
$s = m^{d} \bmod n = m^{d'} \bmod n$. To compute $\lambda(n)$, use the relation
$\lambda(n) = \dfrac{(p1)(q1)}{\gcd(p1, q1)}$.

You might ask if there is a way to find the factors of $n$ given just $d$ and $e$.
This is possible.
For more details, see our page RSA: how to factorize N given d.
Summary of RSA
 $n = pq$, where $p$ and $q$ are distinct primes.
 $\phi = (p1)(q1)$
 $e \lt n$ such that $\gcd(e, \phi)=1$
 $d = e^{1} \bmod \phi$
 $c = m^{e} \bmod n, 1\lt m \lt n$
 $m = c^{d} \bmod n$
For more on the theory and mathematics behind the algorithm, see the RSA Theory page.
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 at least 1024 bits. A key length of 512 bits is no longer considered secure, although cracking it is still not a trivial task for the likes of you and me. The longer your information needs 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 one small 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 = $\lceil \log_{2}(n+1))\rceil $, where $\lceil x\rceil$ is the ceiling function, the least integer greater than or equal to $x$. The other convention, sometimes used, is that the key length is the number of bytes needed to store $n$ multiplied by eight, i.e. $\lceil\log_{256}(n+1)\rceil \times 8$.
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 ABThe 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 [NIST80057]. 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  SHA1  80 
3TDEA  2048  SHA224  112 
AES128  3072  SHA256  128 
AES192  7680  SHA384  192 
AES256  15360  SHA512  256 
Note just how huge (and impractical) an RSA key needs to be for comparable security with AES192 or AES256 (although these two algorithms have had some weaknesses exposed recently; AES128 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 $y = x^{e} \bmod n$ is known as modular exponentiation and one efficient method to carry this out on a computer is the binary lefttoright method. To solve, let $e$ be represented in base 2 as
$e = e_{k1}e_{k2}\ldots e_{1}e_{0}$
where $e_{k1}$ is the most significant nonzero 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 $c = m^{e} \bmod n$. 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 will not be secure.
An alternative method of representing the private key uses the The Chinese Remainder Theorem (CRT).
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 $m = c^{d} \bmod n$. The extra values for the private key are:
dP = (1/e) mod (p1) dQ = (1/e) mod (q1) qInv = (1/q) mod p where p > q
where the (1/e)
notation means the modular inverse (see note 3 above).
These values are precomputed and saved along with $p$ and $q$ 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.
[20080902] 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);
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 55 can probably even do it by hand).
 Select primes p=11, q=3.
 n = pq = 11.3 = 33
phi = (p1)(q1) = 10.2 = 20  Choose e=3
Check gcd(e, p1) = gcd(3, 10) = 1 (i.e. 3 and 10 have no common factors except 1),
and check gcd(e, q1) = gcd(3, 2) = 1
therefore gcd(e, phi) = gcd(e, (p1)(q1)) = gcd(3, 20) = 1  Compute d such that ed ≡ 1 (mod phi)
i.e. compute d = (1/e) mod phi = (1/3) mod 20
i.e. find a value for d such that phi divides (ed1)
i.e. find d such that 20 divides 3d1.
Simple testing (d = 1, 2, ...) gives d = 7
Check: ed1 = 3.7  1 = 20, which is divisible by phi.  Public key = (n, e) = (33, 3)
Private key = (n, d) = (33, 7).
$c = m^e \bmod n = 7^3 \bmod 33 = 343 \bmod 33 = 13$.Hence the ciphertext c = 13.
$m' = c^d \bmod n = 13^7 \bmod 33 = 7$.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
$a = bc \bmod n = (b \bmod n)\cdot (c \bmod n) \bmod n$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.
Note that any number can be expressed as a sum of powers of 2. In particular 7 = 4 + 2 + 1.
So first compute values of $13^{2}, 13^{4}, 13^{8}, \ldots$ by repeatedly squaring successive values modulo 33.
$13^{2} = 169 \equiv 4, 13^{4} = 4.4 = 16, 13^{8} = 16.16 = 256 \equiv 25$.
