Sending and Authenticating Messages with Elliptic Curves

Last time we saw the Diffie-Hellman key exchange protocol, and discussed the discrete logarithm problem and the related Diffie-Hellman problem, which form the foundation for the security of most protocols that use elliptic curves. Let’s continue our journey to investigate some more protocols.

Just as a reminder, the Python implementations of these protocols are not at all meant for practical use, but for learning purposes. We provide the code on this blog’s Github page, but for the love of security don’t actually use them.

Shamir-Massey-Omura

Recall that there are lots of ways to send encrypted messages if you and your recipient share some piece of secret information, and the Diffie-Hellman scheme allows one to securely generate a piece of shared secret information. Now we’ll shift gears and assume you don’t have a shared secret, nor any way to acquire one. The first cryptosystem in that vein is called the Shamir-Massey-Omura protocol. It’s only slightly more complicated to understand than Diffie-Hellman, and it turns out to be equivalently difficult to break.

The idea is best explained by metaphor. Alice wants to send a message to Bob, but all she has is a box and a lock for which she has the only key. She puts the message in the box and locks it with her lock, and sends it to Bob. Bob can’t open the box, but he can send it back with a second lock on it for which Bob has the only key. Upon receiving it, Alice unlocks her lock, sends the box back to Bob, and Bob can now open the box and retrieve the message.

To celebrate the return of Game of Thrones, we’ll demonstrate this protocol with an original Lannister Infographic™.

Assuming the box and locks are made of magically unbreakable Valyrian steel, nobody but Bob (also known as Jamie) will be able to read the message.

Now fast forward through the enlightenment, industrial revolution, and into the age of information. The same idea works, and it’s significantly faster over long distances. Let $C$ be an elliptic curve over a finite field $k$ (we’ll fix $k = \mathbb{Z}/p$ for some prime $p$, though it works for general fields too). Let $n$ be the number of points on $C$.

Alice’s message is going to be in the form of a point $M$ on $C$. She’ll then choose her secret integer $0 < s_A < p$ and compute $s_AM$ (locking the secret in the box), sending the result to Bob. Bob will likewise pick a secret integer $s_B$, and send $s_Bs_AM$ back to Alice.

Now the unlocking part: since $s_A \in \mathbb{Z}/p$ is a field, Alice can “unlock the box” by computing the inverse $s_A^{-1}$ and computing $s_BM = s_A^{-1}s_Bs_AM$. Now the “box” just has Bob’s lock on it. So Alice sends $s_BM$ back to Bob, and Bob performs the same process to evaluate $s_B^{-1}s_BM = M$, thus receiving the message.

Like we said earlier, the security of this protocol is equivalent to the security of the Diffie-Hellman problem. In this case, if we call $z = s_A^{-1}$ and $y = s_B^{-1}$, and $P = s_As_BM$, then it’s clear that any eavesdropper would have access to $P, zP$, and $yP$, and they would be tasked with determining $zyP$, which is exactly the Diffie-Hellman problem.

Now Alice’s secret message comes in the form of a point on an elliptic curve, so how might one translate part of a message (which is usually represented as an integer) into a point? This problem seems to be difficult in general, and there’s no easy answer. Here’s one method originally proposed by Neal Koblitz that uses a bit of number theory trickery.

Let $C$ be given by the equation $y^2 = x^3 + ax + b$, again over $\mathbb{Z}/p$. Suppose $0 \leq m < p/100$ is our message. Define for any $0 \leq j < 100$ the candidate $x$-points $x_j = 100m + j$. Then call our candidate $y^2$-values $s_j = x_j^3 + ax_j + b$. Now for each $j$ we can compute $x_j, s_j$, and so we’ll pick the first one for which $s_j$ is a square in $\mathbb{Z}/p$ and we’ll get a point on the curve. How can we tell if $s_j$ is a square? One condition is that $s_j^{(p-1)/2} \equiv 1 \mod p$. This is a basic fact about quadratic residues modulo primes; see these notes for an introduction and this Wikipedia section for a dense summary.

Once we know it’s a square, we can compute the square root depending on whether $p \equiv 1 \mod 4$ or $p \equiv 3 \mod 4$. In the latter case, it’s just $s_j^{(p+1)/4} \mod p$. Unfortunately the former case is more difficult (really, the difficult part is $p \equiv 1 \mod 8$). You can see Section 1.5 of this textbook for more details and three algorithms, or you could just pick primes congruent to 3 mod 4.

I have struggled to find information about the history of the Shamir-Massey-Omura protocol; every author claims it’s not widely used in practice, and the only reason seems to be that this protocol doesn’t include a suitable method for authenticating the validity of a message. In other words, some “man in the middle” could be intercepting messages and tricking you into thinking he is your intended recipient. Coupling this with the difficulty of encoding a message as a point seems to be enough to make cryptographers look for other methods. Another reason could be that the system was patented in 1982 and is currently held by SafeNet, one of the US’s largest security providers. All of their products have generic names so it’s impossible to tell if they’re actually using Shamir-Massey-Omura. I’m no patent lawyer, but it could simply be that nobody else is allowed to implement the scheme.

Digital Signatures

Indeed, the discussion above raises the question: how does one authenticate a message? The standard technique is called a digital signature, and we can implement those using elliptic curve techniques as well. To debunk the naive idea, one cannot simply attach some static piece of extra information to the message. An attacker could just copy that information and replicate it to forge your signature on another, potentially malicious document. In other words, a signature should only work for the message it was used to sign. The technique we’ll implement was originally proposed by Taher Elgamal, and is called the ElGamal signature algorithm. We’re going to look at a special case of it.

So Alice wants to send a message $m$ with some extra information that is unique to the message and that can be used to verify that it was sent by Alice. She picks an elliptic curve $E$ over $\mathbb{F}_q$ in such a way that the number of points on $E$ is $br$, where $b$ is a small integer and $r$ is a large prime.

Then, as in Diffie-Hellman, she picks a base point $Q$ that has order $r$ and a secret integer $s$ (which is permanent), and computes $P = sQ$. Alice publishes everything except $s$:

Public information: $\mathbb{F}_q, E, b, r, Q, P &fg=000000$

Let Alice’s message $m$ be represented as an integer at most $r$ (there are a few ways to get around this if your message is too long). Now to sign $m$ Alice picks a message specific $k < r$ and computes what I’ll call the auxiliary point $A = kQ$. Let $A = (x, y)$. Alice then computes the signature $g = k^{-1}(m + s x) \mod r$. The signed message is then $(m, A, g)$, which Alice can safely send to Bob.

