RSA – theory and implementation

RSA has been a staple of public key cryptography for over 40 years, and is
still being used today for some tasks in the newest TLS 1.3 standard. This
post describes the theory behind RSA – the math that makes it work, as well as
some practical considerations; it also presents a complete implementation of RSA
key generation, encryption and decryption in Go.

The RSA algorithm

The beauty of the RSA algorithm is its simplicity. You don’t need much more
than some familiarity with elementary number theory to understand it, and the
prerequisites can be grokked in a few hours.

In this presentation M is the message we want to encrypt, resulting in the
ciphertext C. Both M and C are large integers. Refer to the Practical
Considerations section for representing arbitrary data with such integers.

The RSA algorithm consists of three main phases: key generation, encryption and

Key generation

The first phase in using RSA is generating the public/private keys. This is
accomplished in several steps.

Step 1: find two random, very large prime numbers p and q and calculate
n=pq. How large should these primes be? The current recommendation is
for n to be at least 2048 bits, or over 600 decimal digits. We’ll assume that
the message M – represented as a number – is smaller than n (see Practical
Considerations for details on what to do if it’s not).

Step 2: select a small odd integer e that is relatively prime to
phi(n), which is Euler’s totient function. phi(n) is
calculated directly from Euler’s formula (its proof is on Wikipedia):

[phi(n) =n prod_{pmid n} left(1-frac{1}{p}right)]

For n=pq where p and q are primes, we get


In practice, it’s recommended to pick e as one of a set of known prime values,
most notably 65537. Picking this known
number does not diminish the security of RSA, and has some advantages such as
efficiency [1].

Step 3: compute d as the multiplicative inverse of e modulo
phi(n). Lemma 3 in this post guarantees
that d exists and is unique (and also explains what a modular multiplicative
inverse is).

At this point we have all we need for the public/private keys. The public key is
the pair [e,n] and the private key is the pair [d,n]. In
practice, when doing decryption we have access to n already (from the public
key), so d is really the only unknown.

Encryption and decryption

Encryption and decryption are both accomplished with the same modular

formula, substituting different values for x and y:


For encryption, the input is M and the exponent is e:


For decryption, the input is the ciphertext C and the exponent is d:


Why does it work?

Given M, we encrypt it by raising to the power of e modulo n. Apparently,
this process is reversible by raising the result to the power of d modulo
n, getting M back. Why does this work?



Recall that e and d are multiplicative inverses modulo phi(n). That
is, edequiv 1pmod{phi(n)}. This means that for some integer k we have
ed=1+kphi(n) or ed=1+k(p-1)(q-1).

Let’s see what M^{ed} is modulo p. Substituting in the formula for
ed we get:

[M^{ed}equiv M(M^{p-1})^{k(q-1)}pmod{p}]

Now we can use Fermat’s little theorem, which states that
if M is not divisible by p, we have M^{p-1}equiv 1pmod{p}. This
theorem is a special case of Euler’s theorem, the proof of which I wrote about

So we can substitute 1 for M^{p-1} in the latest equation, and raising 1
to any power is still 1:

[M^{ed}equiv Mpmod{p}]

Note that Fermat’s little theorem requires that M is not divisible by p. We
can safely assume that, because if Mequiv 0pmod{p}, then trivially
M^{ed}equiv 0pmod{p} and again M^{ed}equiv Mpmod{p}.

We can similarly show that:

[M^{ed}equiv Mpmod{q}]

So we have M^{ed}equiv M for the prime factors of n. Using
a corollary to the Chinese Remainder Theorem, they are
then equivalent modulo n itself:

[M^{ed}equiv Mpmod{n}]

Since we’ve defined M to be smaller than n, we’ve shown that
Dec(Enc(M))=M ∎

Why is it secure?

Without the private key in hand, attackers only have the result of
M^epmod {n}, as well as n and e (as they’re part of the public
key). Could they infer M from these numbers?

There is no known general way of doing this without factoring
n (see the original RSA paper,
section IX), and factoring is known to be a difficult problem. Specifically,
here we assume that M and e are sufficiently large that M^e>n
(otherwise decrypting would be trivial).

If factoring was easy, we could factor n into p and q, then compute
phi(n) and then finally find d from
edequiv 1pmod{phi(n)} using the extended Euclidean algorithm.

