Victorian public key cryptography

Electronic computers were invented before public key cryptography. Would public key cryptography have been possible before computers?

The security of RSA encryption depends on the ratio of the difficulty of factoring relative to the difficulty of multiplication. This ratio was high, maybe higher, before modern computers.

Suppose the idea of RSA encryption had occurred to Lewis Carroll (1832–1898). What key size would have been necessary for security in his day? Would it have been practical to manually encrypt data using keys of that size?

I imagine if you handed Victorians a pair of public and private RSA keys, they could have manually carried out public key encryption. But coming up with private keys, i.e. finding pairs of large prime numbers, would be harder than carrying out the encryption process.

The largest prime discovered without a computer was 2127 − 1, proved prime by Edouard Lucas in 1876. Such primes would have been large enough—I doubt it was feasible to factor the product of 40-digit primes at the time—but this was a prime of a very special form, a Mersenne prime. Lucas had developed an algorithm for efficiently testing whether a Mersenne number is prime. To this day the largest known primes are Mersenne primes. More on this here.

Lucas would not have been able to produce two 40-digit primes. The largest known prime in 1851 had 12 digits:


Because of the special form of this number, it would seem that even coming up with 12-digit primes was quite an achievement. Euler (1707–1783) had found a 10-digit prime, but it was also a Mersenne prime. Large primes without special structure were unknown.

Perhaps if Lewis Carroll had found a couple moderately large primes, he might have presented them to his queen to be used in public key cryptography. Their product could be published in newspapers, but the factors would be state secrets. Anyone could send Queen Victoria encrypted messages via public communication.

Diffie-Hellman public key encryption might have been more practical. It only requires one large prime, and that prime can be made public. Everyone can share it.

The prime p that Lucas discovered would do, until people realized that p − 1 has a lot of small factors [1], which could be used to break Diffie-Hellman cryptography. I don’t know that any large safe primes were known until more recently.

If someone from the future had given the Victorians a large safe prime, Diffie-Hellman cryptography would have been possible, though laborious. Someone could write a steampunk novel about a time traveler giving the pre-computerized world a big safe prime and teaching them Diffie-Hellman cryptography.


[1] 2126 − 1 = 3³ × 7² × 19 × 43 × 73 × 127 × 337 × 5419 × 92737 × 649657 × 77158673929

See the next post for a theorem that would allow you to find this factorization by hand.

Timing attacks

If you ask someone a question and they say “yes” immediately, that gives you different information than if they pause and slowly say “yes.” The information you receive is not just the response but also the time it took to generate the response.

Encryption can be analogous. The time it takes to encrypt data can leak information about the data being encrypted. It probably won’t reveal the data per se, but it may reveal enough about the data or the encryption process to reduce the effort needed to break the encryption.

There are two ways of thwarting timing attacks. One is to try to make the encryption take the same amount of time, independent of the data. This would prevent an attacker from inferring, for example, which branch of an algorithm was taken if one branch executes faster than the other.

If the encryption process always takes the same amount of time, then the execution time of the encryption process carries no information. But its enough that the execution time carries no useful information.

It may be easier to make execution time uncorrelated with content than to make execution time constant. Also, keeping the execution time of an algorithm constant may require making the program always run as slow as the worst case scenario. You may get faster average execution time by allowing the time to vary in a way that is uncorrelated with any information useful to an attacker.

One example of this would be Garner’s algorithm used in decrypting RSA encoded messages.

Suppose you’re using RSA encryption with a public key e, private key d, and modulus n. You can decrypt cyphertext c to obtain the cleaertext m by computing

m = cd mod n.

An alternative would be to compute a random message r and decrypt rec:

(rec)d = red  cdrm mod n

then multiply by the inverse of r mod n to obtain m. Because r is random, the time required to decrypt rec is uncorrelated with the time required to decrypt c.

Elliptic curve Diffie-Hellman key exchange

I concluded the previous post by saying elliptic curve Diffie-Hellman key exchange (ECDHE) requires smaller keys than finite field Diffie-Hellman (FFDHE) to obtain the same level of security.

