That construction is as follows.

The start is very importantly in the complex plane:

7(175x

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

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

_{2}(x)+ 7)

where the a's are roots of

a

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

^{2}- 14x) = 0

At this point there is just quadratic construction and the information that the a's are kind of bizarre in that they are roots of a quadratic! Not something you see every day I think, but also VERY IMPORTANT that at this point you're in the complex plane. There is a good reason for that being true which comes up soon.

And now normalize the functions, that is, have functions that equal 0, when x=0.

Looking at x=0, gives

a

^{2}+ a = 0, so a

_{1}(0) = 0, or -1, and a

_{2}(0) = -1, or 0.

So I can let a

_{1}(0) = 0, and then introduce a new function b

_{2}(x), where:

b

_{2}(x) = a

_{2}(x) + 1, so a

_{2}(x) = b

_{2}(x) - 1, and making that substitution:

7(175x

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

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

_{2}(x)+ 2)

where now a

_{1}(0) = b

_{2}(0) = 0.

So with normalized functions I have:

7(175x

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

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

_{2}(x)+ 2)

And a lot of the idea here is that it's starting to look clear from what we can see on the complex plane how the 7 distributed, where it looks easy.

So now I have easily that the 7 multiplied in a trivial way--which makes sense as I made up this example, why would I make it too hard?

So now it looks like we need one more functional substitution which is:

a

_{1}(x) = 7b

_{1}(x)

Which gives me:

7(175x

^{2}- 15x + 2) = (5(7)b

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

_{2}(x)+ 2)

And I can divide off the trivial factor of 7 to find:

175x

^{2}- 15x + 2 = (5b

_{1}(x) + 1)(5b

_{2}(x)+ 2)

And in the complex plane, we're done!

Easy.

So at this point I've removed an extraneous factor of 7 in a fairly straightforward way, using basic algebraic techniques, but I need a LOT more, which is what's coming up, where I'm moving from the field of complex numbers as I need a factor result.

But I don't exactly state it upfront which may seem sneaky. I'll explain why in a bit.

But um, we have this result now that one of the a's results by multiplying one of the b's by 7, and the a's are roots of:

a

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

^{2}- 14x) = 0

But now if we want to play with our result, say with x=1, we find that the a's are: 3+sqrt(-26) and 3-sqrt(-26), and in fact if you try:

7(175(1)

^{2}- 15(1) + 2) = (5(3+sqrt(-26)) + 7)(5(3-sqrt(-26)) + 7)

You'll see it is correct! That math worked!

But it implies that one of the a's has 7 as a factor. But established number theory says

*neither*of the a's have 7 as a factor?

And I've made a seemingly HUGE leap! In the complex plane factors are meaningless, so why make this sudden claim about 3+sqrt(-26)?

So what did I do wrong?

Turns out, I didn't do anything wrong. I simply forced you to rely on the field of complex numbers telling you what happened at the top level, so you can figure out what happens if you go to a ring where factors are meaningful in a way that they aren't in the field.

So let's look at a seeming counter-example to this approach!

Consider:

a

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

^{2}- 14x) = 0

Here I've removed a 7 so that I can force a case where we can seemingly see the 7 split up into square roots, as at x=1, I have: a

^{2}+ 35 = 0

Then the a's are sqrt(-35) and -sqrt(-35), and clearly sqrt(7) is a factor, not 7.

But what about the complex plane? The complex plane encompasses lesser rings, right?

But in the complex plane with my earlier example:

7(175x

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

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

_{2}(x)+ 2)

with normalized functions, requiring that the 7 distributed in only one way, and yup, the mathematical linchpin here is the distributive property!

Quite simply there is no way for sqrt(7) to have distributed, as then you'd see something like:

7(175x

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

_{1}(x) + sqrt(7))(5b

_{2}(x)+ 2sqrt(7))

And notice, it's basic math at this point. If the 7 split up into squares then they'd multiply through, and you'd actually

*see*them in the construction.

So how do we maintain mathematical consistency at this point? Or is it time to throw away algebra?

