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August 30, 2014

August 30, 2014

Posted by **nimrod** on Friday, February 18, 2011 at 8:54am.

- math -
**bobpursley**, Friday, February 18, 2011 at 9:03amNo, your premise is not true.

2^3+3^2= a^3

8+27=

35, and 35 is not a cube of a natural number.

- math -
**tchrwill**, Friday, February 18, 2011 at 9:43amFermat's Last Theorem states that there are no solutions to x^n + y^n = z^n for n = 3 or greater. Surprisingly, there are solutions to x^n + y^n = z^(n+1) or z^(n-1).

1--Divide through by z^n resulting in (x/z)^n + (y/z)^n = z.

2--Let (x/z) and (y/z) be any positive numbers other than 1.

3--Solve for z and then x and y.

4--Example: x^3 + y^3 = z^4

5--Let (x/z) = 2 and (y/z) = 5. Then 2^3 + 5^3 = 133 = z.

6--Then x = 2(133) = 266 and y = 5(133) = 665.

7--Checking: 266^3 + 665^3 = 18,821,096 + 294,079,625 = 312,900,721 = 133^4.

It works for any n and (n+1).

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