## Monday, November 04, 2013

### The conditional residue

In formalizing an area I call tautological spaces I have realized I need to formalize a bit further.

I found that I could use identities for mathematical analysis, relying on a special form which I discovered after a bit of effort.

The simplest identity I call a tautological space is:

x+y+vz = x+y+vz

And of course you can do simple operations with it, like the following:

x+y = -vz + (x+y+vz)

And squaring both sides gives:

x2 + 2xy + y2 = v2z2 - 2vx(x+y+vz) + (x+y+vz)2

And that can get complicated quickly so I use instead a modular form:

x2 + 2xy + y2 = v2z2 mod (x+y+vz)

That expression is always true, so it retains that aspect from the identity, but no longer really looks like one, and I call it a tautological space. In this case it is a space where that truth of x+y+vz holds.

While the tautological space is always true, the use of this technique for analysis comes with adding conditions, where I call equations used in this way, conditionals.

For instance, consider the condition that x2 + y2 = z2, which is not always true, as it depends on the values of x, y and z, where two are independent and one is a dependent variable.

We can subtract that conditional from the tautological space, and find:

2xy = (v2-1)z2 mod (x+y+vz)

which is:

(v2-1)z2 - 2xy = 0 mod (x+y+vz)

Which is what I call the conditional residue.

Notice it, unlike the identity, is only true with conditions.

So formally you can say:

(v2-1)z2 - 2xy = 0 mod (x+y+vz) if x2 + y2 = z2

And the conditional residue now has properties that follow from the conditional, which means you can analyze the conditional equation by analyzing the conditional residue.

Now v is a free variable no matter what! Which means that regardless of what x, y and z may be, you can always set v to any value you wish.

To see a less trivial result consider an important conditional residue with which I was able to find a general way to reduce binary quadratic Diophantine equations:

(c4 - c5v - c1v2)z2 + ((c2 -2c1)v - c5 + c6)zy + (c2 - c1 - c3)y2 = 0 (mod x+y+vz)

if

c1x2 + c2xy + c3y2 = c4z2 + c5zx + c6zy

Remarkably the conditional residue in this case can actually be used to get to an explicit equation, by appropriately setting v.

And again, v is a completely free variable which can be set as needed for analysis.

And I found I could use:  v = -(x+y)z-1 mod p

Weirdly enough that allowed me to get to an explicit equation which is important to mention as it shows this process of using identities with modular methods can lead to explicit results!

To see the complete derivation of the conditional residue and the explicit result see: