 
- [Dr Thomas Britz]
Hi, and welcome to
the University of
New South Wales,
and in particular the School
of Mathematics and Statistics.
My name is Thomas Britz,
and I'll be looking
at problem 75,
for the course, MATH3411.
The aim of this problem
is to factorise a number,
in this case N, which is 83411,
or I suppose you
see 8, 3, 4, 1, 1.
A little silly, maybe,
but there you go.
We want to factorise it using
a particular technique, which
is Fermet factorization.
Now, this is sort of similar
just to trial and error.
Where we just try to see where
the small numbers factorise that
and we just keep trying
bigger and bigger numbers.
Except, it's a bit
more intelligent.
So, what we do in
particular, is we try to find
some numbers s and t,
so that n is equal to
(clears throat)
t^2 minus s^2.
If we can find such numbers,
then we can, of course, write
this as
the product of s plus t,
times s minus,
Sorry.
Times t minus s,
and there we factorised
our number n.
So, that's what we hope to be
able to do using this scheme.
Now, it's a lot more efficient
than just trying the
numbers like 2, 3, 5, etc.
and seeing whether they
divide this number.
And lets see how it works.
So
(clears throat)
first of all, we start off
roughly around about the
square root of this number.
So we calculate the
square root of n.
And we round up, and
that gives you 289.
So that's the first
number that we'll look at
in this algorithm here.
Now, we want to calculate s^2,
which is going to be
equal to t^2 minus N.
These numbers here,
these s^2 numbers,
they're not necessarily
square integer numbers.
So they're, of course,
square real numbers.
But we want to find, we
hope to find, at least
one of these numbers
that we'll generate here
to be a square integer number,
and then we can factorise,
as I said before.
So, first of all, we have to
calculate this very first term.
So, t^2, this number here,
minus n.
And that gives you
110.
So there's a tiny
bit of calculation
right here in the beginning.
We also want to
calculate 2t plus 1,
so that's 579.
And,
(clears throat)
at this step, and
every other step,
we're going to ask: is
this number a square?
So, in other words, s^2 is 110.
If we take the
square root of that,
is that an integer?
And the answer is no.
Let's draw a red
cross like that.
So, as long as that's
the case, we continue.
So, the next value of t
is
290.
And here, to get
the next number,
this is just 2t plus 1,
here we've added 1 here.
So, we're adding 2 here.
So, this is 581.
Now, to get this number here,
we just take that number
there, and add it on to here.
So we get 689.
And, the reason for that, that
we just add this number here,
is if we see
(clears throat)
if we have s^2 be
equal to t^2 minus n.
Then if we see what
happens if we have
t plus 1^2 minus n.
That's equal to
t^2 minus n,
plus 2t plus 1.
So, that's exactly that term
there that we have there.
And this was our
original s^2 term.
So we take this
number and this number
add them together, and you
get the next number here.
So it's all very efficient,
these calculations.
So,
(clears throat)
Lets continue.
We first have to check whether
our new s^2 value is
actually an integer square.
And, while you can sort of...
Since the numbers aren't
much bigger than that,
we can recognise that this
is not a square number.
Of course you could use a
calculator or something,
or other methods to check
whether this was square.
But, so far, we
can recognise this
to not be square
just straight off.
So, lets continue.
So we have 291,
583,
and then we get 1270,
which again, is not a square.
So we continue.
Then we have 292
and 585,
and then we get 1853.
That is still not a square.
I hope we get a
square soon, lets see.
Then we get 293, 587, and 2438.
Ah, that is still not a square.
That's very, very close--
well, I suppose it's not
very close to being a square.
But, it's not far from
being fifty squared. But,
it's not quite, so 294,
589,
and that number
there is then 3025.
Ah, okay, that's
a square number.
We recognise that as 65^2.
Oh, sorry, 55^2.
So, we suddenly,
finally we found
one of these numbers
which is an integer square.
And, then we're done.
Now we just have to write up
what the factorization
is equal to.
So, here t was this value here.
So, t is equal to 294.
And s was equal to
(clears throat)
55.
So, if we call a
the sum of these 2,
and we have that that's 349,
and if we say that b is the
difference between these 2.
Then we get 239.
And, then by what
we said before,
we have that n is
equal to a times b.
Or, in other words, equal to
349 times 239.
And we have broken
n up into 2 factors,
and if you look
at those factors,
well, I suppose you'd
look a little harder
than what I can do right now.
But, both of those are
prime, if I'm not mistaken.
So, we actually have a prime
factorisation right there.
And we are done.
So, thank you very much.
 