How much smaller are we talking about? According to NIST recommendations, a 256-bit elliptic curve provides about the same security as working over a 3072-bit finite field. Not only are elliptic curves smaller, they scale better. A 512-bit elliptic curve is believed to be about as secure as a 15360-bit finite field: a factor of 2x for elliptic curves and a factor of 5x for finite fields.

The core idea of Diffie-Hellman is to pick a group G, an element g, and a large number of x. If y is the result of starting with x and applying the group operation x times, it is difficult to recover x from knowing y. This is called the discrete logarithm problem, taking its name from the case of the group operation being multiplication. But the inverse problem is still called the discrete logarithm problem when the group is additive.

In FFDHE the group G is the multiplicative group of a generator g modulo a large prime p. Applying the group operation (i.e. multiplication) to g a number of times x is computing

y = gx

and x is rightly called a discrete logarithm; the process is directly analogous to taking the logarithm of a real number.

In ECDHE the group is given by addition on an elliptic curve. Applying the group operation x times to g, adding g to itself x times, is written xg. The problem of recovering x from xg is still called the discrete logarithm problem, though you could call it the discrete “division” problem.

Some groups are unsuited for Diffie-Hellman cryptography because the discrete logarithm problem is easy. If we let G be the additive group modulo a prime (not the multiplicative group) then it is easy to recover x from xg.

Note that when we talk about applying the group operation a large number of times, we mean a really large number of times, in theory, though not in practice. If you’re working over an elliptic curve with on the order of 2256 elements, and x is on the order of 2256, then xg is the result of adding x to itself on the order of 2256 times. But in practice you’d double g on the order of 256 times. See fast exponentiation.

In the post on FFDHE we said that you have to be careful that your choice of prime and generator doesn’t give the group structure that a cryptanalysist could exploit. This is also true for the elliptic curves used in ECDHE, and even more so because elliptic curves are more subtle than finite fields.

If large-scale quantum computing ever becomes practical, Diffie-Hellman encryption will be broken because a quantum computer can solve discrete logarithm problems efficiently via Schor’s algorithm. This applies equally to finite fields and elliptic curves.

Related posts

Finite field Diffie Hellman primes

Diffie-Hellman key exchange is conceptually simple. Alice and Bob want to generate a shared cryptographic key. They want to use asymmetric (public) cryptography to share a symmetric (private) key.

The starting point is a large prime p and a generator 1 < g < p.

Alice generates a large random number x, her private key, and sends Bob gx mod p.

Similarly, Bob generates a large random number y, his private key, and sends Alice gy mod p.

Alice takes gy and raises it to her exponent x, and Bob takes gx and raises it to the exponent y. They arrive at a common key k because

k = (gy)x = (gx)y mod p.

The security of the system rests on the assumption that the discrete logarithm problem is hard, i.e. given g and gz it is computationally impractical to solve for z. This assumption appears to be true in general, but can fail when the group generated by g has exploitable structure.

You can read more about Diffie-Hellman here.

Recommended primes

The choice of prime p and generator g can matter is subtle ways and so there are lists of standard choices that are believed to be secure.

IETF RFC 7919 recommends five standard primes. These have the form

p = 2^b - 2^{b-64} + 2^{64} \left( \lfloor 2^{b-130} e\rfloor + X \right) - 1

where b is the size of p in bits, e is the base of natural logarithms, and X is the smallest such that p is a safe prime. In every case the generator is g = 2.

The values of b are 2048, 3072, 4096, 6144, and 8192. The values of X and p are given in RFC 7919, but they’re both determined by b.

I don’t imagine there’s anything special about the constant e above. I suspect it’s there to shake things up a bit in a way that doesn’t appear to be creating a back door. Another irrational number like π or φ would probably do as well, but I don’t know this for sure.


The recommended primes have names of the form “ffdhe” followed by b. For b = 2048, the corresponding value is X is 560316.

