Lecture 18

Public Key Cryptography

In this section, we’ll show off a major modern application of all the ideas above. The aim is to talk about (one small but key part of) modern cryptography.

Cryptography is the study of how to send messages in a form which cannot be read except by the intended recipients. To encrypt the messages is to put them in a form which cannot be read easily; to decrypt the messages is to take such messages and recover them in readable form.

The literature of cryptography usually talks about three people:

• Alice who wishes to send a private message to Bob,

• Bob who wishes to receive a private message from Alice, and

• Eve who wishes to find out what Alice is telling Bob.

Alice and Bob are of course named so that the message goes from $A$ to $B$. Eve is so named because she is an eavesdropper, or perhaps because she is evil.

In the olden days, Alice and Bob would have come up with some kind of system depending on a shared secret key with which you could encrypt and decrypt a message. Perhaps you’ve seen many of these techniques already! For example, you could substitute the letters of the alphabet according to some agreed system: then the key would describe that system and would be a list of facts like $A\mapsto Q$, $B\mapsto J$, etc.

The big disadvantage with that is that Alice and Bob have to exchange the key somehow at the beginning: if Eve can spy on that conversation, she has the key and can decrypt Alice’s message just as easily as Bob can.

The problem with this old-time approach is that the same secret is used to encrypt and decrypt the message, so needs exchanging.

Suppose instead there was a type of encryption with a key for encryption and another key for decryption, such that, even if you know exactly how to encrypt a message, it is very hard indeed to work out how to decrypt it.

That suggests the following plan:

1. Bob comes up with a system of encrypting and decrypting of that sort.

2. Bob takes the key which tells you how to decrypt messages, the private key, and keeps it to himself, never telling anyone.

3. Bob takes the key which tells you how to encrypt messages, the public key, and shares it with everyone who wants it, with no secrecy whatsoever. In particular, he sends Alice a postcard telling her his public key. Of course Eve finds it out quickly, but Bob doesn’t care.

4. Alice uses Bob’s public key to encrypt a message for Bob.

5. Alice sends Bob the encrypted message.

6. Bob uses his private key to decrypt it, and read Alice’s message.

So the only question is, how can we come up with such a system, where being able to encrypt things doesn’t help you decrypt things very much?

The approach we’ll describe was the first one to be thought of, in the 1970s. It is known as RSA after its American inventors Rivest, Shamir and Adleman. (A British mathematician, Cocks, invented it a few years earlier, but he was working in secret for the government, so this was not known for many years). RSA is still in very widespread use on the internet.

The secret of RSA is to work modulo $pq$, where $p$ and $q$ are (different) primes. We’re going to need to do modular arithmetic mod $pq$, including exponentiation. So we’ll need to see what Fermat-Euler says:

Proposition

Let $p$ and $q$ be different primes. Then the number $\varphi(pq)$, of integers from $1$ to $pq$ coprime to $pq$, is given by $$\varphi(pq) = (p-1)(q-1).$$

Proof

There are $pq$ integers $a$ between $1$ and $pq$. Those that are not coprime to $pq$ are either multiples of $p$ or of $q$.

Of these, $q$ of them are multiples of $p$ (namely $p,2p,\ldots,pq$).

Also, $p$ of them are multiples of $q$ (namely $q,2q,\ldots,pq$).

Lastly, one of them (namely $pq$) is a multiple of $p$ and of $q$.

Hence $q+p-1$ are not coprime to $pq$, and so $$\varphi(pq) = pq-q-p+1 = (p-1)(q-1).$$

As a result of that, we know (from the Fermat-Euler Theorem that, for all $a$ coprime to $pq$, we have $$a^{(p-1)(q-1)} \equiv 1 \pmod{pq},$$ and indeed $$a^{k(p-1)(q-1)} \equiv 1 \pmod{pq}.$$ for all $k$.

So, Bob chooses two fairly large primes $p$ and $q$, and keeps them secret. He also chooses a number $e$ which is coprime to $(p-1)(q-1)$.

He also calculates the inverse $d$ to $e$, modulo $(p-1)(q-1)$, by using Euclid’s algorithm.

His public key consists of $pq$ and $e$, so he sends that to Alice (and Eve); his private key consists of $pq$ and $d$. He shreds any evidence of what $p$ and $q$ are.

Alice represents her message as a number $m$ between $1$ and $pq$. It is overwhelmingly likely that her choice will be coprime to $pq$. She calculates $$m^e\pmod{pq}$$ and sends it on to Bob.

Bob receives this number $m^e$ from Alice, and raises it to the power $d$ modulo $pq$. He thus obtains something congruent to $$(m^e)^d = m^{de}.$$

Now, because $de\equiv 1\pmod{\varphi(pq)}$, we have $de=1+k\varphi(pq)$ for some $k$. As a result, $$(m^e)^d {}= m^{de} {}= m^{1+k\varphi(pq)} {}= m(m^{\varphi(pq)})^k {}\equiv m1^k {}\equiv m\pmod{pq}.$$

Hence, using his private key, Bob can recover what $m$ was from being told $m^e$.

The idea is that it should be very hard for anyone else to work out $d$ from $pq$ and $e$; we did this using Euclid’s algorithm, but we needed to know more than just $pq$: we needed to know $(p-1)(q-1)$.