Practical considerations

The algorithm described so far is sometimes called textbook RSA (or
schoolbook RSA). That’s because it deals entirely in numbers, ignoring all
kinds of practical matters. In fact, textbook RSA is susceptible to several clever

and has to be enhanced with random padding schemes for practical use.

A simple padding scheme called PKCS #1 v1.5 has been used for many years and is
defined in RFC 2313. These days more
advanced schemes like OAEP are
recommended instead, but PKCS #1 v1.5 is very easy to explain and therefore I’ll
use it for didactic purposes.

Suppose we have some binary data D to encrypt. The approach works for data of
any size, but we will focus on just encrypting small pieces of data. In
practice this is sufficient because RSA is commonly used to only encrypt a
symmetric encryption key, which is much smaller than the RSA key size [2]. The
scheme can work well enough for arbitrary sized messages though – we’ll just
split it to multiple blocks with some pre-determined block size.

From D we create a block for encryption – the block has the same length as our
RSA key:

Here PS is the padding, which should occupy all the bytes not taken by the
header and D in the block, and should be at least 8 bytes long (if it’s
shorter, the data may be broken into two separate blocks). It’s a sequence
of random non-zero bytes generated separately for each encryption. Once we
have this full block of data, we convert it to a number treating the bytes
as a big-endian encoding [3]. We end up with a large number x, which we then
perform the RSA encryption step on with Enc(x)=x^epmod{n}. The result
is then encoded in binary and sent over the wire.

Decryption is done in reverse. We turn the received byte stream into a number,
perform Dec(C)=C^dpmod{n}, then strip off the padding (note that the
padding has no 0 bytes and is terminated with a 0, so this is easy) and get our
original message back.

The random padding here makes attacks on textbook RSA impractical, but the
scheme as a whole may still be vulnerable to
more sophisticated attacks
in some cases. Therefore, more modern schemes like OAEP should be used in

Implementing RSA in Go

I’ve implemented a simple variant of RSA encryption and
decryption as described in this post, in Go. Go makes it particularly easy to
implement cryptographic algorithms because of its great support for
arbitrary-precision integers with the stdlib big package. Not only does
this package support basics of manipulating numbers, it also supports several
primitives specifically for cryptography – for example the Exp method
supports efficient modular exponentiation, and the ModInverse method
supports finding modular multiplicative modular inverses. In addition, the
crypto/rand contains randomness primitives specifically designed for
cryptographic uses.

Go has a production-grade crypto implementation in the standard library. RSA is
in crypto/rsa, so for anything real please use that [4]. The code shown
and linked here is just for educational purposes.

The full code, with some tests, is available on GitHub. We’ll start by
defining the types to hold public and private keys:

type PublicKey struct {
N *big.Int
E *big.Int

type PrivateKey struct {
N *big.Int
D *big.Int

The code also contains a GenerateKeys function that will randomly generate
these keys with an appropriate bit length. Given a public key, textbook
encryption is simply:

func encrypt(pub *PublicKey, m *big.Int) *big.Int {
c := new(big.Int)
c.Exp(m, pub.E, pub.N)
return c

And decryption is:

func decrypt(priv *PrivateKey, c *big.Int) *big.Int {
m := new(big.Int)
m.Exp(c, priv.D, priv.N)
return m

You’ll notice that the bodies of these two functions are pretty much the same,
except for which exponent they use. Indeed, they are just typed wrappers around
the Exp method.

Finally, here’s the full PKCS #1 v1.5 encryption procedure, as described above:

// EncryptRSA encrypts the message m using public key pub and returns the
// encrypted bytes. The length of m must be <= size_in_bytes(pub.N) – 11,
// otherwise an error is returned. The encryption block format is based on
// PKCS #1 v1.5 (RFC 2313).
func EncryptRSA(pub *PublicKey, m []byte) ([]byte, error) {
// Compute length of key in bytes, rounding up.
keyLen := (pub.N.BitLen() + 7) / 8
if len(m) > keyLen11 {
return nil, fmt.Errorf(“len(m)=%v, too long”, len(m))

// Following RFC 2313, using block type 02 as recommended for encryption:
// EB = 00 || 02 || PS || 00 || D
psLen := keyLen len(m) 3
eb := make([]byte, keyLen)
eb[0] = 0x00
eb[1] = 0x02

// Fill PS with random non-zero bytes.
for i := 2; i < 2+psLen; {
_, err := rand.Read(eb[i : i+1])
if err != nil {
return nil, err
if eb[i] != 0x00 {
eb[2+psLen] = 0x00

