FLT, The Series Approach

We need an alternate approach to describe the problem!

Diagrams arriving soon to clarify things here ...

The cn = an + bn 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 cn. 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) = cn - (c-1)n


The series formula for area of a 2D square

f(c) = c2 - (c-1)2, simplifying to: f(c)= 2c-1

E.g. the series for a 2D square of side length 5

(2(1)-1) + (2(2)-1) + (2(3)-1) + (2(4)-1) + (2(5)-1)

  = 1 + 3 + 5 + 7 + 9

  = 25


The series formula for a 3D cube

f(c) = c3 - (c-1)3

  = c3 - (c2-2c+1)(c-1)

  = c3 - (c3-2c2+c-c2+2c-1)

  = 3c2-3c+1

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 an + bn. The following hold:

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 an + bn = cn where n=2:

n=2, pythagorean triplet {a=3,b=4,c=5}
rowc2delta
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, c2=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
rowc3delta
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)
rowc2delta
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, c2=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
rowc3delta
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 cn is made up of (m - 1)n + j + mn - k, and since j-k = mn-(m-1)n -2k,

cn = 2(mn - k)

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

e.g. c=5,c2 = 25, 2*42 - 2*3.5 = 25

e.g. c=13,c2 = 169, 2*102 - 2*15.5 = 169

n = 3, condensed

A few more notes for next evening ...

rowc3delta
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