Then, since 7 = 4 + 2 + 1, we have $m' = 13^{7} = 13^{(4+2+1)} = 13^{4}.13^{2}.13^{1}$
$\equiv 16 \times 4 \times 13 = 832 \equiv 7 \pmod{33}$
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 n1 will always do this for any $n$, no matter how large. But in practice, these 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 $m_{1}, m_{2}, \ldots$
(9,6,13,13,16,24,16,19,13,5)Using our table above, we obtain ciphertext integers $c_{1}, c_{2}, \ldots$
(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 $m^{e} \bmod n$ is easy, but calculating the inverse $c^{e} \bmod n$ 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. Please note that this method is not secure in any way. It just shows another example of the mechanism of RSA with small numbers.
For this example, to keep things simple, we'll limit our characters to the letters A to Z and the space character.
ATTACK AT SEVEN = ATT ACK _AT _SE VEN
In the same way that any decimal number can be represented uniquely as the sum of powers of ten, e.g.
$135 = 1 \times 10^{2} + 3 \times 10^{1} + 5$,
we can represent our blocks of three characters as the sum of powers of 27 using
SPACE=0, A=1, B=2, C=3, .. E=5, .. K=11, .. N=14, .. S=19, T=20, .. V=22, ..., Z=26.
ATT ⇒ 1 x 27^2 + 20 x 27^1 + 20 = 1289 ACK ⇒ 1 x 27^2 + 3 x 27^1 + 11 = 821 _AT ⇒ 0 x 27^2 + 1 x 27^1 + 20 = 47 _SE ⇒ 0 x 27^2 + 19 x 27^1 + 5 = 518 VEN ⇒ 22 x 27^2 + 5 x 27^1 + 14 = 16187
Using this system of integer representation, the maximum value of a block (ZZZ) is $27^{3}1 = 19682$, so we require a modulus n greater than this value.
 We choose e = 3
 We select primes p=173 and q=149 and check
 gcd(e, p1) = gcd(3, 172) = 1 ⇒ OK
 gcd(e, q1) = gcd(3, 148) = 1 ⇒ OK.
 Thus we have $n = pq = 173 \times 149 = 25777$, and
$\phi = (p1)(q1) = 172 \times 148 = 25456$.  We compute $d = e^{1} \bmod \phi = 3^{1} \bmod 25456 = 16971$.
 Note that $ed = 3 \times 16971 = 50913 = 2 \times 25456 + 1$
 That is, $ed \equiv 1 \bmod 25456 \equiv 1 \bmod \phi$
 Hence our public key is (n, e) = (25777, 3) and our private key is (n, d) = (25777, 16971). We keep the values of p, q, d and φ secret.
To encrypt the first integer that represents "ATT", we have
$c = m^{e} \bmod n = 1289^{3} \bmod 25777 = 18524$.
Overall, our plaintext ATTACK AT SEVEN
is represented by the sequence of five integers
$m_{1}, m_{2}, m_{3}, m_{4}, m_{5}$:
m_i = (1289, 821, 47, 518, 16187)
We compute corresponding ciphertext integers $c_{i} = m_{i}^{e} \bmod n$, (which is still possible by using a calculator, honest):
$c_{1} = 1289^{3} \bmod 25777 = 18524 \\ c_{2} = 821^{3} \bmod 25777 = 7025 \\ c_{3} = 47^{3} \bmod 25777 = 715 \\ c_{4} = 518^{3} \bmod 25777 = 2248 \\ c_{5} = 16187^{3} \bmod 25777 = 24465 $
We can send this sequence of integers, $c_{i}$, to the person who has the private key.
c_i = (18524, 7025, 715, 2248, 24465)
We can compute the inverse of these ciphertext integers using $m = c^{d} \bmod n$ to verify that the RSA algorithm still holds. However, this is now outside the realm of hand calculations.
To help you carry out these modular arithmetic calculations, download our free modular arithmetic command line programs.