// Copy the message m into the rest of the encryption block.
copy(eb[3+psLen:], m)

// Now the encryption block is complete; we take it as a m-byte big.Int and
// RSA-encrypt it with the public key.
mnum := new(big.Int).SetBytes(eb)
c := encrypt(pub, mnum)

// The result is a big.Int, which we want to convert to a byte slice of
// length keyLen. It’s highly likely that the size of c in bytes is keyLen,
// but in rare cases we may need to pad it on the left with zeros (this only
// happens if the whole MSB of c is zeros, meaning that it’s more than 256
// times smaller than the modulus).
padLen := keyLen len(c.Bytes())
for i := 0; i < padLen; i++ {
eb[i] = 0x00
copy(eb[padLen:], c.Bytes())
return eb, nil

There’s also DecryptRSA, which unwraps this:

// DecryptRSA decrypts the message c using private key priv and returns the
// decrypted bytes, based on block 02 from PKCS #1 v1.5 (RCS 2313).
// It expects the length in bytes of the private key modulo to be len(eb).
// Important: this is a simple implementation not designed to be resilient to
// timing attacks.
func DecryptRSA(priv *PrivateKey, c []byte) ([]byte, error) {
keyLen := (priv.N.BitLen() + 7) / 8
if len(c) != keyLen {
return nil, fmt.Errorf(“len(c)=%v, want keyLen=%v”, len(c), keyLen)

// Convert c into a bit.Int and decrypt it using the private key.
cnum := new(big.Int).SetBytes(c)
mnum := decrypt(priv, cnum)

// Write the bytes of mnum into m, left-padding if needed.
m := make([]byte, keyLen)
copy(m[keyLenlen(mnum.Bytes()):], mnum.Bytes())

// Expect proper block 02 beginning.
if m[0] != 0x00 {
return nil, fmt.Errorf(“m[0]=%v, want 0x00”, m[0])
if m[1] != 0x02 {
return nil, fmt.Errorf(“m[1]=%v, want 0x02”, m[1])

// Skip over random padding until a 0x00 byte is reached. +2 adjusts the index
// back to the full slice.
endPad := bytes.IndexByte(m[2:], 0x00) + 2
if endPad < 2 {
return nil, fmt.Errorf(“end of padding not found”)

return m[endPad+1:], nil

Digital signatures with RSA

RSA can be also used to perform digital signatures. Here’s how it works:

Key generation and distribution remains the same. Alice has a public key and
a private key. She publishes her public key online.
When Alice wants to send Bob a message and have Bob be sure that only she
could have sent it, she will encrypt the message with her private key,
that is S=Sign(M)=M^dpmod{n}. The signature is attached to the
When Bob receives a message, he can decrypt the signature with Alice’s
public key: Check(S)=S^epmod{n} and if he gets the original message
back, the signature was correct.

The correctness proof would be exactly the same as for encryption. No one else
could have signed the message, because proper signing would require having the
private key of Alice, which only she possesses.

This is the textbook signature algorithm. One difference between the practical
implementation of signing and encryption is in the padding protocol used. While
OAEP is recommended for encryption, PSS is recommended
for signing [5]. I’m not going to implement signing for this post, but the
Go standard library has great code for this – for example rsa.SignPKCS1v15
and rsa.SignPSS.

For two reasons: one is that we don’t have to randomly find another large
number – this operation takes time; another is that 65537 has only two
bits “on” in its binary representation, which makes modular
exponentiation algorithms faster

A strong AES key is 256 bits, while RSA is commonly 2048 or more. The
reason RSA encrypts a symmetric key is efficiency – RSA encryption is
much slower than block ciphers, to the extent that it’s often impractical
to encrypt large streams of data with it. A hybrid scheme – wherein a
strong AES key is first encrypted with RSA, and then AES is used to
encrypt large data – is very common. This is the general idea behind what
TLS and similar secure protocols use.

Note that the first 8 bits of the data block are 0, which makes it easy
to ensure that the number we encrypt is smaller than n.

The stdlib implementation is resilient to common kinds of side-channel
attacks, such as using algorithms whose run time is independent of
certain characteristics of the input, which makes timing attacks less

The reason for a different protocol is that the attacks on
encrypted messages and on signatures tend to be different. For example,
while for encrypted messages it’s unthinkable to let attackers know any
characteristics of the original message (the base in the
exponentiation), in signing it’s usually plainly available.

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