Before we see how Bob verifies the message, notice that the signature integer involves everything: Alice’s secret key, the message-specific secret integer $k$, and most importantly the message. Remember that this is crucial: we want the signature to work only for the message that it was used to sign. If the same $k$ is used for multiple messages then the attacker can find out your secret key! (And this has happened in practice; see the end of the post.)

So Bob receives $(m, A, g)$, and also has access to all of the public information listed above. Bob authenticates the message by computing the auxiliary point via a different route. First, he computes $c = g^{-1} m \mod r$ and $d = g^{-1}x \mod r$, and then $A’ = cQ + dP$. If the message was signed by Alice then $A’ = A$, since we can just write out the definition of everything:

Now to analyze the security. The attacker wants to be able to take any message $m’$ and produce a signature $A’, g’$ that will pass validation with Alice’s public information. If the attacker knew how to solve the discrete logarithm problem efficiently this would be trivial: compute $s$ and then just sign like Alice does. Without that power there are still a few options. If the attacker can figure out the message-specific integer $k$, then she can compute Alice’s secret key $s$ as follows.

Given $g = k^{-1}(m + sx) \mod r$, compute $kg \equiv (m + sx) \mod r$. Compute $d = gcd(x, r)$, and you know that this congruence has only $d$ possible solutions modulo $r$. Since $s$ is less than $r$, the attacker can just try all options until they find $P = sQ$. So that’s bad, but in a properly implemented signature algorithm finding $k$ is equivalently hard to solving the discrete logarithm problem, so we can assume we’re relatively safe from that.

On the other hand one could imagine being able to conjure the pieces of the signature $A’, g’$ by some method that doesn’t involve directly finding Alice’s secret key. Indeed, this problem is less well-studied than the Diffie-Hellman problem, but most cryptographers believe it’s just as hard. For more information, this paper surveys the known attacks against this signature algorithm, including a successful attack for fields of characteristic two.

Signature Implementation

We can go ahead and implement the signature algorithm once we’ve picked a suitable elliptic curve. For the purpose of demonstration we’ll use a small curve, $E: y^2 = x^3 + 3x + 181$ over $F = \mathbb{Z}/1061$, whose number of points happens to have the a suitable prime factorization ($1047 = 3 \cdot 349$). If you’re interested in counting the number of points on an elliptic curve, there are many theorems and efficient algorithms to do this, and if you’ve been reading this whole series something then an algorithm based on the Baby-Step Giant-Step idea would be easy to implement. For the sake of brevity, we leave it as an exercise to the reader.

Note that the code we present is based on the elliptic curve and finite field code we’re been implementing as part of this series. All of the code used in this post is available on this blog’s Github page.

The basepoint we’ll pick has to have order 349, and $E$ has plenty of candidates. We’ll use $(2, 81)$, and we’ll randomly generate a secret key that’s less than $349$ (eight bits will do). So our setup looks like this:

if __name__ == &quot;__main__&quot;:
F = FiniteField(1061, 1)

# y^2 = x^3 + 3x + 181
curve = EllipticCurve(a=F(3), b=F(181))
basePoint = Point(curve, F(2), F(81))
basePointOrder = 349
secretKey = generateSecretKey(8)
publicKey = secretKey * basePoint


Then so sign a message we generate a random key, construct the auxiliary point and the signature, and return:

def sign(message, basePoint, basePointOrder, secretKey):
modR = FiniteField(basePointOrder, 1)
oneTimeSecret = generateSecretKey(len(bin(basePointOrder)) - 3) # numbits(order) - 1

auxiliaryPoint = oneTimeSecret * basePoint
signature = modR(oneTimeSecret).inverse() *
(modR(message) + modR(secretKey) * modR(auxiliaryPoint[0]))

return (message, auxiliaryPoint, signature)


So far so good. Note that we generate the message-specific $k$ at random, and this implies we need a high-quality source of randomness (what’s called a cryptographically-secure pseudorandom number generator). In absence of that there are proposed deterministic methods for doing it. See this draft proposal of Thomas Pornin, and this paper of Daniel Bernstein for another.

Now to authenticate, we follow the procedure from earlier.

def authentic(signedMessage, basePoint, basePointOrder, publicKey):
modR = FiniteField(basePointOrder, 1)
(message, auxiliary, signature) = signedMessage

sigInverse = modR(signature).inverse() # sig can be an int or a modR already
c, d = sigInverse * modR(message), sigInverse * modR(auxiliary[0])

auxiliaryChecker = int(c) * basePoint + int(d) * publicKey
return auxiliaryChecker == auxiliary


Continuing with our example, we pick a message represented as an integer smaller than $r$, sign it, and validate it.

&gt;&gt;&gt; message = 123
&gt;&gt;&gt; signedMessage = sign(message, basePoint, basePointOrder, secretKey)
&gt;&gt;&gt; signedMessage
(123, (220 (mod 1061), 234 (mod 1061)), 88 (mod 349))
&gt;&gt;&gt; authentic(signedMessage, basePoint, basePointOrder, publicKey)
True


So there we have it, a nice implementation of the digital signature algorithm.

When Digital Signatures Fail

As we mentioned, it’s extremely important to avoid using the same $k$ for two different messages. If you do, then you’ll get two signed messages $(m_1, A_1, g_1), (m_2, A_2, g_2)$, but by definition the two $g$’s have a ton of information in common! An attacker can recognize this immediately because $A_1 = A_2$, and figure out the secret key $s$ as follows. First write

$\displaystyle g_1 – g_2 \equiv k^{-1}(m_1 + sx) – k^{-1}(m_2 + sx) \equiv k^{-1}(m_1 – m_2) \mod r$.

Now we have something of the form $\text{known}_1 \equiv (k^{-1}) \text{known}_2 \mod r$, and similarly to the attack described earlier we can try all possibilities until we find a number that satisfies $A = kQ$. Then once we have $k$ we have already seen how to find $s$. Indeed, it would be a good exercise for the reader to implement this attack.

The attack we just described it not an idle threat. Indeed, the Sony corporation, producers of the popular Playstation video game console, made this mistake in signing software for Playstation 3. A digital signature algorithm makes sense to validate software, because Sony wants to ensure that only Sony has the power to publish games. So Sony developers act as one party signing the data on a disc, and the console will only play a game with a valid signature. Note that the asymmetric setup is necessary because if the console had shared a secret with Sony (say, stored as plaintext within the hardware of the console), anyone with physical access to the machine could discover it.

