Lecture 13

We’ve developed techniques to find one solution. Euclid’s algorithm gives us that gcd(54,39)=gcd(1×39+15,39)=gcd(15,39)=gcd(39,15)=gcd(2×15+9,15)=gcd(9,15)=gcd(15,9)=gcd(1×9+6,9)=gcd(6,9)=gcd(9,6)=gcd(1×6+3,6)=gcd(3,6)=gcd(6,3)=gcd(2×3+0,3)=gcd(0,3)=3.\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=339x+54y = 3: 3=96=9(159)=2×915=2×(392×15)15=2×395×15=2×395×(5439)=7×395×54.\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 39×7+54×(5)=3,39\times 7 + 54\times (-5) = 3, and we multiply both sides by 4040 to get 39×280+54×(200)=120,39\times280 + 54\times(-200) = 120, or in other words, that x=280x=280, y=200y=-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=120and39x+54y=120.39x + 54y=120\qquad\text{and}\qquad 39x'+54y'=120. Subtracting, we get 39(xx)+54(yy)=0.39(x-x') + 54(y-y')=0. Dividing out by the greatest common divisor, we get 13(xx)+18(yy)=0,13(x-x') + 18(y-y')=0, or 13(xx)=18(yy).13(x-x') = -18(y-y'). This means that, as 1818 divides the right-hand side, then we also have 1813(xx)18\mid 13(x-x'). But since 1313 and 1818 are coprime, we have 18(xx)18\mid(x-x') by So we can write xx=18kx-x'=18k. But then we can solve to get yy=13ky-y'=-13k, and it’s easy to check that any such kk works.

Hence the general solution is x=28018k,y=13k200.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 aa and bb be positive integers. Any common divisor of aa and bb is a divisor of the greatest common divisor.

Proof

If dad\mid a and dbd\mid b, then d(aqb)d\mid(a-qb) for any qq. Hence dd 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\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:

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


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

We use the notation ab(modm)a\equiv b\pmod{m} to indicate that aa and bb are congruent modulo mm.

For example, 3167267(mod100);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 100100.

We can now say that an even number is a number congruent to 00 (modulo 22), and an odd number is a number congruent to 11 (modulo 22).

Rather than saying that “nn is of the form 18k44018k-440”, we can say that “nn is congruent to 440-440, modulo 1818”.

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 77.

Notice that saying that aa is congruent to 00, modulo mm, is exactly the same as saying that aa is a multiple of mm (since it’s saying that m(a0)m\mid(a-0)).

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

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