The exponent of a number indicates how many times the number is multiplied by itself. For example:

\( 4^{3} = 4 \cdot 4 \cdot 4 = 64 \)

Here, 3 is the exponent (or power), and 4 is the base. In words, we say "4 to the power of 3."

Here are the laws of exponents:

You already use modulo computation when you look at a clock and, for example, need to figure out what time it will be 3 hours after 11 o'clock, which is 2 o'clock. In mathematics, we write this as:

\( (11 + 3) \: mod \: 12 = 2 \)

Here, 12 is the modulus because we want the time as an integer between 0 and 11 (in this case, 12 o'clock is denoted by 0). In words, we say "11 plus 3 modulo 12 equals 2." The result of a modulo computation is an integer between 0 and the modulus minus 1. For example, with modulus 3, we have:

• \( 1 \: mod \: 3 = 1 \)
• \( 2 \: mod \: 3 = 2 \)
• \( 3 \: mod \: 3 = 0 \)
• \( 4 \: mod \: 3 = 1 \)
• \( 5 \: mod \: 3 = 2 \)
• \( 6 \: mod \: 3 = 0 \)
• etc.

For example, if we look at \( 27 \: mod \: 5 \), modulo computes how many times 5 divides into 27 and then returns the remainder, which is 2 in this case. That is, \( 27 \: mod \: 5 = 2 \). But how do we get this result?

First, we compute how many times we can multiply 5 by an integer \( x \) so that the result is as close as possible to 27 without exceeding it. In other words, we find the maximum value of \( x \) such that \( 5 \cdot x \leq 27 \). In this case, \( x = 5 \) because \( 5 \cdot 5 = 25 \leq 27 \). Then, by subtracting 25 from 27, we get the answer: \( 27 - 25 = 2 \).

If the integer is negative, for example \( -27 \: mod \: 5 \), we have to do it slightly differently, and the answer is \( -27 \: mod \: 5 = 3 \). In this case, the integer \( x \) is negative and should be the closest integer such that \( 5 \cdot x \) does not exceed -27. That is, we find the minimum value of \( -x \) such that \( 5 \cdot -x \geq -27 \). Here, \( -x = -6 \) because \( 5 \cdot -6 = -30 \geq -27 \). Then, by subtracting -30 from -27, we get the answer: \( -27 - (-30) = -27 + 30 = 3 \).

It is important that \( x \) or \( -x \) is an integer such as \( -14, 3, 17 \), etc., and NOT a fraction or decimal such as \( \frac{1}{4}, \frac{-3}{7}, 2.5, 5.1 \), etc.

If two integers \( a \) and \( b \) modulo the same modulus \( c \) return the same remainder \( r \), then we say that \( a \) and \( b \) are congruent modulo \( c \). That is, if \( a \: mod \: c = r \) and \( b \: mod \: c = r \), then \( a \equiv b \: (mod \: c) \). Also, notice that if the modulus \( c \) is greater than the integer \( a \), i.e., \( c > a \), the result will always be \( a \: mod \: c = a \).

A prime number is an integer greater than 1 that can only be divided evenly by 1 and itself. "Divided evenly" means the result is an integer, not a float. For example, if you divide 13 by 3, you get the float \( \frac{13}{3} = 4.333 \). We see that 13 is a prime number because it can only be divided evenly by 1 and itself: \( \frac{13}{1} = 13 \) and \( \frac{13}{13} = 1 \).

If an integer is not a prime number, it is called a composite number. For example, the integer 6 is a composite number because it can be divided evenly by 1, 2, 3, and 6:

\( \frac{6}{1} = 6 \), \( \frac{6}{2} = 3 \), \( \frac{6}{3} = 2 \), and \( \frac{6}{6} = 1 \)

The term "composite" means "something made by combining things." So, a composite number is an integer made by multiplying prime numbers:

• 2 is a prime
• 3 is a prime
• 4 is composite: \( 2 \cdot 2 = 4 \)
• 5 is a prime
• 6 is composite: \( 2 \cdot 3 = 6 \)
• 7 is a prime
• 8 is composite: \( 2 \cdot 2 \cdot 2 = 8 \)
• 9 is composite: \( 3 \cdot 3 = 9 \)
• etc.

Therefore, every integer greater than 1 is either a prime number or a product of prime numbers (a composite number).

The famous Greek mathematician Euclid proved that there are infinitely many prime numbers.