Now here come the cringing part. Sony made the mistake of using the same $k$ to sign every game! Their mistake was discovered in 2010 and made public at a cryptography conference. This video of the humorous talk includes a description of the variant Sony used and the attacker describe how the mistake should have been corrected. Without a firmware update (I believe Sony’s public key information was stored locally so that one could authenticate games without an internet connection), anyone could sign a piece of software and create games that are indistinguishable from something produced by Sony. That includes malicious content that, say, installs software that sends credit card information to the attacker.

So here we have a tidy story: a widely used cryptosystem with a scare story of what will go wrong when you misuse it. In the future of this series, we’ll look at other things you can do with elliptic curves, including factoring integers and testing for primality. We’ll also see some normal forms of elliptic curves that are used in place of the Weierstrass normal form for various reasons.

Until next time!

Elliptic Curve Diffie-Hellman

So far in this series we’ve seen elliptic curves from many perspectives, including the elementary, algebraic, and programmatic ones. We implemented finite field arithmetic and connected it to our elliptic curve code. So we’re in a perfect position to feast on the main course: how do we use elliptic curves to actually do cryptography?

History

As the reader has heard countless times in this series, an elliptic curve is a geometric object whose points have a surprising and well-defined notion of addition. That you can add some points on some elliptic curves was a well-known technique since antiquity, discovered by Diophantus. It was not until the mid 19th century that the general question of whether addition always makes sense was answered by Karl Weierstrass. In 1908 Henri Poincaré asked about how one might go about classifying the structure of elliptic curves, and it was not until 1922 that Louis Mordell proved the fundamental theorem of elliptic curves, classifying their algebraic structure for most important fields.

While mathematicians have always been interested in elliptic curves (there is currently a million dollar prize out for a solution to one problem about them), its use in cryptography was not suggested until 1985. Two prominent researchers independently proposed it: Neal Koblitz at the University of Washington, and Victor Miller who was at IBM Research at the time. Their proposal was solid from the start, but elliptic curves didn’t gain traction in practice until around 2005. More recently, the NSA was revealed to have planted vulnerable national standards for elliptic curve cryptography so they could have backdoor access. You can see a proof and implementation of the backdoor at Aris Adamantiadis’s blog. For now we’ll focus on the cryptographic protocols themselves.

The Discrete Logarithm Problem

Koblitz and Miller had insights aplenty, but the central observation in all of this is the following.

Adding is easy on elliptic curves, but undoing addition seems hard.

What I mean by this is usually called the discrete logarithm problem. Here’s a formal definition. Recall that an additive group is just a set of things that have a well-defined addition operation, and the that notation $ny$ means $y + y + \dots + y$ ($n$ times).

Definition: Let $G$ be an additive group, and let $x, y$ be elements of $G$ so that $x = ny$ for some integer $n$. The discrete logarithm problem asks one to find $n$ when given $x$ and $y$.

I like to give super formal definitions first, so let’s do a comparison. For integers this problem is very easy. If you give me 12 and 4185072, I can take a few seconds and compute that $4185072 = (348756) 12$ using the elementary-school division algorithm (in the above notation, $y=12, x=4185072$, and $n = 348756$). The division algorithm for integers is efficient, and so it gives us a nice solution to the discrete logarithm problem for the additive group of integers $\mathbb{Z}$.

The reason we use the word “logarithm” is because if your group operation is multiplication instead of addition, you’re tasked with solving the equation $x = y^n$ for $n$. With real numbers you’d take a logarithm of both sides, hence the name. Just in case you were wondering, we can also solve the multiplicative logarithm problem efficiently for rational numbers (and hence for integers) using the square-and-multiply algorithm. Just square $y$ until doing so would make you bigger than $x$, then multiply by $y$ until you hit $x$.

But integers are way nicer than they need to be. They are selflessly well-ordered. They give us division for free. It’s a computational charity! What happens when we move to settings where we don’t have a division algorithm? In mathematical lingo: we’re really interested in the case when $G$ is just a group, and doesn’t have additional structure. The less structure we have, the harder it should be to solve problems like the discrete logarithm. Elliptic curves are an excellent example of such a group. There is no sensible ordering for points on an elliptic curve, and we don’t know how to do division efficiently. The best we can do is add $y$ to itself over and over until we hit $x$, and it could easily happen that $n$ (as a number) is exponentially larger than the number of bits in $x$ and $y$.

What we really want is a polynomial time algorithm for solving discrete logarithms. Since we can take multiples of a point very fast using the double-and-add algorithm from our previous post, if there is no polynomial time algorithm for the discrete logarithm problem then “taking multiples” fills the role of a theoretical one-way function, and as we’ll see this opens the door for secure communication.

Here’s the formal statement of the discrete logarithm problem for elliptic curves.

Problem: Let $E$ be an elliptic curve over a finite field $k$. Let $P, Q$ be points on $E$ such that $P = nQ$ for some integer $n$. Let $|P|$ denote the number of bits needed to describe the point $P$. We wish to find an algorithm which determines $n$ and has runtime polynomial in $|P| + |Q|$. If we want to allow randomness, we can require the algorithm to find the correct $n$ with probability at least 2/3.

So this problem seems hard. And when mathematicians and computer scientists try to solve a problem for many years and they can’t, the cryptographers get excited. They start to wonder: under the assumption that the problem has no efficient solution, can we use that as the foundation for a secure communication protocol?

The Diffie-Hellman Protocol and Problem

Let’s spend the rest of this post on the simplest example of a cryptographic protocol based on elliptic curves: the Diffie-Hellman key exchange.

A lot of cryptographic techniques are based on two individuals sharing a secret string, and using that string as the key to encrypt and decrypt their messages. In fact, if you have enough secret shared information, and you only use it once, you can have provably unbreakable encryption! We’ll cover this idea in a future series on the theory of cryptography (it’s called a one-time pad, and it’s not all that complicated). All we need now is motivation to get a shared secret.

Because what if your two individuals have never met before and they want to generate such a shared secret? Worse, what if their only method of communication is being monitored by nefarious foes? Can they possibly exchange public information and use it to construct a shared piece of secret information? Miraculously, the answer is yes, and one way to do it is with the Diffie-Hellman protocol. Rather than explain it abstractly let’s just jump right in and implement it with elliptic curves.

As hinted by the discrete logarithm problem, we only really have one tool here: taking multiples of a point. So say we’ve chosen a curve $C$ and a point on that curve $Q$. Then we can take some secret integer $n$, and publish $Q$ and $nQ$ for the world to see. If the discrete logarithm problem is truly hard, then we can rest assured that nobody will be able to discover $n$.

