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January 28th, 2017, 11:52 PM  #1 
Senior Member Joined: Dec 2012 Posts: 951 Thanks: 23  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 $M_2=(2X1)$. Or the rule: a^2 = 1+3+5+.... (2a1) 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 $(2^{n1} 1)$ = 1 x $2^{n1}$ then.... 
February 11th, 2017, 01:40 AM  #2 
Senior Member Joined: Dec 2012 Posts: 951 Thanks: 23 
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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