P(x) = (g

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

_{2}(x) + 2)

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

_{1}(0) = g

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

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

(Updating October 26, 2017, to correct wrong beliefs about existence in ring of algebraic integers. Was able to simply remove from this earlier section so was easy.)

I pushed the envelope by imagining the quadratic is what's called primitive and irreducible over Q, which means that it does NOT factor into polynomials.

And I showed stepping through some simple algebraic manipulations where I multiplied by 7, but this time I will multiply by k, where k is a nonzero integer, and not 1 or -1:

Introduce new functions f

_{1}(x), and f

_{2}(x), where I'll use the second one first, as let

g

_{2}(x) = f

_{2}(x) + k-2.

Now make the substitution:

P(x) = (g

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

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

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

_{2}(x) + k)

Multiply both sides by k:

k*P(x) = (kg

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

_{2}(x) + k)

Now, let g

_{1}(x) = f

_{1}(x)/k, and make that substitution to get:

k*P(x) = (f

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

_{2}(x) + k)

Now multiply it out:

k*P(x) = f

_{1}(x)*f

_{2}(x) + k(f

_{1}(x) + f

_{2}(x)) + k

^{2}

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: f

_{2}(x) = -f

_{1}(x) + H(x),

and make that substitution, to get:

k*P(x) = f

_{1}(x)*(-f

_{1}(x) + H(x)) + kH(x) + k

^{2}

So you have:

k*P(x) = -f

_{1}

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

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

^{2}

Which means:

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

^{2}(x) + 4kH(x) + 4k

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

which is:

f

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

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

And we can see when that is rational.

Looking at our trivial example again, H(x) = kx + x - k + 2 = (k+1)x - k + 2

And inside that square root then is:

((k+1)x + k + 2)

^{2}- 4k(x

^{2}+ 3x + 2)

which is:

(k+1)

^{2}x

^{2}+ 2(k+1)(k+2)x + (k+2)

^{2}- 4kx

^{2}- 12kx - 8k

which is:

(k-1)

^{2}x

^{2}+ 2(k-1)(k-2)x + (k-2)

^{2}= ((k-1)x + k-2)

^{2}

And making that substitution:

f

_{1}(x) = ((k+1)x - k + 2 +/- ((k-1)x + k-2))/2 = kx or x - k + 2 as required.

So only polynomial factorizations can remove that square root for all integer x.

Here is a non-polynomial factorization example though.

Where I use:

P(x) = 175x

^{2}- 15x + 2, and k = 7

And further I have:

7*P(x) = 7(175x

^{2}- 15x + 2) = (5a

_{1}(x) + 7)(5a

_{2}(x) + 7)

So f

_{1}(x) = 5a

_{1}(x), and f

_{2}(x) = 5a

_{2}(x), where the a's are roots of:

a

^{2}- (7x-1)a + (49x

^{2}- 14x) = 0

Where notice H(x) = 5(7x - 1), as you can see the a's summed in there, and just multiply that by 5, to get the sum of the f's.

That non-polynomial factorization though from which I worked out what the value for H(x) must be comes from my use of a special construction.

James Harris

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