How can we use this to established a shared secret? This is where Diffie-Hellman comes in. Take our two would-be communicators, Alice and Bob. Alice and Bob each pick a binary string called a secret key, which in interpreted as a number in this protocol. Let’s call Alice’s secret key $s_A$ and Bob’s $s_B$, and note that they don’t have to be the same. As the name “secret key” suggests, the secret keys are held secret. Moreover, we’ll assume that everything else in this protocol, including all data sent between the two parties, is public.

So Alice and Bob agree ahead of time on a public elliptic curve $C$ and a public point $Q$ on $C$. We’ll sometimes call this point the base point for the protocol.

Bob can cunningly do the following trick: take his secret key $s_B$ and send $s_B Q$ to Alice. Equally slick Alice computes $s_A Q$ and sends that to Bob. Now Alice, having $s_B Q$, computes $s_A s_B Q$. And Bob, since he has $s_A Q$, can compute $s_B s_A Q$. But since addition is commutative in elliptic curve groups, we know $s_A s_B Q = s_B s_A Q$. The secret piece of shared information can be anything derived from this new point, for example its $x$-coordinate.

If we want to talk about security, we have to describe what is public and what the attacker is trying to determine. In this case the public information consists of the points $Q, s_AQ, s_BQ$. What is the attacker trying to figure out? Well she really wants to eavesdrop on their subsequent conversation, that is, the stuff that encrypt with their new shared secret $s_As_BQ$. So the attacker wants find out $s_As_BQ$. And we’ll call this the Diffie-Hellman problem.

Diffie-Hellman Problem: Suppose you fix an elliptic curve $E$ over a finite field $k$, and you’re given four points $Q, aQ, bQ$ and $P$ for some unknown integers $a, b$. Determine if $P = abQ$ in polynomial time (in the lengths of $Q, aQ, bQ, P$).

On one hand, if we had an efficient solution to the discrete logarithm problem, we could easily use that to solve the Diffie-Hellman problem because we could compute $a,b$ and them quickly compute $abQ$ and check if it’s $P$. In other words discrete log is at least as hard as this problem. On the other hand nobody knows if you can do this without solving the discrete logarithm problem. Moreover, we’re making this problem as easy as we reasonably can because we don’t require you to be able to compute $abQ$. Even if some prankster gave you a candidate for $abQ$, all you have to do is check if it’s correct. One could imagine some test that rules out all fakes but still doesn’t allow us to compute the true point, which would be one way to solve this problem without being able to solve discrete log.

So this is our hardness assumption: assuming this problem has no efficient solution then no attacker, even with really lucky guesses, can feasibly determine Alice and Bob’s shared secret.

Python Implementation

The Diffie-Hellman protocol is just as easy to implement as you would expect. Here’s some Python code that does the trick. Note that all the code produced in the making of this post is available on this blog’s Github page.

def sendDH(privateKey, generator, sendFunction):
return sendFunction(privateKey * generator)

return privateKey * receiveFunction()


And using our code from the previous posts in this series we can run it on a small test.

import os

def generateSecretKey(numBits):
return int.from_bytes(os.urandom(numBits // 8), byteorder='big')

if __name__ == &quot;__main__&quot;:
F = FiniteField(3851, 1)
curve = EllipticCurve(a=F(324), b=F(1287))
basePoint = Point(curve, F(920), F(303))

aliceSecretKey = generateSecretKey(8)
bobSecretKey = generateSecretKey(8)

alicePublicKey = sendDH(aliceSecretKey, basePoint, lambda x:x)
bobPublicKey = sendDH(bobSecretKey, basePoint, lambda x:x)

sharedSecret1 = receiveDH(bobSecretKey, lambda: alicePublicKey)
sharedSecret2 = receiveDH(aliceSecretKey, lambda: bobPublicKey)
print('Shared secret is %s == %s' % (sharedSecret1, sharedSecret2))


Pythons os module allows us to access the operating system’s random number generator (which is supposed to be cryptographically secure) via the function urandom, which accepts as input the number of bytes you wish to generate, and produces as output a Python bytestring object that we then convert to an integer. Our simplistic (and totally insecure!) protocol uses the elliptic curve $C$ defined by $y^2 = x^3 + 324 x + 1287$ over the finite field $\mathbb{Z}/3851$. We pick the base point $Q = (920, 303)$, and call the relevant functions with placeholders for actual network transmission functions.

There is one issue we have to note. Say we fix our base point $Q$. Since an elliptic curve over a finite field can only have finitely many points (since the field only has finitely many possible pairs of numbers), it will eventually happen that $nQ = 0$ is the ideal point. Recall that the smallest value of $n$ for which $nQ = 0$ is called the order of $Q$. And so when we’re generating secret keys, we have to pick them to be smaller than the order of the base point. Viewed from the other angle, we want to pick $Q$ to have large order, so that we can pick large and difficult-to-guess secret keys. In fact, no matter what integer you use for the secret key it will be equivalent to some secret key that’s less than the order of $Q$. So if an attacker could guess the smaller secret key he wouldn’t need to know your larger key.

The base point we picked in the example above happens to have order 1964, so an 8-bit key is well within the bounds. A real industry-strength elliptic curve (say, Curve25519 or the curves used in the NIST standards*) is designed to avoid these problems. The order of the base point used in the Diffie-Hellman protocol for Curve25519 has gargantuan order (like $2^{256}$). So 256-bit keys can easily be used. I’m brushing some important details under the rug, because the key as an actual string is derived from 256 pseudorandom bits in a highly nontrivial way.

So there we have it: a simple cryptographic protocol based on elliptic curves. While we didn’t experiment with a truly secure elliptic curve in this example, we’ll eventually extend our work to include Curve25519. But before we do that we want to explore some of the other algorithms based on elliptic curves, including random number generation and factoring.

Why do we use elliptic curves for this? Why not do something like RSA and do multiplication (and exponentiation) modulo some large prime?

Well, it turns out that algorithmic techniques are getting better and better at solving the discrete logarithm problem for integers mod $p$, leading some to claim that RSA is dead. But even if we will never find a genuinely efficient algorithm (polynomial time is good, but might not be good enough), these techniques have made it clear that the key size required to maintain high security in RSA-type protocols needs to be really big. Like 4096 bits. But for elliptic curves we can get away with 256-bit keys. The reason for this is essentially mathematical: addition on elliptic curves is not as well understood as multiplication is for integers, and the more complex structure of the group makes it seem inherently more difficult. So until some powerful general attacks are found, it seems that we can get away with higher security on elliptic curves with smaller key sizes.

I mentioned that the particular elliptic curve we chose was insecure, and this raises the natural question: what makes an elliptic curve/field/basepoint combination secure or insecure? There are a few mathematical pitfalls (including certain attacks we won’t address), but one major non-mathematical problem is called a side-channel attack. A side channel attack against a cryptographic protocol is one that gains additional information about users’ secret information by monitoring side-effects of the physical implementation of the algorithm.

The problem is that different operations, doubling a point and adding two different points, have very different algorithms. As a result, they take different amounts of time to complete and they require differing amounts of power. Both of these can be used to reveal information about the secret keys. Despite the different algorithms for arithmetic on Weierstrass normal form curves, one can still implement them to be secure. Naively, one might pad the two subroutines with additional (useless) operations so that they have more similar time/power signatures, but I imagine there are better methods available.

But much of what makes a curve’s domain parameters mathematically secure or insecure is still unknown. There are a handful of known attacks against very specific families of parameters, and so cryptography experts simply avoid these as they are discovered. Here is a short list of pitfalls, and links to overviews:

1. Make sure the order of your basepoint has a short facorization (e.g., is $2p, 3p,$ or $4p$ for some prime $p$). Otherwise you risk attacks based on the Chinese Remainder Theorem, the most prominent of which is called Pohlig-Hellman.
2. Make sure your curve is not supersingular. If it is you can reduce the discrete logarithm problem to one in a different and much simpler group.
3. If your curve $C$ is defined over $\mathbb{Z}/p$, make sure the number of points on $C$ is not equal to $p$. Such a curve is called prime-field anomalous, and its discrete logarithm problem can be reduced to the (additive) version on integers.
4. Don’t pick a small underlying field like $\mathbb{F}_{2^m}$ for small $m$. General-purpose attacks can be sped up significantly against such fields.
5. If you use the field $\mathbb{F}_{2^m}$, ensure that $m$ is prime. Many believe that if $m$ has small divisors, attacks based on some very complicated algebraic geometry can be used to solve the discrete logarithm problem more efficiently than any general-purpose method. This gives evidence that $m$ being composite at all is dangerous, so we might as well make it prime.

This is a sublist of the list provided on page 28 of this white paper.

The interesting thing is that there is little about the algorithm and protocol that is vulnerable. Almost all of the vulnerabilities come from using bad curves, bad fields, or a bad basepoint. Since the known attacks work on a pretty small subset of parameters, one potentially secure technique is to just generate a random curve and a random point on that curve! But apparently all respected national agencies will refuse to call your algorithm “standards compliant” if you do this.

Next time we’ll continue implementing cryptographic protocols, including the more general public-key message sending and signing protocols.

Until then!

Elliptic Curves as Python Objects

Last time we saw a geometric version of the algorithm to add points on elliptic curves. We went quite deep into the formal setting for it (projective space $\mathbb{P}^2$), and we spent a lot of time talking about the right way to define the “zero” object in our elliptic curve so that our issues with vertical lines would disappear.

With that understanding in mind we now finally turn to code, and write classes for curves and points and implement the addition algorithm. As usual, all of the code we wrote in this post is available on this blog’s Github page.

Points and Curves

Every introductory programming student has probably written the following program in some language for a class representing a point.

class Point(object):
def __init__(self, x, y):
self.x = x
self.y = y


It’s the simplest possible nontrivial class: an x and y value initialized by a constructor (and in Python all member variables are public).

We want this class to represent a point on an elliptic curve, and overload the addition and negation operators so that we can do stuff like this:

p1 = Point(3,7)
p2 = Point(4,4)
p3 = p1 + p2


But as we’ve spent quite a while discussing, the addition operators depend on the features of the elliptic curve they’re on (we have to draw lines and intersect it with the curve). There are a few ways we could make this happen, but in order to make the code that uses these classes as simple as possible, we’ll have each point contain a reference to the curve they come from. So we need a curve class.

It’s pretty simple, actually, since the class is just a placeholder for the coefficients of the defining equation. We assume the equation is already in the Weierstrass normal form, but if it weren’t one could perform a whole bunch of algebra to get it in that form (and you can see how convoluted the process is in this short report or page 115 (pdf p. 21) of this book). To be safe, we’ll add a few extra checks to make sure the curve is smooth.

class EllipticCurve(object):
def __init__(self, a, b):
# assume we're already in the Weierstrass form
self.a = a
self.b = b

self.discriminant = -16 * (4 * a*a*a + 27 * b * b)
if not self.isSmooth():
raise Exception(&quot;The curve %s is not smooth!&quot; % self)

def isSmooth(self):
return self.discriminant != 0

def testPoint(self, x, y):
return y*y == x*x*x + self.a * x + self.b

def __str__(self):
return 'y^2 = x^3 + %Gx + %G' % (self.a, self.b)

def __eq__(self, other):
return (self.a, self.b) == (other.a, other.b)


And here’s some examples of creating curves

&gt;&gt;&gt; EllipticCurve(a=17, b=1)
y^2 = x^3 + 17x + 1
&gt;&gt;&gt; EllipticCurve(a=0, b=0)
Traceback (most recent call last):
[...]
Exception: The curve y^2 = x^3 + 0x + 0 is not smooth!


So there we have it. Now when we construct a Point, we add the curve as the extra argument and a safety-check to make sure the point being constructed is on the given elliptic curve.

class Point(object):
def __init__(self, curve, x, y):
self.curve = curve # the curve containing this point
self.x = x
self.y = y

if not curve.testPoint(x,y):
raise Exception(&quot;The point %s is not on the given curve %s&quot; % (self, curve))


Note that this last check will serve as a coarse unit test for all of our examples. If we mess up then more likely than not the “added” point won’t be on the curve at all. More precise testing is required to be bullet-proof, of course, but we leave explicit tests to the reader as an excuse to get their hands wet with equations.

Some examples:

&gt;&gt;&gt; c = EllipticCurve(a=1,b=2)
&gt;&gt;&gt; Point(c, 1, 2)
(1, 2)
&gt;&gt;&gt; Point(c, 1, 1)
Traceback (most recent call last):
[...]
Exception: The point (1, 1) is not on the given curve y^2 = x^3 + 1x + 2


Before we go ahead and implement addition and the related functions, we need to be decide how we want to represent the ideal point $[0 : 1 : 0]$. We have two options. The first is to do everything in projective coordinates and define a whole system for doing projective algebra. Considering we only have one point to worry about, this seems like overkill (but could be fun). The second option, and the one we’ll choose, is to have a special subclass of Point that represents the ideal point.

class Ideal(Point):
def __init__(self, curve):
self.curve = curve

def __str__(self):
return &quot;Ideal&quot;


Note the inheritance is denoted by the parenthetical (Point) in the first line. Each function we define on a Point will require a 1-2 line overriding function in this subclass, so we will only need a small amount of extra bookkeeping. For example, negation is quite easy.

class Point(object):
...
def __neg__(self):
return Point(self.curve, self.x, -self.y)

class Ideal(Point):
...
def __neg__(self):
return self


Note that Python allows one to override the prefix-minus operation by defining __neg__ on a custom object. There are similar functions for addition (__add__), subtraction, and pretty much every built-in python operation. And of course addition is where things get more interesting. For the ideal point it’s trivial.

class Ideal(Point):
...
return Q


Why does this make sense? Because (as we’ve said last time) the ideal point is the additive identity in the group structure of the curve. So by all of our analysis, $P + 0 = 0 + P = P$, and the code is satisfyingly short.

