- #1
chuck klasky
- 5
- 0
I have been studying the Pythagorean theorem and Fermat’s Last theorem for fun and to brush up on my algebra.
I think I’ve come up with an equation for redefining An in terms of B and C. But my math is too rusty to verify if its correct.
I hope you look at it and see if there is a mistake somewhere. Thanks
C^n – B^n = A^n
C – B = Q so C/Q – B/Q = Q/Q =1.
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[C^n-B^n]/Q^n = C^n/Q^n – B^n/Q^n = (C/Q)^n- (B/Q)^n = (C/Q)^n – ((C-Q)/Q)^n
= (C/Q)^n – (C/Q – Q/Q)^n = (C/Q)^n – (C/Q -1)^n { C-B = Q }
= (C / C-B)^n – ( (C/ C-B) – 1)^n and so
[Cn-Bn]/(C-B)^n = (C / C-B)^n – ( (C/ C-B) – 1)^n and therefore
Cn-B^n = (C-B)^n [(C / C-B)^n – ( (C/ C-B) – 1)^n] = A^n
I think I’ve come up with an equation for redefining An in terms of B and C. But my math is too rusty to verify if its correct.
I hope you look at it and see if there is a mistake somewhere. Thanks
C^n – B^n = A^n
C – B = Q so C/Q – B/Q = Q/Q =1.
__________________________________________
[C^n-B^n]/Q^n = C^n/Q^n – B^n/Q^n = (C/Q)^n- (B/Q)^n = (C/Q)^n – ((C-Q)/Q)^n
= (C/Q)^n – (C/Q – Q/Q)^n = (C/Q)^n – (C/Q -1)^n { C-B = Q }
= (C / C-B)^n – ( (C/ C-B) – 1)^n and so
[Cn-Bn]/(C-B)^n = (C / C-B)^n – ( (C/ C-B) – 1)^n and therefore
Cn-B^n = (C-B)^n [(C / C-B)^n – ( (C/ C-B) – 1)^n] = A^n
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