Answer is, easy, as look at:

175x

^{2}- 15x + 2 = (5b

_{1}(x) + 1)(5b

_{2}(x)+ 2)

Here I don't see a 7 at all. And I've used the assumption:

a

_{1}(x) = 7b

_{1}(x)

So if we extrapolate from here to something that results from removing the 7, if we say the first of the a's is sqrt(-35) we'd have a solution for one of the b's of:

b

_{1}(1) = sqrt(-35)/7 = sqrt(-5)/sqrt(7)

So a 1/sqrt(7) forced its way into the picture.

And note that the above is not direct from my original construction but would follow if I made one with a key 7 removed, as I'm talking about what happens if you try to force a factor of 7 to move around, when the field of complex numbers tells you it can't. The math simply shifts, as it IS consistent. It has no choice.

The field of complex numbers says how the 7 distributed which includes all lesser rings. If you have an example where that appears to not be true, then you are wrong, as the field of complex numbers is right!

Here is a giant area and it's critical to hammer home the point that the field of complex numbers tells you how the 7 distributed and it MUST be correct. From there certain things logically follow--or you lose something called mathematical consistency.

That is, otherwise I've demonstrated that human mathematics falls apart here.

(I actually considered that carefully for some time. Whether our mathematics is consistent or not is actually a really big deal.)

The construction is built using special techniques designed to make it valid in what I call the ring of objects which I figured out to handle this problem.

So now it's time to focus on how I built my construction!!!

It's suddenly of HUMONGOUS importance, as the field of complex numbers will only allow one thing, but we broke my construction by just removing a 7, which

*forced*1/sqrt(7) to be included in whatever ring or field, or where are we now?

Traditionally to talk about factors with non-rationals I'd be using the ring of algebraic integers.

However, it turns out in that ring you can prove that neither of the a's can have 7 as a factor!!!

Here things get a little more complicated as if you don't remember your number theory or never had much of it--like I hadn't as I was a physics student--you might wish to see that proof. And I have to say, go look for it.

For the math people who are up on their number theory though, it's a standard result which is well established, and they can see my conundrum: I'd proven a factor result using the complex plane by the distributive property, which MUST include lesser rings, only to find that the ring of algebraic integers didn't wish to agree, with the field of complex numbers!!!

And I thought about it, and thought about, and thought about it some more and came up with the ring of objects which doesn't have the problem. In the ring of objects 7 is indeed a factor of 3+sqrt(-26).

One thing that emerges though is that you cannot

*see*the 7, as it's permanently entangled, because the square root cannot be resolved.

Like imagine you couldn't resolve sqrt(9), how would you know with the following?

It's kind of like 10 + sqrt(9) has 7 as a factor for one of its two solutions.

If you think 10 + sqrt(9) is 13, you're wrong, as it's 13 or 7 because sqrt(9) = 3 or -3.

For the truly curious and mathematically sophisticated, you may now wonder how the ring of algebraic integers pulls off the trick of not allowing 7 to be a factor. Turns out I puzzled that one out years ago, where the answer is, it wraps it up in unit factors, which are units in the ring of objects but NOT in the ring of algebraic integers!

Tricky, but not really. Unit factors CAN move in the field of complex numbers with the original example, with ease, without being forced to show themselves in the same way as the 7 is forced.

What's wild is that these are not unit factors within the ring of algebraic integers, but must be in the ring of objects.

And notice I rely heavily on the result from the field of complex numbers because there is where you can

*see*what actually happens with the 7.

And the weird thing for those who'd like to try and start from the ring of algebraic integers is, it won't allow you to see the result. In fact, it won't even let you divide the 7 off, blocking entirely the final result:

175x

^{2}- 15x + 2 = (5b

_{1}(x) + 1)(5b

_{2}(x)+ 2)

with normalized functions.

So could this have been figured out before? Well, read through all of the above and it's easy to see why it's hard to figure out. The ring of algebraic integers works you can say, to hide the correct result. Only by use of the field of complex numbers and belief in the distributive property was I able to tease out the correct answer, and then needed to figure out a ring which worked ok!

Whew! That was actually quite a bit of work. Had never figured out my own ring before, but I had no choice.

James Harris