For example, to compute $18524^{16971} \bmod 25777$, use the bd_modexp
command:
bd_modexp 18524 16971 25777 18524^16971 mod 25777 = 1289
$ m_{1} = 18524^{16971} \bmod 25777 = 1289 \\ m_{2} = 7025^{16971} \bmod 25777 = 821 \\ m_{3} = 715^{16971} \bmod 25777 = 47 \\ m_{4} = 2248^{16971} \bmod 25777 = 518 \\ m_{5} = 24465^{16971} \bmod 25777 = 16187 \\ $
16187 ÷ 27^2 = 16187 ÷ 729 = 22 rem 149, 22 → 'V' 149 ÷ 27^1 = 149 ÷ 27 = 5 rem 14, 5 → 'E' 14 ÷ 27^0 = 14 ÷ 1 = 14 rem 0, 14 → 'N'Hence the integer m = 16187 represents the string "VEN".
47 ÷ 27^2 = 0 rem 47, 0 → SPACE; 47 ÷ 27^1 = 1 rem 20, 1 → 'A'; 20 ÷ 27^0 = 20 rem 0, 20 → 'T'giving the string "_AT".
Question: Why can't we use this method of encoding integers into blocks of three letters to encode the ciphertext?
A caution about this example
Note that this example is a very insecure method of encryption and should not be used in practice. We can easily factorize the modulus and hence break the cipher.
Factorising a small RSA modulus
Starting with the knowledge that the modulus 25777 is the product of exactly two distinct prime numbers, and that one of these must be less than its integer square root (why?), a little testing of suitable candidates from a table of prime numbers will get you the answer pretty quickly.
Given n = 25777, compute √25777 = 160.55, and then work downwards through the prime numbers < 160, i.e. (157, 151, 149, 139, ...),
and try to divide into $n$ in turn:
157: 25777 / 157 = 164 remainder 29, so not a factor;
151: 25777 / 151 = 170 remainder 107, so not a factor;
149: 25777 / 149 = 173 exactly, so we have it.
You could also write a simple computer program to factor n that just divides it by every odd number starting from 3 until it either finds an exact factor or stops when it reaches a number greater than the square root of n.
A real example
In practice, we use a modulus of size in the order of 1024 bits. That is a number over 300 decimal digits long. One example is
n = 11929413484016950905552721133125564964460656966152763801206748195494305685115033 38063159570377156202973050001186287708466899691128922122454571180605749959895170 80042105263427376322274266393116193517839570773505632231596681121927337473973220 312512599061231322250945506260066557538238517575390621262940383913963
This is composed of the two primes
p = 10933766183632575817611517034730668287155799984632223454138745671121273456287670 008290843302875521274970245314593222946129064538358581018615539828479146469 q = 10910616967349110231723734078614922645337060882141748968209834225138976011179993 394299810159736904468554021708289824396553412180514827996444845438176099727
With a number this large, we can encode all the information we need in one big integer. We put our message into an octet string 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 AES128. 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 conventionallyencrypted 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 oneway hash function like SHA1 or SHA256 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 508bit 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<=k11
.
OUTPUT: Encrypted data block of length k bytes
 Form the kbyte encoded message block, EB,
EB = 00  02  PS  00  D
where
denotes concatenation and PS is a string ofkD3
nonzero randomlygenerated bytes (i.e. at least eight random bytes).  Convert the byte string, EB, to an integer, m, most significant byte first,
m = StringToInteger(EB)
 Encrypt with the RSA algorithm
c = m^e mod n
 Convert the resulting ciphertext, $c$, to a $k$byte output block, OB^{‡}
OB = IntegerToString(c, 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. See What is the difference between a bit string and an integer?
‡20120523: Thanks to "dani torwS" for pointing out a typo in the formula.