Michael Rabin and Gary Miller developed an algorithm that determines whether an integer is a prime or a composite number by testing the integer with multiple bases, denoted by \( a \). The algorithm is called the Rabin-Miller primality test.

Before we describe what the greatest common divisor of two integers is, we first define what we mean by a divisor. In this context, a divisor of an integer \( x \) is any integer that divides \( x \) evenly, meaning the result is not a fraction. For example, if you divide \( x=12 \) by 5, you get the fraction \( \frac{12}{5} = 2.4 \), so 5 is not a divisor of \( x=12 \). For \( x=12 \), the divisors are 1, 2, 3, 4, 6, and 12 because \( \frac{12}{1} = 12 \), \( \frac{12}{2} = 6 \), \( \frac{12}{3} = 4 \), \( \frac{12}{4} = 3 \), \( \frac{12}{6} = 2 \), and \( \frac{12}{12} = 1 \).

Similarly, the divisors of 16 are 1, 2, 4, 8, and 16 because \( \frac{16}{1} = 16 \), \( \frac{16}{2} = 8 \), \( \frac{16}{4} = 4 \), \( \frac{16}{8} = 2 \), and \( \frac{16}{16}=1 \).

The greatest common divisor of 12 and 16 is therefore 4, because it is the largest integer among their common divisors. In mathematics, we write this as \( \gcd(12, 16) = 4 \).

Two integers whose greatest common divisor is 1 are called relatively prime numbers or co-primes. For example, 15 and 28 are relatively prime because \( \gcd(15, 28) = 1 \) (note that 28 is not a prime number).

If one of the two integers is a prime number, the greatest common divisor will always be 1, i.e., \( \gcd(a, p) = 1 \), where \( a \) is any integer (either prime or composite) and \( p \) is a prime number.

One method to compute the greatest common divisor of two integers is by using the Euclidean algorithm, developed by the famous Greek mathematician Euclid. See "The extended Euclidean algorithm" for more information about how to compute the greatest common divisor of two integers.

The extended Euclidean algorithm is an extension of the Euclidean algorithm, which only returns the greatest common divisor of two integers. Given two integers \( a \) and \( b \), the extended Euclidean algorithm returns the integers \( a \), \( b \), \( \lambda \), and \( \mu \) such that:

\( a \cdot \lambda + b \cdot \mu = \gcd(a, b) \)

Here, \( \lambda \) and \( \mu \) are called the Bézout coefficients for \( a \) and \( b \). Only if \( a \) and \( b \) are relatively prime, i.e., \( \gcd(a, b) = 1 \), then:

\( a \cdot \lambda + b \cdot \mu = 1 \)

In this case, \( \lambda \; mod \; b \) is the inverse of \( a \), denoted \( a^{-1} = \lambda \; mod \; b \), and \( \mu \: mod \: a \) is the inverse of \( b \), denoted \( b^{-1} = \mu \: mod \: a \) (see "Modulo computation" for more information about the \( mod \) operator). One useful property of an integer and its inverse is that \( a \cdot a^{-1} \; mod \; b = 1 \) and \( b \cdot b^{-1} \; mod \; a = 1 \).

You can easily compute \( \gcd(a, b) \), \( \lambda \), and \( \mu \) for example with \( a=5 \) and \( b=39 \) using a simple table. First, let us create a table with three columns (we do not yet know how many rows there will be). Let us denote the entry in the first row and first column as [1,1], the entry in the first row and second column as [1,2], the entry in the second row and first column as [2,1], and so on.

Next, we write \( b=39 \) in entry [1,1] and \( a=5 \) in entry [2,1]. Then we try to find the largest integer \( q_{1} \) such that \( q_{1} \cdot a \leq b \). We have \( q_{1}=7 \), which we write in entry [2,2], because \( 7 \cdot 5 = 35 \leq 39 \), and a remainder of \( r_{1}=4 \), which we write in entry [3,1].

Again, we try to find the largest integer \( q_{2} \) such that \( q_{2} \cdot r_{1} \leq a \). We have \( q_{2}=1 \), which we write in entry [3,2], because \( 1 \cdot 4 = 4 \leq 5 \), and a remainder of \( r_{2}=1 \), which we write in entry [4,1]. Notice that we are repeating the same process as before, just with the numbers in the next row.

The next computation returns a remainder of \( r_{3} = 0 \) because \( q_{3} \cdot r_{2} = 4 \cdot 1 = 4 \leq 4 = r_{1} \). We have now computed \( \gcd(5, 39)=r_{2}=1 \) since \( r_{3} = 0 \). Because 5 and 39 are relatively prime, we know that \( \lambda \) and \( \mu \) exist, and we can start using the last column.

