## 数学代写|密码学代写cryptography theory代考|Galois Fields

Galois fields are also known as finite fields. You will learn more about Galois himself in the history section of this chapter. These types of fields are very important in cryptography, and you will see them used in later chapters. These are called finite fields because they have a finite number of elements (Vinogradov 2016). If you think about some of the groups, rings, and fields we have previously discussed, all were infinite. The sets of integers, rational numbers, and real numbers are all infinite. Galois fields are finite.

In a Galois field, there are some integers such that $\mathrm{n}$ repeated terms equal zero. Put another way, there is some boundary on the field, thus making it finite. The smallest $n$ that satisfies this is some prime number, and that prime number is referred to as the characteristic of the field. You will often see a Galois field defined as follows:
$$\mathrm{GF}\left(p^n\right)$$

In this case, the GF does not denote a function. Rather, this statement is saying that there is a Galois field with $\mathrm{p}$ as the prime number (the characteristic we mentioned previously in this section), and the field has $p^n$ elements. A Galois field is some finite set of numbers (from 0 to $p^n-1$ ) and some mathematical operations (usually addition and multiplication) along with the inverses of those operations.
Now you may immediately be thinking, how can a finite field even exist? If the field is indeed finite, would it not be the case that addition and multiplication operations could lead to answers that don’t exist within the field? To understand how this works, think back to the modulus operation we discussed in Chap. 4. Essentially, operations “wrap around” the modulus. If we consider the classic example of a clock, then any operation whose answer would be more than 12 simply wraps around. The same thing occurs in a Galois field. Any answer greater than $\mathrm{p}^{\mathrm{n}}$ simply wraps around.

Let us examine an example, a rather trivial example but one that makes the point. Consider the Galois field GF $\left(3^1\right)$. First, I hope you realize that $3^1$ is the same as 3 and most texts would simply write this as GF(3). Thus, we have a Galois field defined by 3 . In any case, addition or multiplication of the elements would cause us to exceed the number 3 . We simply wrap around. This example is easy to work with because it only has three elements: 1,2 , and 3 . Considering operations with those elements, several addition operations pose no problem at all:
$$1+1=2 \quad 1+2=3$$
However, others would be a problem $2+2=4$, except that we are working within a Galois field, which utilizes the modulus operations so $2+2=1$ (we wrapped around at three).
Similarly, $2+3=2$.
The same is true with multiplication.
$2 \times 2=1$ (we wrap around at 3 )
$2 \times 3=0$ (again we wrap around at 3 )
You can, I hope, see how this works. Now in cryptography we will deal with Galois fields that are larger than the trivial example we just examined, but the principles are exactly the same.

## 数学代写|密码学代写cryptography theory代考|Diophantine Equations

A Diophantine equation is any equation for which you are interested only in the integer solutions to the equation. Thus, a linear Diophantine equation is a linear equation $a x+b y=c$ with integer coefficients for which you are interested only in finding integer solutions. There are two types of Diophantine equations. Linear Diophantine equations have elements that are of degree 1 or zero. Exponential Diophantine equations have at least one term that has an exponent greater than 1 .

The word Diophantine comes from Diophantus, a third-century C.E. mathematician who studied such equations. You have actually seen Diophantine equations before, though you might not have realized it. The traditional Pythagorean theorem can be a Diophantine equation, if you are interested in only integer solutions as can be the case in many practical applications.
The simplest Diophantine equations are linear and are of the form
$$a x+b y=c$$
where $a, b$, and $c$ are all integers. Now there is an integer solution to this problem if and only if $\mathrm{c}$ is a multiple of the greatest common divisor of and $\mathrm{b}$. For example: the Diophantine equation $3 x+6 y=18$ does have solutions (in integers) since $\operatorname{gcd}(3,6)=3$ which does, indeed, evenly divide 18 .

# 密码学代考

## 数学代写|密码学代写cryptography theory代考|Galois Fields

$$\mathrm{GF}\left(p^n\right)$$

$$1+1=2 \quad 1+2=3$$

$2 \times 2=1$ (我们在 3 点结束)
$2 \times 3=0$ (我们再次在 3 点结束)

## 数学代写|密码学代写cryptography theory代考|Diophantine Equations

$$a x+b y=c$$

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