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January 28th, 2017, 11:52 PM   #1
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Math Crystallization Grains

Equalities in integers / rationals can be solved using the

Math Crystallization Grains method:

Starting from some simple example I'm sure will be immediately clear the concept:

- I transform 1 in 1x1 area (or cube or hypercube)

than any equality, if true, is an identity like:

1 = 1

We can represent it on a plane as a square 1x1 equal to a square 1x1.

This is what I call: "Crystallization Grain", and is the first level of definition of our problem: we state there is an identity in the Integers or in the Rational,

Using this definition we can go over using the

2) Second level of definition of the problem: how many grains we have in the right and in the left hand.

3) We have a 3th level of definition of the problem: the property we fix to one, or both, side of the identity, or to each grain.

4) each grain is the littlest indivisible in our problem

a typicall colection of "hard" problems is the one fix for both side be a Power of Integers and we have two ( Beal for example and Fermat too) or more in the left hand and on just in the right hand.

Example: if our problem must be in the integers, and starts from:

1x1 + 1x1 = 1x2

Than we can ask how we can modify the first two left hand terms to be equal to a Power of an Integers (a square for example).

So now we have to apply the rules we know for Square:

they can be represented in the form $a^2$ or in the form of Sum of Gnomons $Base= 1$, Heigh $M_2=(2X-1)$.

Or the rule: a^2 = 1+3+5+.... (2a-1)

Therefore one trasformation can be:

- let the first left hand term equal to 1x1

- add the 2th term 1x2 ( so 1+2=3)

the third becomes a square 2x2=4 since:

1x1 + 1x3 = 1x2 + 1x2

The general rule for more grains on the left rest the same. If we start from:

1x1 + 1x1 + 1x1= 1x3

then we can transform:

1x1 + (1x1 + 1x2) + (1x1+ 1x4)= 1x3 + ( 1x2 +1x4) = 3^2

Beal as example:

1) to transform all the grains in Powers we can for example take:

1x1 + 1x1 = 1x2

than see that the problem becomes in base 2, we can write:

1x1 x $2^n$ + 1x1 x $2^n$ = 1x2 x $2^n$ = $2^{n+1}$

2) If we start from:

1x1 + 1x $2^3$ = 1 x $3^2$

And we ask all the grains becomes Powers we can apply:

1x1 x $3^3$ + 1x $2^3$x$3^3$ = 1 x $3^2$x $3^3$ = $3^5$

3) If we start from:

1x1 + 1x $(2^{n-1} -1)$ = 1 x $2^{n-1}$

then....
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February 11th, 2017, 01:40 AM   #2
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Here the result of the Grain involving 2^n:

For n=2 (the good triplet has $Check \in \mathbb{N^+}$)



For n=3,4,5:



... and for any bigger n.

Last edited by complicatemodulus; February 11th, 2017 at 01:45 AM.
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