Posted by Jason Polak on 27. July 2018 · Write a comment · Categories: number-theory · Tags: , , ,

This is the final post on the Jacobi symbol. Recall that the Jacobi symbol $(m/n)$ for relatively prime integers $m$ and $n$ is defined to be the sign of the permutation $x\mapsto mx$ on the ring $\Z/n$. In the introductory post we saw this definition, some examples, and basic properties for calculation purposes.

In Part 2 we saw that for an odd prime $p$ and an integer $a$ that is relatively prime to $p$, the Jacobi symbol $(a/p) = 1$ if and only if $a$ is a square modulo $p$ (a "quadratic residue"). The basic properties of the Jacobi symbol then give the classic law of quadratic reciprocity.

Now, we're going to see one last application of the Jacobi symbol: primality testing in what's called the Solovay-Strassen primality test. How does it work? It starts with an observation we saw before: in the ring $\Z/p$, there exists a primitive element $g\in \Z/p$. It is an element that generates the multiplicative cyclic group $\Z/p^\times\cong \Z/(p-1)$.
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Posted by Jason Polak on 25. July 2018 · Write a comment · Categories: number-theory · Tags: , ,

In the last post, we examined the Jacobi symbol: for two relatively prime integers $m$ and $n$, we defined the Jacobi symbol $(m/n)$ to be the sign of the permutation $x\mapsto mx$ on the ring $\Z/n$.

It turns out that the Jacobi symbol plays a part in the theory of quadratic residues. For a number $n$, we say that an element $a\in \Z/n$ is a quadratic residue if it's a square in $\Z/n$.

When $n = p$ is a prime number, the question of which integers are quadratic residues goes back to the ancient days of Gauss. He realized that for an odd prime $p$, half the numbers in the list $1,2,\dots, p-1$ are quadratic residues, and the other half are not quadratic residues (a.k.a. quadratic nonresidues). Indeed, if $x^2 = y^2$ in the field $\Z/p$, then $(x+y)(x-y) = 0$. Therefore, if $x\not= y$, we must have $x=-y$. So it is clear that amongst the numbers $1^2,2^2,\dots,(p-1)^2$, there are exactly $(p-1)/2$ distinct numbers.

This guy means business!

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Posted by Jason Polak on 21. July 2018 · Write a comment · Categories: number-theory · Tags: ,

If $m$ and $n$ are relatively prime integers, the Jacobi symbol $(m/n)$ is defined as the sign of the permutation $x\mapsto mx$ on the set $\Z/n$. Let's give a simple example: $(7/5)$. The permutation on $\{1,2,3,4\}$ is given by $(1 2 4 3) = (1 2)(2 4)(4 3)$ which has an odd number of transpositions. Therefore, $(7/5) = -1$.

Note that as in this example, it is sufficient to compute the sign of the permutation on $\Z/n – \{0\}$, since multiplication always leaves zero fixed.

But what if we wanted to compute something like $(3/412871)$? These numbers aren't so big, so a computer could do it directly. However, there is a better way to do the computation of the Jacobi symbol $(m/n)$ if one of $m$ or $n$ is much larger than the other one. This method is good for computers too when one of the numbers is so large, that a direct computation even by a fast computer would be hopeless.
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Posted by Jason Polak on 08. July 2018 · Write a comment · Categories: number-theory · Tags:

A perfect number is a positive integer $n$ such that
$$\sum_{d|n} d = 2n.$$Put another way, $n$ is the sum of its proper divisors. Check out a a quick intro to perfect numbers that I wrote last November. The first three perfect numbers are $6, 28,$ and $496$. Currently, the largest perfect number, corresponding to the largest Mersenne prime is
$$(2^{77232917} – 1)\cdot 2^{77232916}$$This perfect number is over 46 million digits long!
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Posted by Jason Polak on 13. May 2018 · Write a comment · Categories: number-theory

We've been talking about the Miller-Rabin randomized primality test, which is one of the easiest to implement and most effective tests that, given a number, will either prove it to be composite or state that it is most likely prime. As good as it is for practical applications, the Miller-Rabin test leaves something to be desired.

