**Proof:**

In the complex plane given:

P(x) = (g

_{1}(x) + 1)(g

_{2}(x) + 2)

where P(x) is a primitive quadratic with integer coefficients, g

_{1}(0) = g

_{2}(0) = 0, but g

_{1}(x) does not equal 0 for all x.

Introduce k, where k is a nonzero integer, and not 1 or -1, and new functions f

_{1}(x), and f

_{2}(x), where:

g

_{2}(x) = f

_{2}(x) + k-2 and g

_{1}(x) = f

_{1}(x)/k,

multiply both sides by k, and substitute for the g's, which gives me the now symmetrical form:

k*P(x) = (f

_{1}(x) + k)(f

_{2}(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:

f

_{1}(x) + f

_{2}(x) = H(x)

So I can find:

f

_{1}

^{2}(x) - H(x)f

_{1}(x) - kH(x) - k

^{2}+ k*P(x) = 0

And you can solve for f

_{1}(x) using the quadratic formula:

f

_{1}(x) = (H(x) +/- sqrt[(H(x) + 2k)

^{2}- 4k*P(x)])/2

And sqrt[(H(x) + 2k)

^{2}- 4k*P(x)] will only resolve if it is to a linear function, which proves that the f's are only polynomials when H(x) is, which means the g's are then as well.

And H(x) is a handle for every possible factorization with the g's. While it has a key constraint, H(0) = -k + 2.

By smoothly transitioning H(x) through all possible equations with the given constraint can encompass all possible equations for the g's with the given requirements at x=0, and in no case can they

*both*be equations that produce algebraic integers with algebraic integer x, except the polynomial case, when k is not zero or a unit.

Not really part of the main argument, but notice if we

*relax*the restriction on k and let it be a unit, like try k=1, then necessarily the g's give algebraic integers. And using k =1 or -1 and using all possible H(x) in that ring covers ALL algebraic integer solutions for the g's.

But with the condition that k not be zero or a unit, there can be values of x for which you get algebraic integers, for instance x=0 will always work. But such exceptions don't change the typical case.

But regardless of whether k is a unit or not, the f's must be in the ring of algebraic integers if H(x) is. So it is shown that in those cases ANY k with the requirements given that it be a nonzero integer, not 1 or -1, will give algebraic integer f's.

Therefore there must exist numbers which I will call objects o, such that k*o must be an algebraic integer, when o is not an algebraic integer, for any k with the requirements.

*Proof complete.*

But what about fields? Does this impact them?

And it is easy to show why fields are not impacted, as consider o

_{2}/o

_{1}, 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:

2o

_{2}/2o

_{1}

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

These numbers are such that k*o will give an algebraic integer, regardless of k as long as not 0, 1 or -1, with it an integer. So for a particular o not an algebraic integer, 3o and 5o will both be algebraic integers.

With these new numbers needed a ring to contain them as well as algebraic integers.

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.

That link goes to the first post on this blog. Posted back in 2005.

These numbers were the biggest reason for the existence of this blog.

But I hadn't discovered this simple approach back then. Wonder if I had, would it have made a difference? Will it make one now?

These results change so much in mathematics. Resistance has been so fierce for over a decade now.

James Harris

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