Worked Example
Bob's 1024bit RSA encryption key in hex format:
n= A9E167983F39D55FF2A093415EA6798985C8355D9A915BFB1D01DA197026170F BDA522D035856D7A986614415CCFB7B7083B09C991B81969376DF9651E7BD9A9 3324A37F3BBBAF460186363432CB07035952FC858B3104B8CC18081448E64F1C FB5D60C4E05C1F53D37F53D86901F105F87A70D1BE83C65F38CF1C2CAA6AA7EB e=010001 d= 67CD484C9A0D8F98C21B65FF22839C6DF0A6061DBCEDA7038894F21C6B0F8B35 DE0E827830CBE7BA6A56AD77C6EB517970790AA0F4FE45E0A9B2F419DA8798D6 308474E4FC596CC1C677DCA991D07C30A0A2C5085E217143FC0D073DF0FA6D14 9E4E63F01758791C4B981C3D3DB01BDFFA253BA3C02C9805F61009D887DB0319A randomlygenerated oneoff session key for AES128 might be
D=4E636AF98E40F3ADCFCCB698F4E80B9F
The encoded message block, EB, after encoding but before encryption, with random padding bytes shown in green,
0002257F48FD1F1793B7E5E02306F2D3228F5C95ADF5F31566729F132AA12009 E3FC9B2B475CD6944EF191E3F59545E671E474B555799FE3756099F044964038 B16B2148E9A2F9C6F44BB5C52E3C6C8061CF694145FAFDB24402AD1819EACEDF 4A36C6E4D2CD8FC1D62E5A1268F496004E636AF98E40F3ADCFCCB698F4E80B9FAfter RSA encryption, the output is
3D2AB25B1EB667A40F504CC4D778EC399A899C8790EDECEF062CD739492C9CE5 8B92B9ECF32AF4AAC7A61EAEC346449891F49A722378E008EFF0B0A8DBC6E621 EDC90CEC64CF34C640F5B36C48EE9322808AF8F4A0212B28715C76F3CB99AC7E 609787ADCE055839829E0142C44B676D218111FFE69F9D41424E177CBA3A435B
The above hex data in C format.
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=48656C6C6F20776F726C6421Then, using the 128bit session key from above,
KY=4E636AF98E40F3ADCFCCB698F4E80B9Fand the uniquelygenerated 128bit initialization vector (IV)
IV=5732164B3ABB6C4969ABA381C1CA75BAthe ciphertext using AES128 in CBC mode with PKCS#5 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: aes128cbc IV: 5732164B3ABB6C4969ABA381C1CA75BA Encrypted Content: 67290EF00818827C777929A56BC3305B
The usual formats used for such a message are either a CMS envelopeddata 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 lessambiguous version of the earlier PKCS#7 standard (also of the same name) and is designed to be used in S/MIME messages. CMS envelopeddata objects (yes, that's enveloped not encrypted) use ASN.1 and are encoded using either DER or BERencoding. (DERencoding is a stricter subset of BER).
The terminology for CMS and ASN.1 may sound messy, but the end results are welldefined and universallyaccepted. 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
SHA1 30 21 30 09 06 05 2b 0e 03 02 1a 05 00 04 14  H
SHA224 30 2d 30 0d 06 09 60 86 48 01 65 03 04 02 04 05 00 04 1c  H
SHA256 30 31 30 0d 06 09 60 86 48 01 65 03 04 02 01 05 00 04 20  H
SHA384 30 41 30 0d 06 09 60 86 48 01 65 03 04 02 02 05 00 04 30  H
SHA512 30 51 30 0d 06 09 60 86 48 01 65 03 04 02 03 05 00 04 40  H
 Form the kbyte 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 thatEB=k
.  Convert the byte string, EB, to an integer m, most significant byte first,
m = StringToInteger(EB)
 Sign with the RSA algorithm
s = m^d mod n
 Convert the resulting signature value, s, to a kbyte output block, OB
OB = IntegerToString(s, k)
 Output OB.