For distinct points we have to follow the algorithm we used last time. Remember that the trick was to form the line $L(x)$ passing through the two points being added, substitute that line for $y$ in the elliptic curve, and then figure out the coefficient of $x^2$ in the resulting polynomial. Then, using the two existing points, we could solve for the third root of the polynomial using Vieta’s formula.

In order to do that, we need to analytically solve for the coefficient of the $x^2$ term of the equation $L(x)^2 = x^3 + ax + b$. It’s tedious, but straightforward. First, write

$\displaystyle L(x) = \left ( \frac{y_2 – y_1}{x_2 – x_1} \right ) (x – x_1) + y_1$

The first step of expanding $L(x)^2$ gives us

$\displaystyle L(x)^2 = y_1^2 + 2y_1 \left ( \frac{y_2 – y_1}{x_2 – x_1} \right ) (x – x_1) + \left [ \left (\frac{y_2 – y_1}{x_2 – x_1} \right ) (x – x_1) \right ]^2$

And we notice that the only term containing an $x^2$ part is the last one. Expanding that gives us

$\displaystyle \left ( \frac{y_2 – y_1}{x_2 – x_1} \right )^2 (x^2 – 2xx_1 + x_1^2)$

And again we can discard the parts that don’t involve $x^2$. In other words, if we were to rewrite $L(x)^2 = x^3 + ax + b$ as $0 = x^3 – L(x)^2 + ax + b$, we’d expand all the terms and get something that looks like

$\displaystyle 0 = x^3 – \left ( \frac{y_2 – y_1}{x_2 – x_1} \right )^2 x^2 + C_1x + C_2$

where $C_1, C_2$ are some constants that we don’t need. Now using Vieta’s formula and calling $x_3$ the third root we seek, we know that

$\displaystyle x_1 + x_2 + x_3 = \left ( \frac{y_2 – y_1}{x_2 – x_1} \right )^2$

Which means that $x_3 = \left ( \frac{y_2 – y_1}{x_2 – x_1} \right )^2 – x_2 – x_1$. Once we have $x_3$, we can get $y_3$ from the equation of the line $y_3 = L(x_3)$.

Note that this only works if the two points we’re trying to add are different! The other two cases were if the points were the same or lying on a vertical line. These gotchas will manifest themselves as conditional branches of our add function.

class Point(object):
...
if isinstance(Q, Ideal):
return self

x_1, y_1, x_2, y_2 = self.x, self.y, Q.x, Q.y

if (x_1, y_1) == (x_2, y_2):
# use the tangent method
...
else:
if x_1 == x_2:
return Ideal(self.curve) # vertical line

# Using Vieta's formula for the sum of the roots
m = (y_2 - y_1) / (x_2 - x_1)
x_3 = m*m - x_2 - x_1
y_3 = m*(x_3 - x_1) + y_1

return Point(self.curve, x_3, -y_3)



First, we check if the two points are the same, in which case we use the tangent method (which we do next). Supposing the points are different, if their $x$ values are the same then the line is vertical and the third point is the ideal point. Otherwise, we use the formula we defined above. Note the subtle and crucial minus sign at the end! The point $(x_3, y_3)$ is the third point of intersection, but we still have to do the reflection to get the sum of the two points.

Now for the case when the points $P, Q$ are actually the same. We’ll call it $P = (x_1, y_1)$, and we’re trying to find $2P = P+P$. As per our algorithm, we compute the tangent line $J(x)$ at $P$. In order to do this we need just a tiny bit of calculus. To find the slope of the tangent line we implicitly differentiate the equation $y^2 = x^3 + ax + b$ and get

$\displaystyle \frac{dy}{dx} = \frac{3x^2 + a}{2y}$

The only time we’d get a vertical line is when the denominator is zero (you can verify this by taking limits if you wish), and so $y=0$ implies that $P+P = 0$ and we’re done. The fact that this can ever happen for a nonzero $P$ should be surprising to any reader unfamiliar with groups! But without delving into a deep conversation about the different kinds of group structures out there, we’ll have to settle for such nice surprises.

In the other case $y \neq 0$, we plug in our $x,y$ values into the derivative and read off the slope $m$ as $(3x_1^2 + a)/(2y_1)$. Then using the same point slope formula for a line, we get $J(x) = m(x-x_1) + y_1$, and we can use the same technique (and the same code!) from the first case to finish.

There is only one minor wrinkle we need to smooth out: can we be sure Vieta’s formula works? In fact, the real problem is this: how do we know that $x_1$ is a double root of the resulting cubic? Well, this falls out again from that very abstract and powerful theorem of Bezout. There is a lot of technical algebraic geometry (and a very interesting but complicated notion of dimension) hiding behind the curtain here. But for our purposes it says that our tangent line intersects the elliptic curve with multiplicity 2, and this gives us a double root of the corresponding cubic.

And so in the addition function all we need to do is change the slope we’re using. This gives us a nice and short implementation

def __add__(self, Q):
if isinstance(Q, Ideal):
return self

x_1, y_1, x_2, y_2 = self.x, self.y, Q.x, Q.y

if (x_1, y_1) == (x_2, y_2):
if y_1 == 0:
return Ideal(self.curve)

# slope of the tangent line
m = (3 * x_1 * x_1 + self.curve.a) / (2 * y_1)
else:
if x_1 == x_2:
return Ideal(self.curve)

# slope of the secant line
m = (y_2 - y_1) / (x_2 - x_1)

x_3 = m*m - x_2 - x_1
y_3 = m*(x_3 - x_1) + y_1

return Point(self.curve, x_3, -y_3)


What’s interesting is how little the data of the curve comes into the picture. Nothing depends on $b$, and only one of the two cases depends on $a$. This is one reason the Weierstrass normal form is so useful, and it may bite us in the butt later in the few cases we don’t have it (for special number fields).

Here are some examples.