First, we write \( x_{1}=0 \) in entry [4,3] and \( x_{2}=1 \) in entry [3,3]. Then we write \( x_{3}=q_{2} \cdot x_{2} + x_{1} = 1 \cdot 1 + 0 = 1 \) in entry [2,3]. For entry [1,3], we compute as before, just with numbers from the row above, i.e., \( x_{4}=q_{1} \cdot x_{3} + x_{2} = 7 \cdot 1 + 1 = 8 \).

Finally, we have that \( a \cdot x_{4} \pm b \cdot x_{3} = r_{2} \), where we need to decide whether it should be plus or minus between the two terms. Because \( a \cdot x_{4} = 5 \cdot 8 = 40 \), \( b \cdot x_{3} = 39 \cdot 1 \), and \( 40 \geq 39 \), we have \( 5 \cdot 8 - 39 \cdot 1 = 1 \) (which is the same as \( 5 \cdot 8 + 39 \cdot (-1) = 1 \)), so the Bézout coefficients are \( \lambda=8 \) and \( \mu=-1 \). Notice that \( a^{-1} = \lambda \; mod \; b = 8 \; mod \; 39 = 8\) and \( b^{-1} = \mu \; mod \; a = -1 \: mod \: 5 = 4\), where \( a \cdot a^{-1} \; mod \; b = 5 \cdot 8 \; mod \; 39 = 1 \) and \( b \cdot b^{-1} \; mod \; a = 39 \cdot 4 \; mod \; 5 = 1 \).

The table for computing \( 5 \cdot \lambda + 39 \cdot \mu = \gcd(5, 39) \) is:

The set of integers \( \{ \dots, -2, -1, 0, 1, 2, \dots \} \) is denoted by the symbol \( \mathbb{Z} \), i.e., \( \mathbb{Z} = \{ \dots, -2, -1, 0, 1, 2, \dots \} \), and is called the ring of integers or an additive group. The group of integers modulo \( n \) is a subset of \( \mathbb{Z} \) and is denoted by \( \mathbb{Z}/n\mathbb{Z} \), but we use the shorthand \( \mathbb{Z}_{n} \). This subset contains the following elements (because we compute modulo \( n \)):

\( \mathbb{Z}_{n} = \{ 0, 1, 2, \dots, n - 1 \}\)

Notice that whenever we perform addition or multiplication in \( \mathbb{Z}_{n} \), we always compute the result modulo \( n \) to obtain an integer in \( \mathbb{Z}_{n} \). To illustrate this, let us look at \( n = 5 \):

\( \mathbb{Z}_{5} = \{ 0, 1, 2, 3, 4 \}\)

When adding \( 3 \) and \( 4 \) in \( \mathbb{Z}_{5} \), we do the following: \( (3 + 4) \: mod \: 5 = 7 \: mod \: 5 = 2 \). Likewise, when multiplying \( 3 \) by \( 4 \) in \( \mathbb{Z}_{5} \), we have: \( (3 \cdot 4) \: mod \: 5 = 12 \: mod \: 5 = 2 \).

An integer \( a \) in \( \mathbb{Z}_{n} \) has an inverse if the greatest common divisor of \( a \) and \( n \) is 1, i.e., \( \gcd(a, n) = 1 \) (see "The extended Euclidean algorithm" for more information). The integer \( a \) is then called a unit. The set of units (all integers with an inverse in \( \mathbb{Z}_{n} \)) is denoted by \( (\mathbb{Z}/n\mathbb{Z})^{*} \). Again, we use the shorthand \( \mathbb{Z}_{n}^{*} \). If \( a_{1} \) and \( a_{2} \) are units, then their product \( (a_{1} \cdot a_{2}) \: mod \: n \) is also always a unit (i.e., \( a_{1} \cdot a_{2} \) is in the group \( \mathbb{Z}_{n}^{*} \)), but the sum \( (a_{1} + a_{2}) \: mod \: n \) may NOT be a unit (i.e., \( a_{1} + a_{2} \) is in \( \mathbb{Z}_{n} \) but may not be in \( \mathbb{Z}_{n}^{*} \)). We see the difference in the two sets \( \mathbb{Z}_{8} \) and \( \mathbb{Z}_{8}^{*} \):