I wrote a little Python code to verify that this value of X does produce a safe prime and that smaller values of X do not.

    #!/usr/bin/env python3
    from sympy import isprime, E, N, floor
    b = 2048
    e = N(E, 1000)
    c = floor(2**(b-130) * e)
    d = 2**b - 2**(b-64) + 2**64*c - 1
    def candidate(b, x):
        p = d + 2**64*x
        return p
    for x in range(560316, 0, -1):
        p = candidate(b, x)
        if isprime(p) and isprime((p-1)//2):

This took about an hour to run. It only printed 560316, verifying the claim in RFC 7919.

Other groups

Finite field Diffie-Hellman is so called because the integers modulo a prime form a finite field. We don’t need a field per se; we’re working in the group formed by the orbit of g within that field. Such groups need to be very large in order to provide security.

It’s possible to use Diffie-Hellman over any group for which the discrete logarithm problem is intractable, and the discrete logarithm problem is harder over elliptic curves than over finite fields. The elliptic curve groups can be smaller and provide the same level of security. Smaller groups mean smaller keys to exchange. For this reason, elliptic curve Diffie-Hellman is more commonly used than finite field Diffie-Hellman.

Gaining efficiency by working modulo factors

Suppose m is a large integer that you are able to factor. To keep things simple, suppose m = pq where p and q are distinct primes; everything in this post generalizes easily to the case of m having more than two factors.

You can carry out calculations mod m more efficiently by carrying out the same calculations mod p and mod q, then combining the results. We analyze m into its remainders by p and q, carry out our calculations, then synthesize the results to get back to a result mod m.

The Chinese Remainder Theorem (CRT) says that this synthesis is possible; Garner’s algorithm, the subject of the next post, shows how to compute the result promised by the CRT.

For example, if we want to multiply xy mod m, we can analyze x and y as follows.

\begin{align*} x &\equiv x_p \pmod p \\ x &\equiv x_q \pmod q \\ y &\equiv y_p \pmod p \\ y &\equiv y_q \pmod q \\ \end{align*}


\begin{align*} xy &\equiv x_py_p \pmod p \\ xy &\equiv x_qy_q \pmod q \\ \end{align*}

and by repeatedly multiplying x by itself we have

\begin{align*} x^e &\equiv x_p^e \pmod p \\ x^e &\equiv x_q^e \pmod q \\ \end{align*}

Now suppose p and q are 1024-bit primes, as they might be in an implementation of RSA encryption. We can carry out exponentiation mod p and mod q, using 1024-bit numbers, rather than working mod n with 2048-bit numbers.

Furthermore, we can apply Euler’s theorem (or the special case Fermat’s little theorem) to reduce the size of the exponents.

\begin{align*} x^e &\equiv x_p^e \equiv x_p^{e \bmod p-1}\pmod p \\ x^e &\equiv x_q^e \equiv x_q^{e \bmod q-1}\pmod q \end{align*}

Assuming again that p and q are 1024-bit numbers, and assuming e is a 2048-bit number, by working mod p and mod q we can use exponents that are 1024-bit numbers.

We still have to put our pieces back together to get the value of xe mod n, but that’s the subject of the next post.

The trick of working modulo factors is used to speed up RSA decryption. It cannot be used for encryption since p and q are secret.

The next post shows that is in fact used in implementing RSA, and that a key file contains the decryption exponent reduced mod p-1 and mod q-1.

Group theory and RSA encryption

RSA encryption a map from numbers mod n to numbers mod n where n is a public key. A message is represented as an integer m and is encrypted by computing

c = me mod n

where e is part of the public key. In practice, e is usually 65537 though it does not have to be.

Multiplicative group

As we discussed in the previous post, not all messages m can be decrypted unless we require m to be relatively prime to n. In practice this is almost certainly the case: discovering a message m not relatively prime to n is equivalent to finding a factor of n and breaking the encryption.

If we limit ourselves to messages which can be encrypted and decrypted, our messages come not from the integers mod n but from the multiplicative group of integers mod n: the integers less than and relatively prime to n form a group G under multiplication.