&gt;&gt;&gt; C = EllipticCurve(a=-2,b=4)
&gt;&gt;&gt; P = Point(C, 3, 5)
&gt;&gt;&gt; Q = Point(C, -2, 0)
&gt;&gt;&gt; P+Q
(0.0, -2.0)
&gt;&gt;&gt; Q+P
(0.0, -2.0)
&gt;&gt;&gt; Q+Q
Ideal
&gt;&gt;&gt; P+P
(0.25, 1.875)
&gt;&gt;&gt; P+P+P
Traceback (most recent call last):
...
Exception: The point (-1.958677685950413, 0.6348610067618328) is not on the given curve y^2 = x^3 + -2x + 4!

&gt;&gt;&gt; x = -1.958677685950413
&gt;&gt;&gt; y = 0.6348610067618328
&gt;&gt;&gt; y*y - x*x*x + 2*x - 4
-3.9968028886505635e-15


And so we crash headfirst into our first floating point arithmetic issue. We’ll vanquish this monster more permanently later in this series (in fact, we’ll just scrap it entirely and define our own number system!), but for now here’s a quick fix:

&gt;&gt;&gt; import fractions
&gt;&gt;&gt; frac = fractions.Fraction
&gt;&gt;&gt; C = EllipticCurve(a = frac(-2), b = frac(4))
&gt;&gt;&gt; P = Point(C, frac(3), frac(5))
&gt;&gt;&gt; P+P+P
(Fraction(-237, 121), Fraction(845, 1331))


Now that we have addition and negation, the rest of the class is just window dressing. For example, we want to be able to use the subtraction symbol, and so we need to implement __sub__

def __sub__(self, Q):
return self + -Q


Note that because the Ideal point is a subclass of point, it inherits all of these special functions while it only needs to override __add__ and __neg__. Thank you, polymorphism! The last function we want is a scaling function, which efficiently adds a point to itself $n$ times.

class Point(object):
...
def __mul__(self, n):
if not isinstance(n, int):
raise Exception(&quot;Can't scale a point by something which isn't an int!&quot;)
else:
if n &lt; 0:
return -self * -n
if n == 0:
return Ideal(self.curve)
else:
Q = self
R = self if n &amp; 1 == 1 else Ideal(self.curve)

i = 2
while i &lt;= n:
Q = Q + Q

if n &amp; i == i:
R = Q + R

i = i &lt;&lt; 1
return R

def __rmul__(self, n):
return self * n

class Ideal(Point):
...
def __mul__(self, n):
if not isinstance(n, int):
raise Exception(&quot;Can't scale a point by something which isn't an int!&quot;)
else:
return self


The scaling function allows us to quickly compute $nP = P + P + \dots + P$ ($n$ times). Indeed, the fact that we can do this more efficiently than performing $n$ additions is what makes elliptic curve cryptography work. We’ll take a deeper look at this in the next post, but for now let’s just say what the algorithm is doing.

Given a number written in binary $n = b_kb_{k-1}\dots b_1b_0$, we can write $nP$ as

$\displaystyle b_0 P + b_1 2P + b_2 4P + \dots + b_k 2^k P$

The advantage of this is that we can compute each of the $P, 2P, 4P, \dots, 2^kP$ iteratively using only $k$ additions by multiplying by 2 (adding something to itself) $k$ times. Since the number of bits in $n$ is $k= \log(n)$, we’re getting a huge improvement over $n$ additions.

The algorithm is given above in code, but it’s a simple bit-shifting trick. Just have $i$ be some power of two, shifted by one at the end of every loop. Then start with $Q_0$ being $P$, and replace $Q_{j+1} = Q_j + Q_j$, and in typical programming fashion we drop the indices and overwrite the variable binding at each step (Q = Q+Q). Finally, we have a variable $R$ to which $Q_j$ is added when the $j$-th bit of $n$ is a 1 (and ignored when it’s 0). The rest is bookkeeping.

Note that __mul__ only allows us to write something like P * n, but the standard notation for scaling is n * P. This is what __rmul__ allows us to do.

We could add many other helper functions, such as ones to allow us to treat points as if they were lists, checking for equality of points, comparison functions to allow one to sort a list of points in lex order, or a function to transform points into more standard types like tuples and lists. We have done a few of these that you can see if you visit the code repository, but we’ll leave flushing out the class as an exercise to the reader.

Some examples:

&gt;&gt;&gt; import fractions
&gt;&gt;&gt; frac = fractions.Fraction
&gt;&gt;&gt; C = EllipticCurve(a = frac(-2), b = frac(4))
&gt;&gt;&gt; P = Point(C, frac(3), frac(5))
&gt;&gt;&gt; Q = Point(C, frac(-2), frac(0))
&gt;&gt;&gt; P-Q
(Fraction(0, 1), Fraction(-2, 1))
&gt;&gt;&gt; P+P+P+P+P
(Fraction(2312883, 1142761), Fraction(-3507297955, 1221611509))
&gt;&gt;&gt; 5*P
(Fraction(2312883, 1142761), Fraction(-3507297955, 1221611509))
&gt;&gt;&gt; Q - 3*P
(Fraction(240, 1), Fraction(3718, 1))
&gt;&gt;&gt; -20*P
(Fraction(872171688955240345797378940145384578112856996417727644408306502486841054959621893457430066791656001, 520783120481946829397143140761792686044102902921369189488390484560995418035368116532220330470490000), Fraction(-27483290931268103431471546265260141280423344817266158619907625209686954671299076160289194864753864983185162878307166869927581148168092234359162702751, 11884621345605454720092065232176302286055268099954516777276277410691669963302621761108166472206145876157873100626715793555129780028801183525093000000))


As one can see, the precision gets very large very quickly. One thing we’ll do to avoid such large numbers (but hopefully not sacrifice security) is to work in finite fields, the simplest version of which is to compute modulo some prime.

So now we have a concrete understanding of the algorithm for adding points on elliptic curves, and a working Python program to do this for rational numbers or floating point numbers (if we want to deal with precision issues). Next time we’ll continue this train of thought and upgrade our program (with very little work!) to work over other simple number fields. Then we’ll delve into the cryptographic issues, and talk about how one might encode messages on a curve and use algebraic operations to encode their messages.

Until then!

Elliptic Curves as Elementary Equations

Finding solutions to systems of polynomial equations is one of the oldest and deepest problems in all of mathematics. This is broadly the domain of algebraic geometry, and mathematicians wield some of the most sophisticated and abstract tools available to attack these problems.

