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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. "
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. A1 = A0 + 2*D
3. B1 = A1 + A0 + B0
4. C1 = B1 + D
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These equations are used with the following
three initial conditions:
a. (C, B, A)
b. (C, A, B)
c. (C, -A, B)
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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)
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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)
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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)
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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.
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