The encryption function that maps m to me is an invertible function on G. Its inverse is the function that maps c to cd where d is the private key. Encryption is an automorphism of G because

(m1 m2) e = m1e m2e mod n.

Totient functions

Euler’s theorem tells us that

mφ(n) = 1 mod n

for all m in G. Here φ is Euler’s totient function. There are φ(n) elements in G, and so we could see this as a consequence of Lagrange’s theorem: the order of an element divides the order of a group.

Now the order of a particular m might be less than φ(n). That is, we know that if we raise m to the exponent φ(n) we will get 1, but maybe a smaller exponent would do. In fact, maybe a smaller exponent would do for all m.

Carmichael’s totient function λ(n) is the smallest exponent k such that

mk = 1 mod n

for all m. For some values of n the two totient functions are the same, i.e. λ(n) = φ(n). But sometimes λ(n) is strictly less than φ(n). And going back to Lagrange’s theorem, λ(n) always divides φ(n).

For example, there are 4 positive integers less than and relatively prime to 8: 1, 3, 5, and 7. Since φ(8) = 4, Euler’s theorem says that the 4th power of any of these numbers will be congruent to 1 mod 8. That is true, but its also true that the square of any of these numbers is congruent to 1 mod 8. That is, λ(8) = 2.

Back to RSA

Now for RSA encryption, n = pq where p and q are large primes and pq. It follows that

φ(pq) = φ(p) φ(q) = (p − 1)(q − 1).

On the other hand,

λ(pq) = lcm( λ(p), λ(q) ) = lcm(p − 1, q − 1).

Since p − 1 and q − 1 at least share a factor of 2,

λ(n) ≤ φ(n)/2.


It’s possible that λ(n) is smaller than φ(n) by more than a factor of 2. For example,

φ(7 × 13) = 6 × 12 = 72


λ(7 × 13) = lcm(6, 12) = 12.

You could verify this last calculation with the following Python code:

>>> from sympy import gcd
>>> G = set(n for n in range(1, 91) if gcd(n, 91) == 1)
>>> set(n**12 % 91 for n in s)

This returns {1}.


The significance of Carmichael’s totient to RSA is that φ(n) can be replaced with λ(n) when finding private keys. Given a public exponent e, we can find d by solving

ed = 1 mod λ(n)

rather than

ed = 1 mod φ(n).

This gives us a smaller private key d which might lead to faster decryption.

OpenSSL example

I generated an RSA key with openssl as in this post

    openssl genpkey -out fd.key -algorithm RSA \
      -pkeyopt rsa_keygen_bits:2048 -aes-128-cbc

and read it using

    openssl pkey -in  fd.key -text -noout

The public exponent was 65537 as noted above. I then brought the numbers in the key over to Python.

    from sympy import lcm

    modulus = xf227d5...a9
    prime1 = 0xf33514...d9
    prime2 = 0xfee496...51
    assert(prime1*prime2 == modulus)

    publicExponent = 65537
    privateExponent = 0x03896d...91

    phi = (prime1 - 1)*(prime2 - 1)
    lamb = lcm(prime1 - 1, prime2 - 1)
    assert(publicExponent*privateExponent % lamb == 1)
    assert(publicExponent*privateExponent % phi != 1)

This confirms that the private key d is the inverse of e = 65537 using modulo λ(pq) and not modulo φ(pq).

Related posts

Möbius transformations over a finite field

A Möbius transformation is a function of the form

g(z) = \frac{az + b}{cz + d}

where adbc = 1.

We usually think of z as a complex number, but it doesn’t have to be. We could define Möbius transformations in any context where we can multiply, add, and divide, i.e. over any field. In particular, we could work over a finite field such as the integers modulo a prime. The plot above represents a Möbius transformation over a finite field which we will describe below.

There is a subtle point, however. In the context of the complex numbers, the transformation above doesn’t quite map the complex plane onto the complex plane. It maps the complex plane minus one point to the complex plane minus one point. The domain is missing the point z = −d/c because that value makes the denominator zero. It’s also missing a point in the range, namely a/c.