If you need primes for cryptographic applications, Miller-Rabin practically perfect. But what if you want to generate super large primes and state for sure that the numbers you are generating really are primes?

The Lucas test is a fast test that can prove a number prime. Here it is:

Theorem (Lucas Test). Let $a$ and $n\gt 1$ be integers. If $a^{n-1}\equiv 1\pmod{n}$ and $a^{(n-1)/q}\not\equiv 1\pmod{n}$ for every prime factor $q\mid n-1$ then $n$ is prime.

The proof is quite simple: the congruences that $a$ satisfy show that $a$ generate the multiplicative group $\mathbb{Z}/n^\times$. Therefore, $n-1\mid |\mathbb{Z}/n^\times| = \varphi(n)$. However, if $n$ were composite then $\varphi(n) \lt n-1$, which contradicts $n-1\mid\varphi(n)$. Therefore, $n$ is prime.
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Posted by Jason Polak on 11. May 2018 · Write a comment · Categories: number-theory · Tags:

Last time, I explained the Miller-Rabin probabilistic primality test. Let's recall it:

Theorem. Let $p$ be an odd prime and write $p-1 = 2^kq$ where $q$ is an odd number. If $a$ is relatively prime to $p$ then at least one of the following statements is true:

  1. $a^q\equiv 1\pmod{p}$, or
  2. One of $a^q,a^{2q},a^{4q},\dots,a^{2^{k-1}q}$ is congruent to $-1$ modulo $p$.

The Miller-Rabin randomized test looks for values of $a$ that don't satisfy the conclusions of this theorem. Such an $a$ is called a witness. If such an $a$ can be found, then $n$ is composite. If no such $a$ is found after a bunch of iterations, then we consider $n$ to be probably prime.

For example, a hundred iterations of Miller-Rabin selected

as a number that is most likely prime. A brute-force factor testing to make sure that this number is prime would take many times the age of the universe. That's pretty crazy! The Miller-Rabin test only takes a couple seconds, but how good is it really?

Now what about the percentage of witnesses for different numbers? Is this percentage close to 75%? In fact, although at least 75% of all possible bases are witnesses for any given composite number, often there are far more bases. In fact, only 0.6% of composite integers below 2000 have fewer than 90% of the total possible bases as witnesses. One such number is 1891 = 31*61. About 76% of possible bases are Miller-Rabin witnesses for 1891, which shows that the estimate of 75% is pretty good.

But for cryptographic applications, we're not interested in finding composite numbers. Given a randomly selected odd composite number, we can say that the probability of the Miller-Rabin test failing to find it composite after $k$ iterations is at most $4^{-k}$, which we can make very small very quickly. However, this conditional probability isn't the same thing as the conditional probability that a randomly selected odd number is composite given that $k$-iterations of Miller-Rabin thinks it is a prime!

A priori, it may be possible that the probability a number is composite given that $k$-iterations of Miller-Rabin thinks it's a prime is higher than $4^{-k}$.

But this isn't the case. Beauchemin, Brassard, Crépeau, Goutier, and Pomerance showed in a paper that this probability (that a randomly selected odd number is composite given $k$-iterations of Miller-Rabin thinks it is a prime) is also at most $4^{-k}$, assuming that the randomly selected odd numbers of sufficiently large (which thankfully is less than the typical modern-day cryptographic application).

Posted by Jason Polak on 09. May 2018 · Write a comment · Categories: number-theory · Tags: , ,

Fermat's little theorem states that for a prime number $p$, any $a\in \Z/p^\times$ satisfies $a^{p-1} = 1$. If $p$ is not prime, this may not necessarily be true. For example:
$$2^{402} = 376 \in \Z/403^\times.$$
Therefore, we can conclude that 403 is not a prime number. In fact, $403 = 13\cdot 31$ Fermat's little theorem can be used as a test for compositeness this way: if you can find a number $a$ relatively prime to $p$ such that $a^{p-1} \not\equiv 1\pmod{p}$, then $p$ is actually composite.

