# Summary

Today we started by reviewing how one can go about "canceling" common factors in congruence equations. Afterwards we introduced the notion of a linear congruence equation, giving a theorem which told us exactly when such an equation has solutions (and, indeed, how many solutions exist).

# Cancellation in Modular Equations

Last class period we stated the following

Proposition: $ca \equiv cb \mod{m}$ is equivalent to $a \equiv b \mod{\frac{m}{(c,m)}}$.

This result will play a central role in today's lecture, so we'll start by giving a full proof.

Proof: We'll use the notation $d = (a,m)$ to make our exposition easier to read.

Suppose first that $a \equib \mod{\frac{m}{d}}$. This means that $\frac{m}{d} \mid a - b$, so that there exists an integer *e* with $e\frac{m}{d} = a-b$. Multiplying this equation by *c* then gives $ce\frac{m}{d} = a-b$. Now notice that since $d \mid c$ we can rewrite the left hand side as $\frac{c}{d}em$, where each of these terms are *bona fide* integers. Hence we have

Since the left hand side is a multiple of *m*, we conclude that $ca \equiv cb \mod{m}$.

Now we'll prove the $(\Rightarrow)$ direction. Since we're told that $ca \equiv cb \mod{m}$, this translates to the divisibility statement $m \mid ca-cb$. Hence there is some integer *e* so that $me = ca-cb$. We can divide each of *m* and *c* by *d* (since $d = (c,m)$) and get an equation of integers

Now since this is an equation of integers, we can cancel out the *d* on both sides, and we're left with the div

from which we have $\frac{m}{d} \mid \frac{c}{d}(a-b)$. We know that $(\frac{m}{d},\frac{c}{d}) = 1$ by an old result, and we also know that this relative primality result together with our divisibility condition implies that

(4)from which we find $a \equiv b \mod{\frac{m}{d}}$ as desired.$\square$

# Linear Congruences

Now that we've played around a bit with modular arithmetic, it's time that we take one of our favorite problems in mathematics and give it a modular spin: solving equations. We'll start off at the beginning, dealing with linear equations.

Definition: For integers

a,bandm, the equation $ax \equiv b \mod{m}$ is called a linear congruence.

The goal, of course, is to find all integers *x* which solve this equation. Given that the equation is really a statement about modular congruence, though, you won't be surprised to hear that we're actually most interested in knowing solutions to the system *modulo m*; that is to say, we want to know which congruence classes modulo *m* solve the given equation.

#### Example: Some linear congruence equations

From our multiplication table, we can read off solutions to some equations modulo 6.

(5)$\square$

So we see that our linear congruences can have either no solutions, 1 solution, or several solutions (where by "solutions" we mean more precisely "distinct solutions modulo m"). The question, then, is how to distinguish when an equation does have a solution from when it doesn't. And if it does have a solution, how can we produce all solutions? How many solutions will there be?

Big Theorem on Linear Congruences: The congruence $ax \equiv b \mod{m}$ has integer solutions if and only if $(a,m) \mid b$. If $x_0$ is such a solution, then all other integral solutions take the form $x_0 + n\left(\frac{m}{(a,m)}\right)$, where $n \in \mathbb{Z}$. A complete list of the distinct solutions modulo

mis given by $x_0 + n\left(\frac{m}{(a,m)}\right)$ when $n \in \{0,\cdots,d-1\}$.

Proof: For notational convenience, we'll write *d* for the gcd of *a* and *m*. Now we'll proceed with the proof in steps: ^{(1)} show that solutions exist if and only if $d \mid b$; ^{(2)} show that other solutions can be expressed in terms of one fixed solution; ^{(3)} determine when two integer solutions are congruent modulo *m*. (Note: we didn't get to prove this last part in class today.)

Step 1: First, suppose that a solution $x_0$ exists to the equation $ax \equiv b \mod{m}$. This implies that $m \mid ax_0 - b$, so that there exists an integer *e* with $me = ax_0-b$. Rearranging, we therefore have

Now since *d* is the gcd of *a* and *m* we have $d \mid a$ and $d \mid m$, and therefore *d* divides any integral linear combination of *a* and *m*. In particular,

Hence if our congruence equation has a solution, then $d \mid b$.

