# Lecture 13

We’ve developed techniques to find *one* solution. Euclid’s algorithm gives us that $\begin{aligned}
\gcd(54,39) &{}= \gcd(1\times 39+15,39) {}= \gcd(15,39) {}= \gcd(39,15)\\
&{}= \gcd(2\times 15+9,15) {}= \gcd(9,15) {}= \gcd(15,9)\\
&{}= \gcd(1\times 9+6,9) {}= \gcd(6,9) {}= \gcd(9,6)\\
&{}= \gcd(1\times 6+3,6) {}= \gcd(3,6) {}= \gcd(6,3)\\
&{}= \gcd(2\times 3+0,3) {}= \gcd(0,3) {}= 3.\end{aligned}$ Then, we can work backwards to find a solution to $39x+54y = 3$: $\begin{aligned}
3 &{}= 9-6\\
&{}= 9-(15-9) {}= 2\times9-15\\
&{}= 2\times(39-2\times 15)-15 {}= 2\times39-5\times15\\
&{}= 2\times39-5\times(54-39) {}= 7\times39-5\times54.\end{aligned}$

So $39\times 7 + 54\times (-5) = 3,$ and we multiply both sides by $40$ to get $39\times280 + 54\times(-200) = 120,$ or in other words, that $x=280$, $y=-200$ gives a solution.

Now, you might wonder whether this is the *only* solution.

There’s a way of analysing this. Suppose we have two solutions: $39x + 54y=120\qquad\text{and}\qquad 39x'+54y'=120.$ Subtracting, we get $39(x-x') + 54(y-y')=0.$ Dividing out by the greatest common divisor, we get $13(x-x') + 18(y-y')=0,$ or $13(x-x') = -18(y-y').$ This means that, as $18$ divides the right-hand side, then we also have $18\mid 13(x-x')$. But since $13$ and $18$ are coprime, we have $18\mid(x-x')$ by So we can write $x-x'=18k$. But then we can solve to get $y-y'=-13k$, and it’s easy to check that any such $k$ works.

Hence the general solution is $x=280-18k,\qquad y=13k-200.$

While I haven’t stated (and certainly haven’t proved) any theorems about it, this approach works perfectly well in general, as you can imagine.

## Common divisors and the gcd

Here’s a useful result about common divisors.

#### Proposition

Let $a$ and $b$ be positive integers. Any common divisor of $a$ and $b$ is a divisor of the greatest common divisor.

#### Proof

If $d\mid a$ and $d\mid b$, then $d\mid(a-qb)$ for any $q$. Hence $d$ is a divisor of the numbers obtained after every step of Euclid’s algorithm, and so it is a divisor of the gcd.

We defined the gcd to be the greatest of all common divisors. This property is arguably a more natural one: this says that the gcd is somehow the “best” common divisor.

As an unexpected advantage, if we think of the gcd as being defined in this way, then we can get that $\gcd(0,0)=0$. This was undefined previously.

# Modular arithmetic

## Congruences

Repeatedly over the last few lectures (and the last few problem sheets) we have seen appearances of lots of things like:

odd numbers;

even numbers;

remainders upon division;

numbers of the form $4n+1$ or $18k-440$, and so on.

All these things look pretty similar, and it’s time we got ourselves a language for discussing these things better.

**Definition:** We say that *$a$ is congruent to $b$ modulo $m$* if $m\mid(a-b)$. Often we abbreviate, and say congruent *mod $m$*.

We use the notation $a\equiv b\pmod{m}$ to indicate that $a$ and $b$ are congruent modulo $m$.

For example, $3167\equiv 267\pmod{100};$ indeed, the fact that these two positive integers have the same last two digits means that their difference is a multiple of $100$.

We can now say that an even number is a number congruent to $0$ (modulo $2$), and an odd number is a number congruent to $1$ (modulo $2$).

Rather than saying that “$n$ is of the form $18k-440$”, we can say that “$n$ is congruent to $-440$, modulo $18$”.

Arguments about time frequently involve understandings of congruences. For example, I was born on a Sunday, and the closing ceremony of the 2012 Summer Olympics took place on a Sunday too. So the number of days since the former is congruent to the number of days since the latter, modulo $7$.

Notice that saying that $a$ is congruent to $0$, modulo $m$, is exactly the same as saying that $a$ is a multiple of $m$ (since it’s saying that $m\mid(a-0)$).

As we’ve defined it, a congruence modulo $m$ doesn’t say that two things are equal, just that their difference is a multiple of $m$.

But it does behave suspiciously like an equality, as we’re about to see.