Worked Example
Alice's 1024bit RSA signing key in hex format:
n= E08973398DD8F5F5E88776397F4EB005BB5383DE0FB7ABDC7DC775290D052E6D 12DFA68626D4D26FAA5829FC97ECFA82510F3080BEB1509E4644F12CBBD832CF C6686F07D9B060ACBEEE34096A13F5F7050593DF5EBA3556D961FF197FC981E6 F86CEA874070EFAC6D2C749F2DFA553AB9997702A648528C4EF357385774575F e=010001 d= 00A403C327477634346CA686B57949014B2E8AD2C862B2C7D748096A8B91F736 F275D6E8CD15906027314735644D95CD6763CEB49F56AC2F376E1CEE0EBF282D F439906F34D86E085BD5656AD841F313D72D395EFE33CBFF29E4030B3D05A28F B7F18EA27637B07957D32F2BDE8706227D04665EC91BAF8B1AC3EC9144AB7F21The message to be signed is, of course,
M="abc"that is, the 3 bytes in hex format,
M=616263The message digest algorithm is SHA1, so
H = Hash("abc") = A9993E364706816ABA3E25717850C26C9CD0D89D
The DigestInfo value for SHA1 is
T= 3021300906052B0E03021A05000414A9993E364706816ABA3E25717850C26C9C D0D89DThe encoded message block, EB, after encoding but before signing is
0001FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF00302130 0906052B0E03021A05000414A9993E364706816ABA3E25717850C26C9CD0D89DAfter RSA signing, the output is
60AD5A78FB4A4030EC542C8974CD15F55384E836554CEDD9A322D5F4135C6267 A9D20970C54E6651070B0144D43844C899320DD8FA7819F7EBC6A7715287332E C8675C136183B3F8A1F81EF969418267130A756FDBB2C71D9A667446E34E0EAD 9CF31BFB66F816F319D0B7E430A5F2891553986E003720261C7E9022C0D9F11F
The above hex data in C format.
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.For an example of this, see Cracking RSA on our page on the The Chinese Remainder Theorem.  Small encryption exponent and small message
 If you use $e=3$ and just encrypt a small message $m$ without padding where $m^{3} < n$ then your ciphertext $c$ can easily be broken by simply computing its real cube root. For example, if we have the public key $(n, e) = (25777, 3)$ and just encrypt the small message $m = 10$ then the ciphertext is $c = 1000$. The secure properties of RSA encryption only work if $m^{e} > n$.
 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.
 Using a common modulus for different users
 Do not use the same modulus $n$ with different $(e_{i}, d_{i})$ pairs for different users in a group.
Given his own pair $(e_{1}, d_{1})$, user 1 can factorize the common $n$ into
$p$ and $q$ and hence compute the private exponents $d_{i}$ of all the other users.
For more details, see our page RSA: how to factorize N given d.
 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.
 Don't encrypt or sign a blind message.
 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 message, 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 operations
$c = m^{e} \bmod n, \; m' = c^{d} \bmod n; \quad s = m^{d} \bmod n, \; s' = s^{e} \bmod n$
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.
RSAESOAEP
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. In fact, hardly at all. That said, if you have a choice in the matter, we recommend that you should use OAEP if you can.
RSASSAPSS
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 were active patents associated with this method until recently and so it is still not supported well. There are currently no known weaknesses with the PKCS#1v1.5 signature scheme.
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 IFSPRSA2. 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 SHA1.
The byte string, EB, is converted to an integer value, the message representative, f.
† ISO/IEC 10118 part numbers for other hash functions are: SHA1=0x33, SHA256=0x34, SHA384=0x36, SHA512=0x35, RIPEMD=0x31.
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 $0 \le f < n$ and $f ≡ 12 \pmod{16}$.
OUTPUT: Signature, an integer s where $0 \le s < n/2$, i.e. a value at least one bit shorter than $n$.
 $t = f^{d} \bmod n$
 $s = \min(t, nt)$
 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 $f \equiv 12 \pmod{16}$, or "invalid signature".
 If $s$ is not in $[0,(n1)/2]$, output "invalid signature" and stop.
 Compute $t = s^{e} \bmod n$
 If $t \equiv 12 \pmod{16}$ then let $f = t$.