The holes in the domain and range are a minor irritant, analogous to the pea in The Princess and the Pea. You can work around the holes, though the formalism is a little complicated. But over a finite field, the holes are a big deal. If you’re working over the integers mod 7, for example, then 1/7th of your domain is missing.

In the case of the complex numbers, the usual fix is to replace the complex numbers ℂ with the extended complex numbers ℂ ∪ ∞ and say that g(−d/c) = ∞ and g(∞) = a/c. There are a couple ways to make this more rigorous/elegant. The topological approach is to think of ℂ ∪ ∞ as the Riemann sphere. The algebraic approach is to think of it as a projective space.

Now let’s turn to finite fields, say the integers mod 17, which we will write as ℤ17. For a concrete example, let’s set a = 3, b = 8, c = 6, and d = 5. Then adbc = 1 mod 17. The multiplicative inverse of 6 mod 17 is 3, so we have a hole in the domain when

z = −d/c = −5/6 = −5 × 3 = − 15 = 2 mod 17.

Following the patch used with complex numbers, we define g(2) to be ∞, and we define

g(∞) = a/c = 3/6 = 3 × 3 = 9 mod 17.

That’s all fine, except now we’re not actually working over ℤ17 but rather ℤ17 ∪ ∞. We could formalize this by saying we’re working in a projective space over ℤ17. For this post let’s just say we’re working over set G with 18 elements that mostly follows the rules of ℤ17 but has a couple additional rules.

Now our function g maps G onto G. No holes.

Here’s how we might implement g in Python.

    def g(n):
        if n == 2:
            return 17
        if n == 17:
            return 9
        a, b, c, d = 3, 8, 6, 5
        denom = c*n + d
        denom_inverse = pow(denom, -1, 17)
        return (a*n + b)*denom_inverse % 17

The plot at the top of the post arranges 18 points uniformly around a circle and connects n to g(n).

    from numpy import pi, linspace, sin, cos
    import matplotlib.pyplot as plt

    θ = 2*pi/18
    t = linspace(0, 2*pi)
    plt.plot(cos(t), sin(t), 'b-')

    for n in range(18):
        plt.plot(cos(n*θ), sin(n*θ), 'bo')
        plt.plot([cos(n*θ), cos(g(n)*θ)],
                 [sin(n*θ), sin(g(n)*θ)], 'g-')

Application to cryptography

What use is this? Möbius transformations over finite fields [1] are “higgledy-piggledy” in the words of George Marsaglia, and so they can be used to create random-like permutations. In particular, Möbius transformations over finite fields are used to design S-boxes for use in symmetric encryption algorithms.

Related posts

[1] Technically, finite fields plus an element at infinity.

[2] “If the [pseudorandom] numbers are not random, they are at least higgledy-piggledy.” — RNG researcher George Marsaglia

Generating and inspecting an RSA private key

In principle you generate an RSA key by finding two large prime numbers, p and q, and computing n = pq. You could, for example, generate random numbers by rolling dice, then type the numbers into Mathematica to test each for primality until you find a couple prime numbers of the right size.

In practice you’d use a specialized program to find the primes and to wrap everything up in a format that software using the keys can understand. There are a lot of layers between the numbers p and q and the file that key generating software produces, and this post aims to peel back these layers a bit.

Here’s an example of generating a private key taken from The OpenSSL Cookbook.

    openssl genpkey -out fd.key -algorithm RSA \
      -pkeyopt rsa_keygen_bits:2048 -aes-128-cbc

The genpkey function can be used for generating several kinds of public keys. The option -algorithm RSA tells it that we want an RSA key, but we could have asked for an elliptic curve key. As noted in the previous post, in practice public key encryption is used to transfer symmetric encryption keys, not messages per se. The flag -aes-128-cbc tells the software the we’d like to use AES encryption with a 128-bit key in CBC (cipher block chaining) mode.

