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Physics 3 - Maxwell
Note from the Author
Table of Contents
What is this book
Historical facts
New aspects
Fully erroneous
Incorrectly interpreted

Physics 3 - Chapter 1
Equations
Complex vectors form
The Most General form
The General Solution

Physics 3 - Chapter 2
Solutions
Initial conditions
Non-homogeneous equation
Solution for three-directions
The four laws

Physics 3 - Supplement
Fermat's proof
Beal's conjecture
Pythagorean triples
Inertial mass
Gravity constans big G
What does the Moon look at?

Physics 3 - Final notes
Final notes

Physics 4 - New book
Entry

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3.1.2. Pythagorean triples

Returning to our presented proof – as a “side effect” it allows also for easy way to find possible solutions of (3.1) for N = 2 (a.k.a. Pythagorean triples).
New variable x along with auxiliary variable  e = c - a  which were introduced in our proof lead to the following relations:

b2=((c - a)+x)2=(e+x)2=e2+2‧ e‧ v+x2   and (3.9)
b2=c2- a2=(e+a)2- a2=e2+2‧ e‧ v (3.10)
Comparing (3.9) and (3.10) we obtain:
x2=2‧ (a - x) , (3.11)
where it can be seen that x must be an even number.
We can introduce new integral variable y such that x=2‧ y, then inserting it into (3.11) we obtain:
wzor8b. (3.12)
We can conclude then e and  y  have to be expressed by following relations: e=2‧ re2 and y=re‧ ry , where re is a divisor of e and ry is a second divisor of y, besides re.

If this were not a case, a, e and y would have common divisor, then this would mean that and b were not relatively prime.

In this case we obtain:

wzor8b, (3.13)
where ry cannot be an even number, since  2   is a divisor of e.

It should be then assumed that:
ry=2‧ rj+1 , where rj - is an integer ⩾ 0,
which results in:
e=2‧ re2 ,   x=2‧ re‧ (2‧ rj+1) ,

a=ry‧ (ry+2‧ re)= (2‧ ry+1)‧ (2‧ (rj+re)+1) , (3.15a)
b=2‧ re‧ (2‧ rj+1)+2‧ re2)= 2‧ re‧ (2‧ rj+re+1) , (3.15b)
c=ry‧ (ry+2‧ re)+2‧ re2= (2‧ rj+1)‧ (2‧ (rj+re)+1)+2‧ re2 (3.15c)
So, we can re-write them as:
a=(2‧ ry+1)‧ 2‧ (rj+re)+1)= (2‧ rj+re+1) 2 - re2 ,
b=2‧ re‧ (2‧ rj+1)+2‧ re2)= 2‧ (2‧ rj+re+1)‧ re ,
c=(2‧ rj+1)‧ (2‧ (rj+re)+1)+2‧ re2= (2‧ rj+re+1)2+re2
and then they correspond to well-known publicized relations
a=m2- n2 , b=2‧ m‧ n , c=m2+n2 . (3.16)
Our derived relations (3.15) have additional advantage since there is no restrictions such as in case of relation (3.16) where m and n cannot be odd numbers at the same time.




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