So the security of this approach depends (among other things) on it being difficult to factorise the number $pq$: if factorising large numbers were easy, we could get $p$ and $q$ for ourselves from Bob’s public key. Currently, we know of no way to do this fast enough: we know how to generate primes that are hundreds of digits long, but not to factorise a product of two of them.

Let’s see an example.

Suppose Bob has low opinions of Eve’s calculational skills, and chooses to use the (unrealistically small) primes $p=101$ and $q=103$. Then $pq = 10403$. Suppose also that Bob chooses $e=71$ for the exponent used for encryption.

Bob advertises that his public key is $pq = 10403$, $e=71$. He must work out his private key, by inverting $71$ modulo $(p-1)(q-1)=10200$. A quick use of Euclid’s algorithm will do this for him, and he gets that $71^{-1}\equiv 431$. Indeed, $$71\times 431 = 30601 = 3\times 10200+1.$$ Thus his private key is $pq = 10403$, $d = 431$.

Suppose Alice decides she needs to send Bob message $1245$, which they’ve agreed in advance should mean “please meet me after this lecture”.

Then Alice has to calculate $1245^{71}$ modulo $10403$. This sounds scary, but she can do it fairly quickly if she’s careful: $$\begin{aligned} 1245^{71} {}\equiv 1245\cdot 1245^{70} {}\equiv 1245\cdot (1245^2)^{35}\\ {}\equiv 1245\cdot 10381^{35} {}\equiv 1245\cdot 10381\cdot 10381^{34} {}\equiv 3819\cdot (10381^2)^{17}\\ {}\equiv 3819\cdot 484^{17} {}\equiv 3819\cdot 484\cdot 484^{16} {}\equiv 7065\cdot (484^2)^8\\ {}\equiv 7065\cdot 5390^8 {}\equiv 7065\cdot (5390^2)^4 {}\equiv 7065\cdot 6924^4\\ {}\equiv 7065\cdot (6924^2)^2 {}\equiv 7065\cdot 4752^2\\ {}\equiv 7065\cdot 6994 {}\equiv 8763.\end{aligned}$$

So she sends Bob $8763$.

Bob receives this, and his task then is to calculate $8763^{431}$ modulo $10403$. A similar strategy makes this possible, too, and he finds that $$8763^{431}\equiv 1245\pmod{10403},$$ so he has reconstructed Alice’s message.

The real numbers

Irrational numbers

We’ve spent nine lectures now talking about $\mathbb{N}$, $\mathbb{Z}$ and $\mathbb{Q}$, laying the foundations of number theory. The rest of this course will be about $\mathbb{R}$. Perhaps sensibly enough, the study of $\mathbb{R}$ is called real analysis.

Let’s set ourselves back to a time before $\mathbb{R}$ was invented, and ask: why was it necessary to invent it? Why should we feel that $\mathbb{Q}$ is not enough?

The result that set the ancient Greeks thinking was this:

Proof

We’ll prove this by contradiction; suppose there is such a number $x\in\mathbb{Q}$. Because it’s in $\mathbb{Q}$, it takes the form $x = p/q$ for some integers $p$ and $q$ with $q\neq 0$.

We may as well take $p$ and $q$ to be coprime (“in lowest terms”).

Then we have $p^2/q^2 = x^2 = 2$, so $p^2 = 2q^2$ with $p$ and $q$ coprime.

Now, the right-hand side is even (it’s given as a multiple of $2$, so the left-hand side, $p^2$ must be even too. That means that $p$ itself must be even: so we can write $p = 2r$.

Then we have $(2r)^2 = 2q^2$, which simplifies to $4r^2 = 2q^2$, or $2r^2 = q^2$. Here the left-hand side is even, so $q^2$ must be even. Hence $q$ itself must be even.

This is a contradiction: $p$ and $q$ can’t both be even. So our initial assumption is absurd, and there is no rational $x$ with $x^2 = 2$.

I felt obliged to word the statement of that theorem fairly carefully.

What I wanted to say, of course, was:

The number $\sqrt{2}$ is not in $\mathbb{Q}$.

But I want to flag that up as being possibly inappropriate: our aim in this section is to define the reals. We shouldn’t even be confident that $\sqrt{2}$ exists yet.

However, thanks to this theorem, we can be confident at least that there’s no number inside $\mathbb{Q}$ which deserves to be called $\sqrt{2}$.

This, to the Greeks, was evidence that there was a world beyond $\mathbb{Q}$; a world of irrational numbers (numbers not in $\mathbb{Q}$). They needed a number called $\sqrt{2}$, so they could talk about the diagonal of a unit square:

Over the years, more and more examples were found of numbers which one might want to talk about, but which cannot be in $\mathbb{Q}$: various powers, logarithms, sines, cosines, and other constructions besides.

One high point includes the proof by Lambert in 1761 that $\pi$ and $e$ are irrational.

On the other hand, modern mathematics is still not particularly good, in general, at proving that numbers are irrational. For example, if you want to become famous, simply prove (please…) that any one of the following numbers are irrational: $$\pi+e,{}\quad \pi-e,{}\quad \pi e,{}\quad \pi/e,{}\quad \ln\pi,{}\quad e^e,{}\quad e^{e^e}.$$