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[ March 19, 2005 ]
Equations concerning Pythagorean Theorem and
Fermat's Last Theorem
Bob Hein (email)

" 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. "

Pythagorean Theorem

 1. D = C – B 2. A1 = A0 + 2*D 3. B1 = A1 + A0 + B0 4. C1 = B1 + 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 (C0, B0, A0) = (5, 4, 3), then

 D = 5 – 4 = 1 A1 = 3 + 2*1 = 5 B1 = 5 + 3 + 4 = 12 C1 = 12 + 1 = 13, therefore, (C1, B1, A1) = (13, 12, 5)

b. Let (C0, A0, B0) = (5, 3, 4), then

 D = 5 – 3 = 2 A1 = 4 + 2*2 = 8 B1 = 8 + 4 + 3 = 15 C1 = 15 + 2 = 17, therefore, (C1, B1, A1) = (17, 15, 8)

c. Let (C0, -A0, B0) = (5, -3, 4) then

 D = 5 – (-3) = 8 A1 = 4 + 2*8 = 20 B1 = 20 + 4 + (-3) = 21 C1 = 21 + 8 = 29, therefore, (C1, B1, A1) = (29, 21, 20)

Fermat's Last Theorem (FLT)
Let C > B > A and assume that Cn = An + Bn
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
En = (A + B – C)n, add zero (0) to both sides of the equation,
assuming that there are solutions to
Cn = An + Bn, then Cn – (An + Bn) = 0
En + 0 = (A + B – C)n + (Cn – (An + Bn)),
factor the right hand side (RHS) of the equation
For n = 2, the Pythagorean Theorem
E2 = 2 * (C - B) * (C - A)
For n = 3
E3 = 3 * (C - B) * (C - A) * (A + B)
For n = p which is a prime number then
Ep = p * (C - B) * (C - A) * (A + B) * (Cp-3 + Bp-3 + Ap-3 + ... - ... )

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

 1. D = C – B 2. A1 = A0 + 2*D 3. B1 = A1 + A0 + B0 4. C1 = B1 + D

aa. Let (C0, B0, A0) = (1 ,0, 1), then

 D = 1 – 0 = 1 A1 = 1 + 2*1 = 3 B1 = 3 + 0 + 1 = 4 C1 = 4 + 1 = 5, therefore, (C1, B1, A1) = (5, 4, 3)

bb. Let (C0, B0, A0) = (5, 4, 3), then

 D = 5 – 4 = 1 A1 = 3 + 2*1 = 5 B1 = 5 + 3 + 4 = 12 C1 = 12 + 1 = 13, therefore, (C1, B1, A1) = (13, 12, 5)

cc. 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 = (4n2 + 4n + 1)/2 - 0.5 = (2n2 + 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.  A000164 Prime Curios! Prime Puzzle  Wikipedia 164 Le nombre 164 ```

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