The elliptic curve straddles the elementary and advanced mathematical worlds in an interesting way. On one hand, it’s easy to describe in elementary terms: it’s the set of solutions to a cubic function of two variables. But despite how simple they seem deep theorems govern their behavior, and many natural questions about elliptic curves are still wide open. Since elliptic curves provide us with some of the strongest and most widely used encryption protocols, understanding elliptic curves more deeply would give insight into the security (or potential insecurity) of these protocols.

Our first goal in this series is to treat elliptic curves as mathematical objects, and derive the elliptic curve group as the primary object of study. We’ll see what “group” means next time, and afterward we’ll survey some of the vast landscape of unanswered questions. But this post will be entirely elementary, and will gently lead into the natural definition of the group structure on an elliptic curve.

Elliptic Curves as Equations

The simplest way to describe an elliptic curve is as the set of all solutions to a specific kind of polynomial equation in two real variables, $x,y$. Specifically, the equation has the form:

$\displaystyle y^2 = x^3 + ax + b$

Where $a,b$ are real numbers such that

$\displaystyle -16(4a^3 + 27b^2) \neq 0$

One would naturally ask, “Who the hell came up with that?” A thorough answer requires a convoluted trip through 19th and 20th-century mathematical history, but it turns out that this is a clever form of a very natural family of equations. We’ll elaborate on this in another post, but for now we can give an elementary motivation.

Say you have a pyramid of spheres whose layers are squares, like the one below

We might wonder when it’s the case that we can rearrange these spheres into a single square. Clearly you can do it for a pyramid of height 1 because a single ball is also a 1×1 square (and one of height zero if you allow a 0x0 square). But are there any others?

This question turns out to be a question about an elliptic curve. First, recall that the number of spheres in such a pyramid is given by

$\displaystyle 1 + 4 + 9 + 16 + \dots + n^2 = \frac{n(n+1)(2n+1)}{6}$

And so we’re asking if there are any positive integers $y$ such that

$\displaystyle y^2 = \frac{x(x+1)(2x+1)}{6}$

Here is a graph of this equation in the plane. As you admire it, though, remember that we’re chiefly interested in integer solutions.

The equation doesn’t quite have the special form we mentioned above, but the reader can rest assured (and we’ll prove it later) that one can transform our equation into that form without changing the set of solutions. In the meantime let’s focus on the question: are there any integer-valued points on this curve besides $(0,0)$ and $(1,1)$? The method we use to answer this question comes from ancient Greece, and is due to Diophantus. The idea is that we can use the two points we already have to construct a third point. This method is important because it forms the basis for our entire study of elliptic curves.

Take the line passing through $(0,0)$ and  $(1,1)$, given by the equation $y = x$, and compute the intersection of this line and the original elliptic curve. The “intersection” simply means to solve both equations simultaneously. In this case it’s

\begin{aligned} y^2 &= \frac{x(x+1)(2x+1)}{6} \\ y &= x \end{aligned}

It’s clear what to do: just substitute the latter in for the former. That is, solve

$\displaystyle x^2 = \frac{x(x+1)(2x+1)}{6}$

Rearranging this into a single polynomial and multiplying through by 3 gives

$\displaystyle x^3 – \frac{3x^2}{2} + \frac{x}{2} = 0$

Factoring cubics happens to be easy, but let’s instead use a different trick that will come up again later. Let’s use a fact that is taught in elementary algebra and precalculus courses and promptly forgotten, that the sum of the roots of any polynomial is $\frac{-a_{n-1}}{a_n}$, where $a_{n}$ is the leading coefficient and $a_{n-1}$ is the next coefficient. Here $a_n = 1$, so the sum of the roots is $3/2$. This is useful because we already know two roots, namely the solutions 0 and 1 we used to define the system of equations in the first place. So the third root satisfies

$\displaystyle r + 0 + 1 = \frac{3}{2}$

And it’s $r = 1/2$, giving the point $(1/2, 1/2)$ since the line was $y=x$. Because of the symmetry of the curve, we also get the point $(1/2, -1/2)$.

Here’s a zoomed-in picture of what we just did to our elliptic curve. We used the two pink points (which gave us the dashed line) to find the purple point.

The bad news is that these two new points don’t have integer coordinates. So it doesn’t answer our question. The good news is that now we have more points! So we can try this trick again to see if it will give us still more points, and hope to find some that are integer valued. (It sounds like a hopeless goal, but just hold out a bit longer). If we try this trick again using $(1/2, -1/2)$ and $(1,1)$, we get the equation

$\displaystyle (3x – 2)^2 = \frac{x(x+1)(2x+1)}{6}$

And redoing all the algebraic steps we did before gives us the solution $x=24, y=70$. In other words, we just proved that

$\displaystyle 1^2 + 2^2 + \dots + 24^2 = 70^2$

Great! Here’s another picture showing what we just did.

In reality we don’t care about this little puzzle. Its solution might be a fun distraction (and even more distracting: try to prove there aren’t any other integer solutions), but it’s not the real treasure. The mathematical gem is the method of finding the solution. We can ask the natural question: if you have two points on an elliptic curve, and you take the line between those two points, will you always get a third point on the curve?

Certainly the answer is no. See this example of two points whose line is vertical.

But with some mathematical elbow grease, we can actually force it to work! That is, we can define things just right so that the line between any two points on an elliptic curve will always give you another point on the curve. This sounds like mysterious black magic, but it lights the way down a long mathematical corridor of new ideas, and is required to make sense of using elliptic curves for cryptography.

Shapes of Elliptic Curves

Before we continue, let’s take a little detour to get a good feel for the shapes of elliptic curves. We have defined elliptic curves by a special kind of equation (we’ll give it a name in a future post). During most of our study we won’t be able to make any geometric sense of these equations. But for now, we can pretend that we’re working over real numbers and graph these equations in the plane.

Elliptic curves in the form $y^2 = x^3 + ax + b$ have a small handful of different shapes that we can see as $a,b$ vary:

The problem is when we cross the point at which the rounded part pinches off in the first animation, and the circular component appears in the second. At those precise moments, the curve becomes “non-smooth” (or singular), and for reasons we’ll see later this is bad. The condition from the beginning of the article (that $-16(4a^3 + 27b^2) \neq 0$) ensures that these two cases are excluded from consideration, and it’s one crucial part of our “elbow grease” to ensure that lines behave nicely.

The “canonical” shape of the elliptic curve is given by the specific example $y^2 = x^3 – x + 1$. It’s the example that should pop up whenever you imagine an elliptic curve, and it’s the example we’ll use for all of our pictures.

So in the next post we’ll roll up our sleeves and see exactly how “drawing lines” can be turned into an algebraic structure on an elliptic curve.

Until then!