If there exists an $a\in\{1,\dots,p-1\}$ with $a^{p-1}\not\equiv 1\pmod{p}$, then $a$ is called a Fermat witness to the compositeness of $p$. That is, it proves that $p$ is composite. Notice that we do not require $a$ to be relatively prime to $p$ in this definition of a Fermat witness: if any number $a\lt p$ exists with $a^{p-1}\not\equiv 1\pmod{p}$, then $p$ cannot be prime, because if it were, $a$ would actually be relatively prime to $p$ and this would contradict Fermat's little theorem.

Fermat's little theorem is a pretty good test for compositeness: if $p$ is some odd composite integer with at least one relatively prime Fermat witness, then at least half the numbers in the range $2,3,\dots,p-2$ will also be witnesses. Note: we are using this range because $p-1$ and $1$ aren't witnesses. Therefore, you can use Fermat's little theorem as a randomized prime-testing algorithm: randomly select elements $a$ from $\{2,\dots,p-2\}$ and check if they satisfy $a^{p-1} = 1\pmod{p}$. Therefore, if you have some large number $N$ that is composite and has at least one witness, a randomised Fermat's little theorem algorithm randomly testing $K$ different bases in $\{2,\dots,N-1\}$ will have probability of less than $1/2^K$ at failing to detect that $N$ is composite.
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Posted by Jason Polak on 06. May 2018 · Write a comment · Categories: number-theory · Tags:

Alice wants her friends to send her stuff only she can read. RSA public-key encryption allows her to do that: she chooses huge primes $p$ and $q$ and releases $N = pq$ along with an encryption exponent $e$ such that ${\rm gcd}(e,(p-1)(q-1)) = 1$. If Bob wants to send Alice a message $m$, he sends $c = m^e$ modulo $N$ to Alice.

Computing $m$ from $m^e$ is hard if you only know $N$, but becomes easy with the prime factors $p$ and $q$. And, only Alice knows $p$ and $q$. The word "hard" is relative, and maybe someone will find an easy way to do it in the future, perhaps with quantum computers.

However, if you know $p$ and $q$, then you can compute $d$ such that $de = 1$ modulo $(p-1)(q-1)$. That is, you're computing the inverse of $e$ in the multiplicative group $(\mathbb{Z}/N)^\times$. You can use Fermat's little theorem: $x^{p-1} = 1$ in $\mathbb{Z}/p^\times$ for an $x\in (\mathbb{Z}/p)^\times$ to compute that $c^d = m^{ed} = m$ modulo $N$. Therefore, $c^d$ in $\mathbb{Z}/N$ is the decrypted message.
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Posted by Jason Polak on 03. May 2018 · Write a comment · Categories: number-theory

In a group $G$, the discrete logarithm problem is to solve for $x$ in the equation $g^x =h$ where $g,h\in G$. In Part 1, we saw that solving the discrete log problem for finite fields would imply that we could solve the Diffie-Hellman problem and crack the ElGamal encryption scheme.

Obviously, the main question at this point is: how hard is the discrete logarithm problem (DLP)? In this post, we will look at various ways to approach the problem. We'll focus on solving the DLP in the cyclic group $\F_p^\times$, though some of the ideas generalise to more general groups as well.

The Brute Force Approach

So, we've got this equation: $g^x = h$ in $\F_p^\times$. Of course, we're assuming that there exists an $x$ that solves this equation, because in practice a solution exists.
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Posted by Jason Polak on 30. April 2018 · Write a comment · Categories: number-theory · Tags:

Given a group $G$, and an element $g\in G$ (the "base"), the discrete logarithm $\log_g(h)$ of an $h\in G$ is an element $x\in G$ such that $g^x = h$ if it exists. Its name "discrete logarithm" essentially means that we are only allowed to use integer powers in the group, rather than extending the definition of exponentiation as with the usual logarithm for the real numbers.

Because of cryptographic applications, usually the discrete logarithm is seen in the world of finite fields or elliptic curves. Here, given a finite field $\F_q$, the discrete logarithm is considered in the multiplicative group $\F_q^\times$. It so happens that this group is cyclic, and hence it must be isomorphic to $\Z/(q-1)$. So, given a $g\in \F_q$ that generates $\F_q^\times$ as a multiplicative group, the discrete logarithm always exists and thus defines a function
$$\log_g: \F_q^\times\longrightarrow \Z/(q-1).$$
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