• \( \mathbb{Z}_{8} = \{ 0, 1, 2, 3, 4, 5, 6, 7 \}\)
• \( \mathbb{Z}_{8}^{*} = \{ 1, 3, 5, 7 \} \)

If we choose \( n \) to be a prime number \( p \), then for all integers \( a \) except 0 in \( \mathbb{Z}_{p} \), we have \( \gcd(a, p) = 1 \), which means that \( \mathbb{Z}_{p}^{*} \) contains all integers from \( \mathbb{Z}_{p} \) except 0, i.e.:

\( \mathbb{Z}_{p}^{*} = \{ 1, 2, 3, \dots, p - 1 \}\)

For example, for \( p=7 \), the only difference between the two sets \( \mathbb{Z}_{7} \) and \( \mathbb{Z}_{7}^{*} \) is the integer 0:

• \( \mathbb{Z}_{7} = \{ 0, 1, 2, 3, 4, 5, 6 \}\)
• \( \mathbb{Z}_{7}^{*} = \{ 1, 2, 3, 4, 5, 6 \} \)

The number of elements in \( \mathbb{Z}_{n}^{*} \) is denoted by \( \phi(n) \), named after the famous Swiss mathematician Euler, and is called Euler's phi function. As we saw previously, if \( n \) is a prime number \( p \), then \( \phi(p) = p-1 \). The number of elements in a group \( G \) is also called the order of \( G \) and is written as \( \left| G \right| \). The order of:

• \( \mathbb{Z}_{n} \) is \( \left| \mathbb{Z}_{n} \right| = n \) where \( n \) is a composite number
• \( \mathbb{Z}_{n}^{*} \) is \( \left| \mathbb{Z}_{n}^{*} \right| = \phi(n) \) where \( n \) is a composite number
• \( \mathbb{Z}_{n}^{*} \) is \( \left| \mathbb{Z}_{n}^{*} \right| = \phi(n) = (p-1) \cdot (q-1) \) where \( n = p \cdot q \) and \( p \) and \( q \) are prime numbers
• \( \mathbb{Z}_{p}^{*} \) is \( \left| \mathbb{Z}_{p}^{*} \right| = \phi(p) = p-1 \) where \( p \) is a prime number

If, for example, we choose the elements \( x \), \( a \), and \( b \) from the group \( \mathbb{Z}_{p}^{*} \) and want to compute \( x^{a + b} \: mod \: p \), then in the exponent we actually compute \( a + b \) modulo the order of the group, i.e., \( a + b \: mod \: (p-1) \) because \( \left| \mathbb{Z}_{p}^{*} \right| = \phi(p) = p-1 \). So, what we actually compute is \( x^{a + b \: mod \: (p-1)} \: mod \: p \). The same is true if we had chosen one of the other groups: we always compute modulo the order of the group in the exponent. Therefore, next time you look at an equation from a cryptosystem and wonder why they suddenly compute, for example, modulo \( p-1 \) instead of modulo \( p \), it is because the equation is used in the exponent of some integer.

In the following, we use the group \( \mathbb{Z}_{p} \) as an example for simplicity, but any group could be chosen.

There exists an element \( g \) in the group of integers \( \mathbb{Z}_{p} \) (see "The group of integers and units" for more information), where \( p \) is a prime number, whose powers generate every element in the set of units \( \mathbb{Z}_{p}^{*} \). That is, there exists an integer \( g \) in \( \mathbb{Z}_{p} \) such that:

\( \mathbb{Z}_{p}^{*} = \{ g^{1} \: mod \: p, g^{2} \: mod \: p, g^{3} \: mod \: p, \dots, g^{p-1} \: mod \: p \} \)

where \( g^{p-1} \: mod \: p = 1 \) because \( p-1 \: mod \: (p-1) = 0 \) (remember that the order of \( \mathbb{Z}_{p}^{*} \) is \( p-1 \), and we compute modulo the order of the group in the exponent of \( g \) because \( g \) belongs to the group \( \mathbb{Z}_{p}^{*} \)), and \( g^{0} = 1 \). Such a \( g \) is called a generator or a primitive root of the group, and it's denoted as \( \left< g \right> = \mathbb{Z}_{p}^{*} \). We also say that the order of \( g \) is \( ord(g) = p-1 \) because \( g \) generates the group \( \mathbb{Z}_{p}^{*} \) with \( p-1 \) elements.

