Factoring Integers_Part 3

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Factoring Integers - Part 3

So in Factoring Integers_Part 2 we saw that we want to

The second part of that we'll deal with next time. Firstly, let's generate the QRs.

There's more than one way to do this. The first one we'll look at is now considered out-of-date, but it's instructive.

Of course, if n is itself a square then we have factored it. We can then take the square root, and our challenge is to factor that.
Firstly, consider the square root of n. This is not an integer so we write it as:

where $e_0$ is between 0 and 1. Since $e_0$ is between 0 and 1 we can take its reciprocal, and write that as $q_1+e_1$ ...

... where $q_1$ is the largest possible integer to leave $e_1$ between 0 and 1.

Then we repeat this process, and we can compute

Continued Fractions are cool, and there's lots you can do with them, including a simple proof that $\sqrt{2}$ is irrational.

But I digress ...

We can write this as $[q_0;q_1,q_2,q_3,...]$ and this is called a "Continued Fraction," or "CF" for short.

We can then look at the truncations of this object:



If you look at the sequence of truncated fractions they get closer and closer, first above, then below, bouncing back and forth around, and getting ever closer to, the value of the CF. The sequence converges to the value, and that's why the individual terms are called the "convergents."
These are called the convergents of the CF (Continued Fraction), and they give a sequence of rationals. We can prove that these rationals are in some sense the best possible rational approximations to $\sqrt{n}.$

So we have $\frac{a_i}{b_i}\approx\sqrt{n}$ and that means that $\frac{a_i^2}{b_i^2}\approx{n},$ and so $a_i^2\approx{b_i^2n}.$

Since this is supposed to be a good approximation (although admittedly some of the convergents are better than other) that means that $a_i^2$ is "small" modulo n.

We need to allow the $a^{}_i$ to be negative. If it's close to, and just a little less than, n, we can think of it as a negative number that's close to zero. Again, it's "small."
In other words, the numerator of each convergent is fairly small, compared with n. That probably means it's fairly easy to factor.

And that's what we wanted! We were trying to find a way of creating QRs that are easy to factor (by some definition of "easy"), and we've done that. The numerators of the convergents of the continued fraction of $\sqrt{n}$ are "small" and therefore likely to be easy to factor.

And now we return to an earlier trick. We don't need to consider just the CF of $\sqrt{n},$ we can also look at the CF of $\sqrt{2n},$ and $\sqrt{3n},$ and so on.

So we have a veritable factory producing QRs that are easy to factor.

In Factoring Integers_Part 4 we'll run through an example to see how this works in practice, before moving on to using them.

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