## Tuesday, April 14, 2015

### Summarizing non-polynomial factorization conclusions

Remarkably pushing a simple algebraic factorization beyond polynomials leads to far reaching conclusions, including the find of numbers which escaped prior non-field ring classification schemes. Here I'll summarize by pulling key points from my post which steps through the full argument.

Deliberately I use a very simple start, as simple as I could get with:

P(x) = (g1(x) + 1)(g2(x) + 2)

where P(x) is a quadratic with integer coefficients, g1(0) = g2(0) = 0, but g1(x) does not equal 0 for all x.

Introduce k, where k is a nonzero integer, and not 1 or -1, and new functions f1(x), and f2(x), where:

g2(x) = f2(x) + k-2 and g1(x) = f1(x)/k,

which gives me the now symmetrical form:

k*P(x) =  (f1(x) + k)(f2(x) + k)

And introduce H(x), where I like using the capital letter here for visual reasons, but it is not to signify H(x) must be a polynomial, where:

f1(x) + f2(x) = H(x)

So I can find:

f12(x) - H(x)f1(x) - kH(x) - k2 + k*P(x) = 0

And you can solve for f1(x) using the quadratic formula:

f1(x) = (H(x) +/- sqrt[(H(x) + 2k)2 - 4k*P(x)])/2

This result tells us that the f's are in the ring of algebraic integers if H(x) is.

It also shows that the f's are not linear equations unless the g's are, when k does not equal 1 or -1. So they have this square root which will only resolve in general for all integer x, if the g's are linear with integer coefficients and our original factorization is a polynomial one.

But of course polynomial factorizations are well understood and boring, so it is more interesting to move to non-polynomial ones, where now we can see that the f's will still be algebraic integers, as long as H(x) is, but they will tend to be non-rational for integer x, because the square root will force it.

And H(x) has constraints as it must fit with the given equations, for instance H(0) = -k + 2, is a requirement from them.

Let's look at a non-polynomial factorization example.

P(x) = 175x2 - 15x + 2,

and let k = 7.

And further I have:

7*P(x) = 7(175x2 - 15x + 2) = (5a1(x) + 7)(5a2(x) + 7)

So f1(x) = 5a1(x), and f2(x) = 5a2(x), where the a's are roots of:

a2 - (7x-1)a + (49x2 - 14x) = 0

Which means H(x) = 5(7x - 1), as you can see the a's sum to (7x-1), and just multiply that by 5, to get the sum of the f's, as H(x) =  f1(x) + f2(x).

However, I didn't just make up an H(x), but worked it out from a non-polynomial factorization I had found already from a much more difficult argument, by which I found the a's. Turns out I had that first, which seems to be a functional approach to solving these type of equations, which requires use of a special construction.

That's because things have to be highly particular if you will have g's which are still like integers, even though non-rational. And I've explained that in detail in an earlier post.

So now what do we know? Well with an example I have values for the f's, which are indeed algebraic integers for algebraic integer x. I know the underlying factorization is:

P(x) = (g1(x) + 1)(g2(x) + 2)

And I can pick a value, say x = 1, for which the a's are 3+sqrt(-26) and 3-sqrt(-26), which tells me the f's are:

5(3+sqrt(-26)) and 5(3-sqrt(-26))

Notice at this value, the polynomial equals P(1) = 175 - 15 + 2 = 162, so we've just factored 162, but we can see a factorization by multiplying by 7:

7*162 = (5(3+sqrt(-26)) + 7)(5(3-sqrt(-26)) + 7)

And we know these connect back to the g's, so here is an existence which means the g's must exist, but my choice of 7 was arbitrary. The underlying factorization doesn't know or care what I multiply times it. And it was just a tool to get to something visible at x=1 by forcing symmetry, and in fact we can use k, where k is a nonzero integer not 1 or -1, with the same set of g's, where for each one, the end result will be algebraic integer f's. So since an infinite number of integers would work we have proof that the g's must be integer like themselves.

So let's label a set of these numbers whose existence and integer like properties we now have proven, o1 and o2, where these are solutions for the g's, to put them in a more mathematical statement.

If o1 is the non algebraic integer, then 7o1 and in general, ko1 is an algebraic integer for k, nonzero and not 1 or -1.

So there is NO WAY the o's can be anything like a fraction, as of course if say, there were a 7 in a denominator then multiplying by some other number wouldn't remove it. And no number could have every possible k in its denominator.

So with absolute mathematical certainty we have new numbers, which are like integers, despite being non-rational, like gaussian integers, but which, unlike gaussian integers, are NOT algebraic integers!

We have new ones to add to the realm of integer-like numbers.

But what about fields? Does this impact them?

And it is easy to show why fields are not impacted, as consider o2/o1, which might not seem to be in the field of algebraic numbers, but you just have to multiply top and bottom by some nonzero integer not 1 or -1, like:

2o2/2o1

and you have a ratio of algebraic integers, showing that it is indeed an algebraic number.

With no impact on fields, these numbers take you stepping outside of the boundaries of polynomial factorizations to find them, which we did by pushing an asymmetrical quadratic factorization into non-polynomial factorizations.

So this approach allows studying the underlying asymmetrical form by moving to a symmetrical form using basic algebra.

And after pondering them for some time I came up with a more robust classification scheme with what I decided to call the object ring.