To clarify, let's look at an example with the prime number 5. Recall that \( \mathbb{Z}_{5}^{*} = \{ 1, 2, 3, 4 \} \) (see "The group of integers and units" for the example). The group \( \mathbb{Z}_{5}^{*} \) has 2 as a generator because the powers of 2 generate every element in the group, i.e., \( \left< 2 \right> = \mathbb{Z}_{5}^{*} \):

• \( 2^{1} \: mod \: 5 = 2 \)
• \( 2^{2} \: mod \: 5 = 4 \)
• \( 2^{3} \: mod \: 5 = 3 \)
• \( 2^{4} \: mod \: 5 = 1 \)

The last power is 4 because \( p - 1 = 5 - 1 = 4 \). On the other hand, 4 is NOT a generator of \( \mathbb{Z}_{5}^{*} \) because the powers of 4 only generate the integers 1 and 4:

• \( 4^{1} \: mod \: 5 = 4 \)
• \( 4^{2} \: mod \: 5 = 1 \)
• \( 4^{3} \: mod \: 5 = 4 \)
• \( 4^{4} \: mod \: 5 = 1 \)

If you don't know the factorization of the integer \( p-1 \), then the only way to find a generator is to perform the above computation, i.e., check that it generates every element in the group \( \mathbb{Z}_{p}^{*} \). However, if you know the factorization of \( p-1 \), then for every prime number \( q \) that evenly divides \( p-1 \), we check that \( g^{(p-1)/q} \: mod \: p \neq 1 \) for a random integer \( g \) in the group \( \mathbb{Z}_{p} \). If this is the case, then \( g \) is a generator of \( \mathbb{Z}_{p}^{*} \).

Because the prime factorization problem is hard (the problem of computing the factorization of the integer \( p-1 \)), we use a so-called safe prime \( p \). A safe prime \( p \) is of the form \( p = 2 \cdot q + 1 \) where \( q \) is a prime number. Now, the factorization of \( p-1 \) is always only \( 2 \) and \( q \). For example, if we compute the safe prime \( p = 2 \cdot 5 + 1 = 11 \) with the prime number \( q = 5 \), we see that \( g = 7 \) is a generator of the group \( \mathbb{Z}_{11}^{*} \) because:

• \( 7^{(11-1)/2} \: mod \: 11 = 10 \neq 1 \)
• \( 7^{(11-1)/5} \: mod \: 11 = 5 \neq 1 \)

where 2 and \( q=5 \) are the only prime numbers that divide \( p-1=11-1=10 \) evenly. For the same reason, \( g = 5 \) is not a generator of \( \mathbb{Z}_{11}^{*} \) because:

• \( 5^{(11-1)/2} \: mod \: 11 = 1 \)
• \( 5^{(11-1)/5} \: mod \: 11 = 3 \neq 1 \)

The group \( \mathbb{Z}_{n}^{*} \), where \( n \) is a composite number, may not have a generator, but if \( n \) is a prime number \( p \), then the group \( \mathbb{Z}_{p}^{*} \) has at least one generator.

Let \( g \) be a generator of the group \( G \). Given the values \( g \) and \( H \), the discrete logarithm (DL) problem is to compute the exponent \( a \) in the following equation, which is considered a hard problem:

\( g^{a} = H \)

The exponent \( a \) is also called the discrete logarithm of \( H \) to the base \( g \).

For example, if we use the group \( \mathbb{Z}_{p}^{*} \), where \( p \) is a large prime number and \( g \) is a generator of the group, then given \( g \), \( p \), and \( H \), it is hard to compute \( a \) such that the following equation holds:

\( g^{a} \: mod \: p = H \)

In addition to the DL problem, there are two related problems: the Diffie-Hellman (DH) problem and the Decisional Diffie-Hellman (DDH) problem. Given the values \( g \), \( g^{a} \), and \( g^{b} \), the DH problem is to compute the exponent \( a \cdot b \) in \( g^{a \cdot b} \).

Similarly, given the values \( g \), \( g^{a} \), \( g^{b} \), and \( g^{c} \), the DDH problem is to decide whether \( c = a \cdot b \) or whether \( c \) is a random integer.

You may have noticed that if we can solve the DL problem, that is, compute \( a \) in \( g^{a} = H \), then we can also solve the DH problem: first compute \( a \) from \( g^{a} \), then \( b \) from \( g^{b} \), and finally calculate \( a \cdot b \). This also implies that we can solve the DDH problem: first compute \( a \cdot b \) as described above, then compute \( c \) from \( g^{c} \), and finally check whether \( c = a \cdot b \).