# Lecture 12

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.