 Else let $f = nt$. If $f \not\equiv 12 \pmod{16}$, 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  0xCCIf 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 SHA1. 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 $f \equiv 6 \pmod{16}$, instead of modulo 12. In other words, make sure the last 4 bits are equal to 0x6 instead of 0xC.
RSAKEM
The RSAKEM Key Transport Algorithm encrypts a random integer with the recipient's public key, and then uses a symmetric keywrapping 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 $n1$.
 Encrypt the integer $z$ with the recipient's RSA public key: $c = z^{e} \bmod n$.
 Derive a keyencrypting 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 keyencrypting 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) and a key wrapping algorithm. The encoding of the final data into the recommended ASN.1 format is messy, too. For more details, see the latest version of [CMSRSAKEM].
FergusonSchneier 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. You need to make sure that the modulus $n = pq$ satisfies both $\gcd(3,(p1)(q1))=1$ and $\gcd(5,(p1)(q1))=1$.
Their method uses 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.
They recommend using the function,
Hash(x):=SHA256(SHA256(x))
, for hashing data.
Algorithm: FergusonSchneier Encrypt Random Key with RSA.
INPUT: Recipient's RSA public key, (n, e).
OUTPUT: Content encryption key, CEK; RSAencrypted CEK, $c$.
 Compute the exact bit length of the RSA key, $k = \lceil log_{2}(n+1)\rceil$.
 Choose a random $r$ in the interval $[0, 2^{k}1]$.
 Compute the content encryption key by hashing $r$, CEK=Hash(r).
 $c = r^{e} \bmod n$.
 Output CEK and $c$.
For a plaintext message, $m$, the transmission sent to the recipient is
IntegerToString(c)E_CEK(m)
,
where $E_{CEK}(m)$ 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 FergusonSchneier 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 nonnegative integer n in bytes, $n = \lceil log_{2}(n+1)\rceil$.
IntegerToString(i, n)
is an nbyte encoding of the integer i with the most significant byte first (i.e. in "bigendian" order). So, for example,IntegerToString(1,4)="00000001" IntegerToString(7658,3)="001DEA"
StringToInteger(S)
is the integer represented by the byte string S with the most significant byte first. $\lceil x\rceil$ is the smallest integer, $n$, such that $n \ge x$.
** Strictly speaking, this is the length of the shortest byte string that can encode the integer.
What is the difference between a bit string and an integer?
A string is a contiguous sequence of symbols, so the string "cat" is a sequence of the letters 'c', 'a' and 't'. A bit string is sequence of binary digits (bits) '0' and '1'. A byte string is similar except it consists of bytes, which are in turn sequences of 8 bits. So a bit string and a byte string are the same thing, except the latter is restricted to multiples of 8 bits. For example, using hexadecimal representation, the byte string "8002EA" is a sequence of 3 bytes, 0x80, 0x02 and 0xEA; and is equal to the bit string "100000000000001011101010".
A note on notation: To differentiate between byte strings and integers here, we show byte strings in hexadecimal representation inside quotes, so "8002EA" denotes the string of three consecutive bytes with hexadecimal values 80, 02 and EA, respectively. We use the "0x" prefix to denote integers, so 0x80 denotes the integer with hexadecimal value 80 (decimal 128), and 0x8002EA denotes the integer with hexadecimal value 8002EA (decimal 8389354). Everything here is in "bigendian" order, with the leftmost bits the most significant.
A string can be split into substrings (e.g. "8002" and "EA") and two strings can be concatenated (joined up) to make another string. The order of symbols is important. The usual convention is to write byte strings with the most significant byte first ("bigendian" or network byte order).
An integer is a whole number that obeys the usual rules of integer arithmetic ($1+1=2,\; 52=3,\; 3\times 2=6,\; 6/3=2$) and modular arithmetic ($10+6\equiv 4 \pmod{12})$). There is no limit in theory as to how large an integer can be: you can always add one to any integer. The integer 8,389,354 in decimal is the same as the number 0x8002EA in hexadecimal notation, but is not the same as the byte string "8002EA", even though it looks the same and may well be stored in your computer in the same form.
You can increment the integer 8389354 to get 8389354+1=8389355 (0x8002EB); but you cannot "increment" the byte string "8002EA". On the other hand, you can concatenate the byte strings "8002" and "EA" to make "8002EA"; but the integers 8389 and 354 do not add to make 8389354. The byte string "00008002EA" is strictly not the same as "8002EA" (the former has two extra bytes of value 0x00 at the start); but the integers 0x008002EA and 0x8002EA are equal (leading zeros do not count).
With RSA encryption, we typically want to encrypt a session key which is a bit string, but the RSA operation $c = m^e \bmod n$ is done with integers, so we need to represent the bit string as an integer first (in practice, we usually add some random bytes and other padding, but we'll ignore that for the time being). Once the RSA operation has been completed, we have another integer, c, but we need to store the result as a bit string, so we encode the integer as a bit (byte) string and pass that string onto our recipient.
Example: Suppose we wish to encrypt the 3byte/24bit key bit string "8002EA" using the RSA public key (n=25009997=0x017D9F4D, e=5)^{†}. For simplicity in this insecure example, we will use the basic RSA algorithm with no padding.
 The message block is the byte string "8002EA".
 Compute the message representative
m = StringToInteger("8002EA") = 8389354
 Encrypt with the RSA algorithm
c = 8389354^5 mod 25009997 = 2242555
 Encode the result as a byte string
OB = IntegerToString(2242555, 4) = 002237FB
Note that the maximum length of the output block is 4 bytes, because the largest possible integer result is 0x017D9F4C ($= n1$), which requires 4 bytes to store in encoded form.
† Thanks to "doctorjay" for pointing out that e=3 did not work for the earlier version of this example.
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 DiffieHellman 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.
References
 [AX931] ANSI X9.311998 Digital Signatures using Reversible Public Key Cryptography for the Financial Services Industry (rDSA), Appendix A, American National Standards Institute, 1998.
 [CMS] RFC 5652. Cryptographic Message Syntax (CMS), R. Housley, September 2009 (obsoletes RFC3852, RFC3369, RFC2630).
 [CMSRSAKEM] RFC 5990 Use of the RSAKEM Key Transport Algorithm in the Cryptographic Message Syntax (CMS), J. Randall, B.Kaliski, J. Brainard, S. Turner. September 2010.
 [COCK73] Clifford Cocks. A Note on 'NonSecret Encryption', CESG Research Report, 20 November 1973.
 [EDIFACT] UN/EDIFACT Finance Group D6 SWGF. Recommended Practice For Message Flow And Security For EDIFACT Payments, Version 2v03, 1 October 2000.
 [FERG03] Niels Ferguson and Bruce Schneier, Practical Cryptography, Wiley, 2003. Note that this book has since been reissued in 2010 almost unchanged as Cryptography Engineering by Niels Ferguson, Bruce Schneier and Tadayoshi Kohno.
 [KALI93] Burton Kalinski. Some Examples of the PKCS Standards, RSA Laboratories, 1999, <link>.
 [NIST80057] NIST Special Publication 80057, Recommendation for Key Management  Part 1: General (Rev. 4), Elaine Barker et al, National Institute of Standards and Technology, January 2016, <link>.
 [P1363] IEEE P1363 Standard Specifications for Public Key Cryptography, IEEE, November 1993.
 [PKCS1] RSA Laboratories. PKCS #1 Cryptography Standard Version 2.2, October 2012. Republished as RFC 8017.
 [RIVE78] R. Rivest, A. Shamir and L. Adleman. A Method for Obtaining Digital Signatures and PublicKey Cryptosystems. Communications of the ACM, 21 (2), pp. 120126, February 1978, <link>.
Author
The content of this page is all original work written by David Ireland, who reserves all intellectual rights. You may freely link to this page. You may use parts of the work for fair dealing for the purposes of research or private study as permitted under copyright law, but you may not post any part of this content on another web site without the explicit permission in writing of the author.
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This page last updated 9 June 2018
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