When you press enter you’ll see a flurry of dots and plus signs that show the progress of the software in generating and testing candidates for the primes p and q. Then you’ll be prompted for a password to encrypt the private key you’ve just created.

If you open the fd.key file you won’t see much:

    % cat fd.key

This is just base 64-encoded data.

The data is encoded in two senses. It is encoded in a non-secret way, expressed in a standardized data structure, then encoded in the sense of being encrypted. The openssl command pkey will undo both levels of encoding to let us see the contents of the file.

Here’s what this produces.

    Private-Key: (2048 bit, 2 primes)
    publicExponent: 65537 (0x10001)

The exponent is the default value 65537. (More on that here.)

The large numbers are displayed in hexadecimal with colons separating pairs of hex digits. If you remove the colons and concatenate everything together, you can verify that the number called modulus is indeed the product of the numbers called prime1 and prime2. I verified this for the output above using a little Python code:

    modulus = 0xa7b839...c5
    prime1  = 0xdc8c27...f9
    prime2  = 0xc2ae20...2d
    assert(prime1*prime2 == modulus)

The file also contains four numbers that require more explanation: privateExponent, exponent1, exponent2, and coefficient. The privateExponent is described here. The remaining numbers are not strictly necessary for RSA but are used in Garner’s algorithm for more efficient decryption.

Related posts

RSA encryption in practice

At its core, RSA encryption is modular exponentiation. That is, given a message m, the encrypted form of m is

x = me mod n

where e is a publicly known exponent and n is a product of two large primes. The number n is made public but only the holder of the private key knows the factors of n, and without knowing the factors of n you can’t recover m from x, or so we assume.

You can implement RSA encryption in just a few lines of code as long as you have a way to work with very large integers.

In principle you could divide your message into segments each less than n and encrypt each segment. In practice, that would be inefficient. Instead, asymmetric (public key) cryptography is only used to exchange symmetric cryptography keys. So, for example, someone wishing to send you a long message would use RSA to share the AES key used to encrypt the rest of the transmission.

So RSA is used to transfer keys, but that’s not the whole story. As is often the case, the real world implementation of cryptography is more complicated than the mathematical core.

In 1993 RSA published its PKCS#1 standard specifying that messages should be padded a certain way. That was an improvement, but then in 1998, Daniel Bleichenbacher published what has become known as the “million message attack” against the PKCS#1 standard. There were multiple proposed fixes, but these were complicated and often implemented incorrectly.

Now the standard is RSA-OAEP (Optimal Asymmetric Encryption Padding) which combines the message with random bits before applying the RSA algorithm per se. So there’s a bit of symmetric encryption, before using asymmetric encryption to share symmetric encryption keys!

My point here is not to get into the details of the OAEP protocol, but only to point out that it’s not trivial. It is, however, secure in the sense that you can prove that if someone can break RSA-OAEP then they can break the core RSA algorithm too.

“Cryptography is a mixture of mathematics and muddle, and without the muddle the mathematics can be used against you.” — Ian Cassels

Related posts

Image above CC BY-SA 4.0 by Jm-lemmi

Density of safe primes

Sean Connolly asked in a comment yesterday about the density of safe primes. Safe primes are so named because Diffie-Hellman encryption systems based on such primes are safe from a particular kind of attack. More on that here.

If q and p = 2q + 1 are both prime, q is called a Sophie Germain prime and p is a safe prime. We could phrase Sean’s question in terms of Sophie Germain primes because every safe prime corresponds to a Sophie Germain prime.

It is unknown whether there are infinitely many Sophie Germain primes, so conceivably there are only a finite number of safe primes. But the number of Sophie Germain primes less than N is conjectured to be approximately

1.32 N / (log N)².

See details here.

Sean asks specifically about the density of safe primes with 19,000 digits. The density of Sophie Germain primes with 19,000 digits or less is conjectured to be about

1.32/(log 1019000)² = 1.32/(19000 log 10)² = 6.9 × 10-10.

So the chances that a 19,000-digit number is a safe prime are on the order of one in a billion.