CS 173
Spring 2021

## Number Theory 3

A fun thing to do with congruence mod k is to create a new "modular" number system. You can do this with any value of k, but we'll assume that k=7 for this vidoe.

## Basic notation

We'll use $$[n]_k$$ to represent the set of all numbers congruent to n mod k. Since our base will always be 7, we'll simplify this to [n]. For example, [2] contains all the numbers congruent to 2.

[2] = {2, 9, -5, 16, -12, ...}
[4] = {4, 11, -3, 18, -10, ...}
[9] = {2, 9, -5, 16, -12, ...}

Notice that [2] = [9] = [-12], because these sets contain the same elements. So we have three names for the same set. So these sets are like teams or dorm rooms. E.g. "Brandon's room" and "Marcus's room" might be two names for the same room, because we can name the room after anyone in it. Brandon is one "representative" that we can use to name the room.

To make a modular number system, we're going to treat each of these sets ("congruence classes") as a single object. So our number system (called $$\mathbb{Z}_7$$) contains seven numbers

$$\mathbb{Z}_7 = \{[0],[1],[2],[3],[4],[5],[6]\}$$

If you need to do a lot of computation with modular numbers, e.g. in a later class, most people drop the square brackets. We'll keep the square brackets in this class, to emphasize that these are modular numbers rather than normal integers.

## Arithmetic

To add or multiply modular numbers, you add or multiply the numbers inside the square brackets. There are theorems (e.g. the one proved in the previous lecture) showing that this works as you'd expect.

[4] + [5] = [9]
[4] x [3] = [12]
$$[6]^2 = [36]$$

However, notice that [9] wasn't in our list of elements in $$\mathbb{Z}_7$$. This is because [9] = [2]. It's usually convenient to normalize arithmetic so that we always use a small value to name each modular number, e.g. stick to values between 0 and 6 when k=7. So we'd actually write:

[4] + [5] = [9] = [2]
[4] x [3] = [12] = [5]
$$[6]^2 = [36] = [1]$$

So modular numbers wrap around, like hours on a clock. If you haven't used unsigned numbers in C or C++, you will soon. Arithmetic on these numbers also wraps around. For example, unsigned char values (often used in representing pictures) go from zero up to 255. Adding three to 255 gives you the result 2.

In just a sec, we'll see that normalizing intermediate numbers allows us to do computations with large numbers efficiently. Normalizing can also allow a computer implementation of modular numbers to quickly decide whether two numbers are equal, because the normalized representations will be exactly the same.

Having said that, when you're working by hand, it's sometimes convenient to use small negative values. For example

$$[6]^2 = [-1]^2 = [1]$$

## Big numbers

In cryptographic applications, we often need to raise numbers to massively large powers, mod some base k. To do this, we need to be smart about keeping intermediate results small during the calculation, so we won't blow out the precision of our calculator/programming language. The big idea is to divide the calculation into small steps and remove copies of k at each intermediate step.

One specific way to organize the computation of $$[p^q]_k$$ is to first compute a table of p raised to each power of two. That is, we're computing $$[p^{(2^m)}]_k$$ for each positive integer m.

For example, with k=7, let's first compute $$[5]^2$$.

$$[5]^2 = [25] = [4]$$

Now notice that $$[5]^4 = ([5]^2)^2$$. So we can compute the second line of our table using the result of the first line:

$$[5]^2 = [25] = [4]$$
$$[5]^4 = ([5]^2)^2 = [4]^2 = [16] = [2]$$

We can keep doing this "repeated squaring" to work out the values for bigger powers:

$$[5]^2 = [25] = [4]$$
$$[5]^4 = [4]^2 = [16] = [2]$$
$$[5]^8 = [2]^2 = [4]$$
$$[5]^{16} = [4]^2 = [16] = [2]$$
$$[5]^{32} = [2]^2 = [4]$$

Notice the repeating pattern of [2] and [4]. Remember that these modular number systems contain only a finite number of values, e.g. there are only 7 distinct numbers in $$\mathbb{Z}_7$$. So the table formed by repeated squaring must end up repeating.

To compute $$[5]^q$$ where q is not a power of 2, break up q into powers of 2. For exmaple, $$55 = 32 + 16 + 4 + 2 + 1$$. Now, remember that addition in the exponent can be converted multiplication like this:

$$a^{b+c} = a^b \times a^c$$

So

$$[5]^{55} = [5]^{32} \times [5]^{16} \times [5]^4 \times [5]^2 \times [5]$$

Looking up each power in our table gives us

$$[5]^{55} = [5]^{32} \times [5]^{16} \times [5]^4 \times [5]^2 \times [5] = [4] \times [2] \times [2] \times [4] \times [5]$$

We can now work out the product. Since we're doing this by hand, let's group the factors intelligently. The big trick is to keep normalizing our intermediate numbers, so their magnitude stays small.

$$[5]^{55} = ([4] \times [2]) \times ([2] \times [4]) \times [5] = [8] \times [8] \times [5] = [1] \times [1] \times [5] = [5]$$