FLT, The Series Approach

We need an alternate approach to describe the problem!

* Diagrams arriving soon to clarify things here ...*

The c^{n} = a^{n} + b^{n} problem can be looked at as a problem about numeric series. Some function f is applied to each element in the series 1 ... c. "f" could be f(c) = c^2 or c^3 or c^4 etc. Here is a table showing the starting values for the 1, 2, 3 and 4 dimension spaces.

# | area ^2 | volume ^3 | 4D ^4 | ... |
---|---|---|---|---|

1 | 1 | 1 | 1 | |

2 | 4 | 8 | 16 | |

3 | 9 | 27 | 81 | |

4 | 16 | 64 | 256 | |

5 | 25 | 125 | 625 | |

6 | 36 | 216 | 1296 | |

7 | 49 | 343 | 2401 | |

8 | 64 | 512 | 4096 |

The formula that gets us the space for an n-D shape of side length "c" is of course c^{n}. To create a series that gives us this same result, we create a function f() that tells us what we add at each increment "c", starting from 1:

A function applied to a series:

f(1) + f(2) + ... + f(c-1) + f(c).

A series function that yeilds shape space for any position a in the series, and power n:

f(c) = c^{n} - (c-1)^{n}

Split The Series Up!

Now there are a few other patterns at work. **We have a series whose space is calculated from f(1) to f(c), which exactly contains spaces a ^{n} + b^{n}**. The following hold:

- Take a
^{n}to be the smaller space, b^{n}to be the larger. - Our series f(1)...f(c) can be split into f(1)...f(a) , the space for a
^{n}, and f(a+1)...f(c), the space for b^{n} - Spaces for a and b can switch places in the sequence, so that f(1)...f(b) covers b
^{n}, and f(b+1)...f(c) covers a^{n} - Now we can stop even involving "b" as part of the problem. Instead our goal is to split f(1) ... f(c) into 3 parts,
- the first part of the sequence covering a
^{n} - then a middle space - call it d - spanning f(a+1) to f(b). d = b-a+1
- then finishing off with space a
^{n}again

- the first part of the sequence covering a
- Approaching the problem from the search perspective, we can start the search for a
^{n}exactly in the middle of the volume c^{n}: v_{m}= c^{n}/2. - Call p the starting point of the search; p = int(n
^{th}root of (v_{m}) - Our search must extend exactly the same length in both directions from p. It will encompass and exactly split the range of space d occupies.
- p likely occurs somewhere within the running sum of a series element, call this element m. Let j = the offset into f(m) that p is. let k be the remainder of f(m) after p.
- Now we have:
f(1)...f(a)...f(m-1) + j = k + f(m+1) ...f(b)...f(c)

- The first f(1)...f(a) and last f(b+1)...f(c) portions of this sequence must match in size, so we can drop them.

The equation we finish off with, and indeed the crux of the matter, is:

**f(a+1)...f(m-1) + j = k + f(m+1)...f(b)**

|--------- d ---------| = |-------- d --------|

If we can show by the geometry that integers {a,m,b} and {j,k} (j,k can have halves) are an impossible set, then we have succeeded.

Pythagorean Triplet Example

Let's first check to see if this series formula actually **is** valid for some pythagorian triples. Then perhaps an argument comes forward that n>2 is impossible. The diagram illustrates the pythagorean triplet {3,4,5} as a solution for a^{n} + b^{n} = c^{n} where n=2:

row | c^{2} | delta | |
---|---|---|---|

1 | 1 | 1 | a |

2 | 4 | 3 | a a a |

3=a | 9 | 5 | a a a a a |

4=m,b | 16 | 7 | a a a p b b b |

5=c | 25 | 9 | b b b b b b b b b |

Given c=5, c^{2}=25,

p=12.5

m=4 (row of p)

j=3.5, k=3.5

m+1'th sequence element would be 5, but series ends with b, and b != c, so no series.

j=k, done

n = 2, condensed

row | c^{3} | delta | |
---|---|---|---|

1 | 1 | 1 | 1 |

2 | 4 | 3 | 2*1 + 1 |

3=a | 9 | 5 | 2*2 + 1 |

4=m,b | 16 | 7 | 1*3 + 1 + 1*3 |

5=c | 25 | 9 | 2*4 + 1 |

Triple #2 Example (5, 12, 13)

row | c^{2} | delta | |
---|---|---|---|

1 | 1 | 1 | a |

2 | 4 | 3 | a a a |

3 | 9 | 5 | a a a a a |

4 | 16 | 7 | a a a a a a a |

5=a | 25 | 9 | a a a a a a a a a |

6 | 36 | 11 | a a a a a a a a a a a |

7 | 49 | 13 | a a a a a a a a a a a a a |

8 | 64 | 15 | a a a a a a a a a a a a a a a |

9 | 81 | 17 | a a a a a a a a a a a a a a a a a |

10=m | 100 | 19 | a a a a
p
b b b b b b b b b b b b b b |

11 | 121 | 21 | b b b b b b b b b b b b b b b b b b b b b |

12=b | 144 | 23 | b b b b b b b b b b b b b b b b b b b b b b b |

13=c | 169 | 25 | b b b b b b b b b b b b b b b b b b b b b b b b b |

Given c=13, c^{2}=169,

p=84.5

m=10 (row of p)

j=3.5, k=15.5

f(6)+f(7)+f(8)+f(9) + j = k + f(11)+f(12) , done

n = 3

row | c^{3} | delta | |
---|---|---|---|

1 | 1 | 1 | 1 |

2 | 8 | 7 | 3*1^2 + 3*1 + 1 |

3 | 27 | 19 | 3*2^2 + 3*2 + 1 |

4 | 64 | 37 | 3*3^2 + 3*3 + 1 |

Two other formulas pop out. The space c^{n} is made up of (m - 1)^{n} + j + m^{n} - k, and since j-k = m^{n}-(m-1)^{n} -2k,

c^{n} = 2(m^{n} - k)

Sometimes contradictions can be unearthed as a result of a parity check - determining which variables must be even or odd

- c must be odd
- m must be even
- k must be odd + a half

e.g. c=5,c^{2} = 25, 2*4^{2} - 2*3.5 = 25

e.g. c=13,c^{2} = 169, 2*10^{2} - 2*15.5 = 169

n = 3, condensed

A few more notes for next evening ...

row | c^{3} | delta | |
---|---|---|---|

1 | 1 | 1 | 1 |

2 | 8 | 7 | 3*1*2 + 1 |

3 | 27 | 19 | 3*2*3 + 1 |

4 | 64 | 37 | 3*3*4 + 1 |

5 | 125 | 61 | 3*4*5 + 1 |

6 | 216 | 91 | 3*5*6 + 1 |