Now we'll prove the converse, showing that a solution exists when $d \mid b$. We start by noting that there exists integers *r* and *s* such that

this follows because the gcd of two integers can be expressed as an integral combination of the two integers. Now using the fact that $d \mid b$, we find an integer *e* so that $de = b$. Multiplying the displayed equation by *e* then gives

Taking this equation modulo *m*, we therefore have $b = a(re) \mod{m}$, and hence $x = re$ is an integer solution to the equation $ax \equiv b \mod{m}$.

Step 2: Now suppose we are given two solutions to the equation, $x_0$ and $x_1$, and we wish to show that $x_1 = x_0 + n\left(\frac{m}{d}\right)$. In order to do this, note that we have

(10)This tells us that $ax_0 \equiv ax_1 \mod{m}$, and so it follows that $m \mid a(x_0-x_1)$ — or, if we turn this divisibility statement into an equation, there exists some integer *k* so that $mk = a(x_0-x_1)$. We'll divide this equation on both sides by *d* — a legal move since *d* is a common divisor of *a* and *m* — and we find that $(\frac{m}{d})k = \frac{a}{d}(x_0-x_1)$. This is equivalent to the divisibility condition $\frac{m}{d} \mid \frac{a}{d}(x_0-x_1)$. Notice, however, that $(\frac{m}{d},\frac{a}{d}) = 1$, and hence homework problem 44a tells us that $\frac{m}{d} \mid x_0 -x_1$. This is the same as saying that $x_0 \equiv x_1 \mod{m}{d}$, so that $x_1 = x_0 + n\frac{m}{d}$ as desired.

Step 3: (Note: we didn't get to discuss this proof in class, but I'm including it in the notes for people interested in seeing the full proof.) To find the distinct solutions (modulo *m*), suppose we pick up two solutions $x_1$ and $x_2$ which are the same modulo *m*. Since $x_i = x_0 + n_i \frac{m}{d}$ by the previous step, this means that we have

Getting rid of the $x_0$ that is common to both sides, we turn this divisibility condition into an equation: $me = \frac{m}{d}(n_1-n_2)$. Hence we have

(12)and after canceling the *m*'s on both sides of the equation (a legal move since this is an equation in integers, not a congruence equation) we're left with $de = n_1-n_2$ — i.e., that $n_1 \equiv n_2 \mod{d}$.

This tells us that two solutions $x_0 + n_1\frac{m}{d}$ and $x_0 + n_2\frac{m}{d}$ are distinct if and only if $n_1 \not\equiv n_2 \mod{d}$. Hence the distinct solutions to $ax \equiv b \mod{m}$ are given as $x_0 + n\left(\frac{m}{(a,m)}\right)$ when $n \in \{0,\cdots,d-1\}$. $\square$

#### Example: Solving $6x \equiv 5 \mod{15}$

Suppose you want to solve the equation $6x \equiv 5 \mod{15}$. Notice that the gcd of 6 and 15 is 3, and that $3 \not\mid 5$. Our big theorem tells us that this congruence equation has no solutions. $\square$

#### Example: Solving $4x \equiv 6 \mod{14}$

Let's put these ideas in practice to try to solve $4x \equiv 6 \mod{14}$. To start, we need to decide whether this congruence will have solutions or not. For this, we just notice that $2=(4,14)$, and that $2 \mid 6$. Hence we know there are solutions, and we're expecting that there should be 2 distinct solutions modulo 14.

To find one such solution, we need to do two things:

- we need to express 2 as a linear combination of 4 and 4, and
- we need to express 6 as a multiple of 2.

Toward the first goal, we know that we can to use the Euclidean Algorithm. The algorithm runs like so:

(13)and from this we see that

(14)Now for the second goal, it isn't too hard to see that $2 \cdot 3 = 6$. Finding a solution, then means we should multiply our expression of 2 as a linear combination by 3:

(15)Taking this equation modulo 14 leaves us with

(16)and hence $x = -9$ is one integer solution.

Now that we have one solution $x_0$, we can find all solutions by taking $x_0 + n\left(\frac{14}{2}\right)$ for $n \in \{0,1\}$. Doing so shows that the distinct solutions modulo 14 are given by $x \equiv -9,-2 \mod{14}$.

Notice that if we had been interested in least non-negative solutions, we would write 5 in place of -9 (since $5 \equiv -9 \mod{14}$) and 12 in place of -2 (since $12 \equiv -2 \mod{14}$). $\square$