[ March 19, 2005 ]
Bob Hein's (email) equations concerning the
Pythagorean Theorem and Fermat's Last Theorem.
" Bob Hein has discovered a method by substitution of
variables for the construction of all infinite Pythagorean Triples.
I have listed the discussion of the subject below. "
http://mathworld.wolfram.com/PythagoreanTheorem.html
http://mathworld.wolfram.com/FermatsLastTheorem.html
http://mathworld.wolfram.com/PythagoreanTriple.html
Pythagorean Theorem
1. D = C B
2. A_{1} = A_{0} + 2*D
3. B_{1} = A_{1} + A_{0} + B_{0}
4. C_{1} = B_{1} + D

These equations are used with the following
three initial conditions:
a. (C, B, A)
b. (C, A, B)
c. (C, A, B)

An example of each initial condition is shown:
a. Let (C_{0}, B_{0}, A_{0}) = (5, 4, 3), then
D = 5 4 = 1
A_{1} = 3 + 2*1 = 5
B_{1} = 5 + 3 + 4 = 12
C_{1} = 12 + 1 = 13,
therefore, (C_{1}, B_{1}, A_{1}) = (13, 12, 5)

b. Let (C_{0}, A_{0}, B_{0}) = (5, 3, 4), then
D = 5 3 = 2
A_{1} = 4 + 2*2 = 8
B_{1} = 8 + 4 + 3 = 15
C_{1} = 15 + 2 = 17,
therefore, (C_{1}, B_{1}, A_{1}) = (17, 15, 8)

c. Let (C_{0}, A_{0}, B_{0}) = (5, 3, 4) then
D = 5 (3) = 8
A_{1} = 4 + 2*8 = 20
B_{1} = 20 + 4 + (3) = 21
C_{1} = 21 + 8 = 29,
therefore, (C_{1}, B_{1}, A_{1}) = (29, 21, 20)

Fermats Last Theorem (FLT)
Let C > B > A and assume that C^{n} = A^{n} + B^{n}
is true for some (C,B,A), this is not FLT (assumption)
From equation 1. Let D = C B then
Let E = A D,
therefore 5. E = A + B C,
raise both sides of the equation to the nth power
E^{n} = (A + B C)^{n}, add zero (0) to both sides of the equation,
assuming that there are solutions to
C^{n} = A^{n} + B^{n}, then C^{n} (A^{n} + B^{n}) = 0
E^{n} + 0 = (A + B C)^{n} + (C^{n} (A^{n} + B^{n})),
factor the right hand side (RHS) of the equation
For n = 2, the Pythagorean Theorem
E^{2} = 2 * (C  B) * (C  A)
For n = 3
E^{3} = 3 * (C  B) * (C  A) * (A + B)
For n = p which is a prime number then
E^{p} = p * (C  B) * (C  A) * (A + B) * (C^{p3} + B^{p3} + A^{p3} + ...  ... )
Conclusion, since E = (A + B C), then it cannot be divided by (A + B),
Therefore, the assumption is false and FLT is true.
Pythagorean Theorem
D = C B
A_{1} = A_{0} + 2*D
B_{1} = A_{1} + A_{0} + B_{0}
C_{1} = B_{1} + D

Let (C_{0}, B_{0}, A_{0}) = (1 ,0, 1), then
D = 1 0 = 1
A_{1} = 1 + 2*1 = 3
B_{1} = 3 + 0 + 1 = 4
C_{1} = 4 + 1 = 5,
therefore, (C_{1}, B_{1}, A_{1}) = (5, 4, 3)

Let (C_{0}, B_{0}, A_{0}) = (5, 4, 3), then
D = 5 4 = 1
A_{1} = 3 + 2*1 = 5
B_{1} = 5 + 3 + 4 = 12
C_{1} = 12 + 1 = 13,
therefore, (C_{1}, B_{1}, A_{1}) = (13, 12, 5)

The next triple can be calculated, but is there an easier way ?
Yes there is:
Let A = Odd number, then
B = (O*O)/2 0.5, and C = B + 1
then O = 2*n + 1, then B = (2n+1)*(2n+1)/2  0.5
B = (4n^{2} + 4n + 1)/2  0.5 = (2n^{2} + 2n + 0.5)  0.5 = 2*n*(n+1)
A = n + (n+1)
B = 2*n*(n+1)
C = B + 1 
The rule is: B first, then C, then A.
Example: solve for the 500th odd number, then
B = 2*500*(500 + 1) = 501000,
C = 501001,
A = 500 + (500 + 1) = 1001
then (C, B, A) = (501001, 501000, 1001)
251002002001 = 251001000000 + 1002001
this method is useable for any odd number.