===== Modular arithmetics =====
$\renewcommand{\mod}{\operatorname{mod}}$

Sometimes when you want to calculate something
you might only require the remainder of the answer when divided by some natural number $n$. With help of modular arithmetics finding the required remainder usually happens to be much easier than evaluating the original expression.

Before we start off with examples, let us introduce a handy notation.

<WRAP def>
For integers $a, b$ and a natural number $n$ let
$a \equiv b \ (\mod n)$
denote that $a$ and $b$ have the same remainder when divided by $n$, i.e., that $a \mod n = b \mod n$ holds. 
</WRAP>

In essence, modular arithmetics means counting in a circle. There are many real-world examples:
clock: 19:00 + 7 hours = 02:00,

weekdays: friday (5th day) + 4 days = tuesday (2nd day), 

counting-out games, etc.

{{url>../illustrations/modular.html 300px, 250px noscroll noborder}}

Let us consider a bit more technical example: Let's say someone asks you to check whether $5638 \times 7891 + 2731 \times 3124$ is divisible by $17$.

One way to do this is to go ahead and calculate the expression: 

$5638 \times 7891 + 2731 \times 3124 = 44489458 + 8531644 = 53021102$, 

then check if $17$ divides the answer... yawn. Ok, done.

On the other hand, a modular arithmetic approach would be to note that $\mod 17$ we have: 

$10 = 17 - 7 \equiv -7$,

$100 = 10^2 \equiv (-7)^2 = 49 = 51 - 2 \equiv -2$,

$5638 = (51+5) \cdot 100 + (34 + 4) \equiv 5 \cdot (-2) + 4 = -6$,

$7891 = (85-7)\cdot 100 + (85+6) \equiv (-7)\cdot(-2) + 6 = 20 \equiv 3$,

$2731 \equiv (-7)\cdot(-2) - 3 = 11 \equiv -6$,

$3124 \equiv (-3)\cdot(-2) + 7 = 13 \equiv -4$,

and

$5638 \times 7891 + 2731 \times 3124
\equiv (-6) \cdot 3 + (-6) \cdot (-4) = (-6) \cdot (-1) = 6$.

So the answer is: no, it is not, as the remainder is $6$, not $0$.

Note that in the above example in the arithmetic expressions following the equivalenvce $\equiv$, we allowed ourselves to replace any number with a number having the same remainder modulo $17$. 
This is because the **“equivalence modulo
$n$”** behaves very nicely with respect to arithmetic operations. (Why not try to prove the following bit of theory yourself?)
<WRAP thtask>
If $a \equiv b \ (\mod n)$ and $c \equiv d \ (\mod n)$ then 

(1) $a+c \equiv b+d \ (\mod n)$

and 
(2) $ac \equiv bd \ (\mod n)$.
++++Proof|
Let us denote the quotient and the remainder when dividing $a$ by $n$ by $q_a$ and $r_a$, i.e., 
$a = q_a n + r_a$, and similarly for $b$, $c$, and $d$. By assumption, we have $r_a = r_b$ and $r_c = r_d$. Then

$(a+c) \mod n = \big((q_a+q_c)n + r_a + r_c \big) \mod n = (r_a + r_c) \mod n$

and

$(b+d) \mod n = \big((q_b+q_d)n + r_b + r_d \big) \mod n = (r_b + r_d) \mod n = (r_a + r_c) \mod n$.

Similarly,

$ac \mod n = (q_a n q_c n + q_a n r_c + q_c n r_a + r_a r_b) \mod n  = r_a r_b \mod n = r_b r_d \mod n = bd \mod n$.
++++
</WRAP>

When we say that an arithmetic expression has to be computed modulo $n$, we mean
that we are interested in the remainder of the value of this expression when divided by
$n$. For example, $(2 + 5) \cdot (8 + 6)$ is equal to $10$ modulo $11$, or
$(2 + 5) \cdot (8 + 6) \equiv 10 \ (\mod 11)$.

The proposition above allows us to perform computations modulo $n$, such that the intermediate
values do not become much larger than $n$. 

Actually, when the operations that we’re
applying are only addition, subtraction and multiplication, then no intermediate value
has to be larger than $n^2$ . Namely, we can compute the remainder (by $n$) after each step
of the computation. Thus, we could perform the previous computation by
  *$2+5=7$
  *$8 + 6 = 14 \equiv 3 \ (\mod 11)$
  *$7 \cdot 3 = 21 \equiv 10 \ (\mod 11)$.

In fact, we can do even more simple operations with this equivalence.

Let us look more closely at the proposition.
Note that part (1) is reversible in the sense that
if $a+c \equiv b+d \ (\mod n)$ and 
$c \equiv d \ (\mod n)$ then $a \equiv b \ (\mod n)$. ++Because| then $-c \equiv -d \ (\mod n)$
and $a+c-c \equiv b+d-d \ (\mod n)$.++

You may use this observation to check the exercise:
<WRAP thtask>
Show that $a \equiv b \ (\mod n)$ if and only if 
$n \mid (a-b)$.

++++Solution|
Since $-b \equiv -b (\mod n)$, we have

$a \equiv b \ (\mod n) \Longleftrightarrow 
a -b \equiv b - b = 0 (\mod n) \Longleftrightarrow n \mid (a-b)$.
++++
</WRAP>

Part (2), on the other hand, is not reversible in general: $4 \cdot 2 \equiv 2 \cdot 2 (\mod 4)$
 but $4 \not \equiv 2 (\mod 4)$.

The following exercise explains what we should do when trying to do the reverse of part (2).

<WRAP thtask>
Let $k \in \mathbb N$. Show that
  * $a \equiv b \ (\mod n)$ if and only if $ak \equiv bk \ (\mod nk)$.
  * If $ka \equiv kb \ (\mod n)$, then $a \equiv b \ (\mod \frac{n}{\gcd(k,n)} )$.
++++Solution|
The assertions follow from the previous exercise and the facts that $nk \mid ak$ is equivalent to $a \mid n$ and $n \mid ka$ implies $\frac{n}{\gcd(k,n)} \mid a$. The former is obvious and for the latter consider the equation $ka = qn$. Dividing by $\gcd(k,n)$, we get
$\frac{k}{\gcd(k,n)} a = q \frac{n}{\gcd(k,n)}$. Since $\frac{k}{\gcd(k,n)}$ and $\frac{n}{\gcd(k,n)}$ have no common divisors, all the divisors of $\frac{n}{\gcd(k,n)}$ must also divide $a$, i.e., $\frac{n}{\gcd(k,n)} \mid a$.
++++
</WRAP>

In particular, if $k$ and $n$  are coprime, then
we essentially get the inverse of part (2):
 $ka \equiv kb \ (\mod n)$ implies $a \equiv b \ (\mod n)$. 

In a short while, we will see a deeper algebraic meaning of this
cancelability.

TODO: Examples on the last exercise.

<WRAP task>
Perform the following computations:
  *$(5 \cdot 8 − 3 \cdot 6) \cdot (8 + 4) \ (\mod 9)$,
  *$7 \cdot 7 \cdots 7 \ (\mod 11)$ (here $7$ is taken 15 times),
  *$1 \cdot 2 \cdots 10 \ (\mod 25)$.
</WRAP>
[[Residue classes]]