### Wednesday, September 08, 2010

## JSH: Using composites to understand the arguments

For years I've used 7 as my favorite constant with equations that I use to try and show a remarkable problem with established number theory, which has for years left open the door for posters to claim that primeness is critical, so here's an explanation with composites! The 7 will be replaced by 9.

Consider:

9(g_1(x) + 1)(g_2(x) + 2) = (f_1(x) + 9)(f_2(x) + 9) = 9*P(x)

where g_1(0) = g_2(0) = 0, and one of the f's equals 0, at x=0, and P(x) is a quadratic with integer coefficients.

Those rules force one of the f's to have 9 as a factor when the f's are polynomials.

Trivially if the f's are linear functions 9 is a factor of only one of them, with those rules.

That follows by the distributive property. But if so, then the f's being linear is immaterial, right? So

Posters replying here should explain:

mathematicians in any way shape or form, shouldn't they be able to answer all 3 questions?

They aren't trick questions! There is no hiding of some obfuscation or some weird psychological trick.

Let's see what actually happens then.

Oh, my point is that 9 is the factor, but provably in the ring of algebraic integers you can have f's where NEITHER of the f's can have 9 as a factor. Some posters pointed out primeness with my 7, so I'm using 9. Curious readers might try to split it up, say, into 3 being a factor of each of the f's just to see if you can—with the rules given about zeros.

If I'm right, then I can collapse much of modern number theory, and show that Galois Theory isn't really useful after all, so a lot at stake.

Ok, enough chatter. Let's see what happens…

Consider:

9(g_1(x) + 1)(g_2(x) + 2) = (f_1(x) + 9)(f_2(x) + 9) = 9*P(x)

where g_1(0) = g_2(0) = 0, and one of the f's equals 0, at x=0, and P(x) is a quadratic with integer coefficients.

Those rules force one of the f's to have 9 as a factor when the f's are polynomials.

Trivially if the f's are linear functions 9 is a factor of only one of them, with those rules.

That follows by the distributive property. But if so, then the f's being linear is immaterial, right? So

**whatever**types of functions they are, then 9 should be the factor, by the distributive property.Posters replying here should explain:

- Is 9 the factor as stated above, and if so, why?
- If they believe it can split up, can they demonstrate with integers?
- Why would it matter if the f's are non-polynomials? Shouldn't 9 still be the factor?

mathematicians in any way shape or form, shouldn't they be able to answer all 3 questions?

They aren't trick questions! There is no hiding of some obfuscation or some weird psychological trick.

Let's see what actually happens then.

Oh, my point is that 9 is the factor, but provably in the ring of algebraic integers you can have f's where NEITHER of the f's can have 9 as a factor. Some posters pointed out primeness with my 7, so I'm using 9. Curious readers might try to split it up, say, into 3 being a factor of each of the f's just to see if you can—with the rules given about zeros.

If I'm right, then I can collapse much of modern number theory, and show that Galois Theory isn't really useful after all, so a lot at stake.

Ok, enough chatter. Let's see what happens…