Harristotelian Logic

Imagine you have a factorization of a polynomial

P(x) = (f(x) + 1)*(g(x) + 2)

and you are told to multiply it by 7, and to pick one way, so you
choose

7*P(x) = (7f(x) + 7)*(g(x) + 2)

can the value of the functions f(x) and g(x) invalidate your choice, turning it into something else?

That simplicity of multiplying by a constant lies at the heart of the non-polynomial factorization arguments.

To understand its importance let

P(x) = 175x^2 - 15x + 2

and now multiply that polynomial times 7 and re-group the resultant terms:

7*P(x) = (49x^2 - 14x)5^2 + (7x-1)(7)(5) + 49

to factor it as

7*P(x) = (5a_1(x) + 7)(5a_2(x)+ 7)

where the a's are roots of

a^2 - (7x-1)a + (49x^2 - 14x) = 0.

Notice that equation comes easily enough from multiplying out the factorization and solving for the a's, as you have

a_1(x)*a_2(x) = 49x^2 - 14x and a_1(x) + a_2(x) = 7x-1.

Now divide off 7:

P(x) = (7x^2 - 2x)5^2 + (7x-1)(5) + 7

giving the factorization

P(x) = (5b_1(x) + w_1)(5b_2(x)+ w_2)

where I've used b's for the new functions and w_1 and w_2 for the rest.

Then I have simply enough—notice 5 still brackets out what must go where:

b_1(x)*b_2(x) = 7x^2 - 2x,

w_1*b_2(x) + w_2*b_1(x) = 7x-1

and w_1*w_2 = 7

so I can substitute out b_2(x) and w_2(x), to get

w_1*(7x^2 - 2x)/b_1(x) + 7*b_1(x)/w_1 = 7x-1

and multiplying both sides by w_1*b_1(x) and simplifying a bit, I have

7*b_1(x)^2 - (7x-1)w_1*b_1(x) + (7x^2 - 2x)w_1^2 = 0

so I can solve for b_1(x) using the quadratic formula:

b_1(x) = ((7x - 1) +/- sqrt((7x-1)^2 -4(7x^2 - 2x)))*w_1/14

proving that b_1(x) cannot be an algebraic integer function unless w_1 has 7 as a factor, as notice otherwise the 7 in the denominator cannot divide out in general.

So the dodge of going to some w_1 and w_2 fails to change the result that the underlying factorization is outside of the ring of algebraic integers.

The functions just obscured how your multiplication proceeded, so let x=0, to clear them out, and you find

a^2 + a = 0

proving that your 7 went through as it did in my simple example above as one of the a's is 0 and the other -1 when you clear out the functions with x=0.

Simplicity is at the core where what complexity there is comes from the use of a creative construction:

7*P(x) = (49x^2 - 14x)5^2 + (7x-1)(7)(5) + 49

to get the non-polynomial factorization.

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Minor fix below…

Imagine you have a factorization of a polynomial

P(x) = (f(x) + 1)*(g(x) + 2)

and you are told to multiply it by 7, and to pick one way, so you choose

7*P(x) = (7f(x) + 7)*(g(x) + 2)

can the value of the functions f(x) and g(x) invalidate your choice, turning it into something else?

That simplicity of multiplying by a constant lies at the heart of the non-polynomial factorization arguments.

To understand its importance let

P(x) = 175x^2 - 15x + 2

and now multiply that polynomial times 7 and re-group the resultant terms:

7*P(x) = (49x^2 - 14x)5^2 + (7x-1)(7)(5) + 49

to factor it as

7*P(x) = (5a_1(x) + 7)(5a_2(x)+ 7)

where the a's are roots of

a^2 - (7x-1)a + (49x^2 - 14x) = 0.

Notice that equation comes easily enough from multiplying out the factorization and solving for the a's, as you have

a_1(x)*a_2(x) = 49x^2 - 14x and a_1(x) + a_2(x) = 7x-1.

Now divide off 7:

P(x) = (7x^2 - 2x)5^2 + (7x-1)(5) + 7

giving the factorization

P(x) = (5b_1(x) + w_1)(5b_2(x)+ w_2)

where I've used b's for the new functions and w_1 and w_2 for the rest.

Then I have simply enough—notice 5 still brackets out what must go where:

b_1(x)*b_2(x) = 7x^2 - 2x,

w_1*b_2(x) + w_2*b_1(x) = 7x-1

and w_1*w_2 = 7

so I can substitute out b_2(x) and w_2(x), to get

w_1*(7x^2 - 2x)/b_1(x) + 7*b_1(x)/w_1 = 7x-1

and multiplying both sides by w_1*b_1(x) and simplifying a bit, I have

7*b_1(x)^2 - (7x-1)w_1*b_1(x) + (7x^2 - 2x)w_1^2 = 0

so I can solve for b_1(x) using the quadratic formula:

b_1(x) = ((7x - 1) +/- sqrt((7x-1)^2 -4(7)(7x^2 - 2x)))*w_1/14

proving that b_1(x) cannot be an algebraic integer function unless w_1 has 7 as a factor, as notice otherwise the 7 in the denominator cannot divide out in general.

So the dodge of going to some w_1 and w_2 fails to change the result that the underlying factorization is outside of the ring of algebraic integers.

The functions just obscured how your multiplication proceeded, so let x=0, to clear them out, and you find

a^2 + a = 0

proving that your 7 went through as it did in my simple example above as one of the a's is 0 and the other -1 when you clear out the functions with x=0.

Simplicity is at the core where what complexity there is comes from the use of a creative construction:

7*P(x) = (49x^2 - 14x)5^2 + (7x-1)(7)(5) + 49

to get the non-polynomial factorization.