# 数学代写|密码学代写cryptography theory代考|CS6260

## 数学代写|密码学代写cryptography theory代考|Finding Suitable Primes

The prime number theorem says that in any given range of integers, primes are fairly common (inversely proportional to the number of digits). Since we can efficiently recognise primes, we can fairly quickly find large primes simply by choosing random large numbers until we find a prime.

By Requirement 2.5, we need a cyclic group such that the group order is divisible by a large prime. This means that we are not just looking for primes, we are looking for a prime $p$ such that $p-1$ is divisible by a large prime.
We can do this by first choosing a sufficiently large prime $\ell$ and then choosing random numbers $k$ until $2 k \ell+1$ is prime. In practice, this algorithm performs as well as a search for an arbitrary prime.

Example 2.19. Testing integers sequentially starting at 2000, we find that $\ell=$ 2003 is prime. We then test multiples starting with 101 and find that when $k=103$ the number $p=2 k \ell+1=412619$ is prime.

Sometimes, we want the $p-1$ to be twice a prime $\ell$. In this case, $p$ is called a safe prime and $\ell$ is called a Sophie-Germain prime. This time, the need for $\ell$ and $2 \ell+1$ to be prime simultaneously means that we need to look at many more candidates before we find a suitable prime. This will be slow, but there are techniques to speed up the search.

Example 2.20. Again, testing integers sequentially starting at 2003, we find that $\ell=2039$ is prime at the same time as $p=2 \ell+1=4079$.

Exercise $2.53$. About one third of all candidates will be divisible by 3 . Checking that a number is not divisible by 3 is much faster than using the primality testing algorithms from Section 2.3. Expand on this idea and explain how we can use so-called trial division by small primes to exclude most candidate primes before finally using the algorithms from Section $2.3$.

## 数学代写|密码学代写cryptography theory代考|Index Calculus

All of the algorithms from Section $2.2$ will work for $\mathbb{F}_p^*$. But it turns out that we can do very much better. We shall develop the ideas of index calculus in a general setting, and then show how the properties of prime fields give us a more efficient algorithm for computing discrete logarithms.

We begin with an observation about abstract cyclic groups, which is an extension of (2.1). Let $G$ be a cyclic group of order $n$. Let $g$ be some generator and let $x$ be a group element. Suppose we have $\nu$ pairs of integers $\left(r_1, t_1\right),\left(r_2, t_2\right), \ldots,\left(r_\nu, t_\nu\right)$ and integers $\alpha_1, \alpha_2, \ldots, \alpha_\nu$ (not all congruent to zero modulo $n$ ) such that
$$\prod_{i=1}^\nu\left(x^{t_i} g^{r_i}\right)^{\alpha_i}=1$$
This will give us the equation
$$x \sum_i \alpha_i t_i g^{\sum_i \alpha_i r_i}=1,$$
which is of the same form as (2.1). As long as $\sum_i \alpha_i t_i$ is invertible modulo $n$, we can recover $\log _g x$ from the equation using $2 \nu+2$ arithmetic operations $(\nu+1$ multiplications, $\nu$ additions and one inversion).
We first consider the case when the group order $n$ is prime.
Example 2.21. Consider the prime $p=1019$ with the elements $g=3$, generating a subgroup $G$ of $\mathbb{F}_p^*$ of order 509 . Let $x=11$.
With the relations
\begin{aligned} &y_1=g^{112} x^{239}=576 \quad y_2=g^{477} x^{274}=70 \quad y_3=g^{378} x^{248}=180 \ &y_4=g^{80} x^{66}=42 \quad y_5=g^{331} x^{488}=720 \ & \end{aligned}
and intègers $\alpha_1=145, \alpha_2=436, \alpha_3=72, \alpha_4=73$ and $\alpha_5=1$, wee get that
$$y_1^{145} x_2^{436} y_3^{72} y_4^{73} y_5=1$$

# 密码学代考

## 数学代写|密码学代写cryptography theory代考|Index Calculus

$$\prod_{i=1}^\nu\left(x^{t_i} g^{r_i}\right)^{\alpha_i}=1$$

$$x \sum_i \alpha_i t_i g^{\sum_i \alpha_i r_i}=1,$$

$$y_1=g^{112} x^{239}=576 \quad y_2=g^{477} x^{274}=70 \quad y_3=g^{378} x^{248}=180 \quad y_4=g^{80} x^{66}=42 \quad y_5=g^{331} x^{488}$$

$$y_1^{145} x_2^{436} y_3^{72} y_4^{73} y_5=1$$

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