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

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:

• Take aⁿ to be the smaller space, bⁿ to be the larger.
• Our series f(1)...f(c) can be split into f(1)...f(a) , the space for aⁿ, and f(a+1)...f(c), the space for bⁿ
• Spaces for a and b can switch places in the sequence, so that f(1)...f(b) covers bⁿ, and f(b+1)...f(c) covers aⁿ
• 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 aⁿ
  • then a middle space - call it d - spanning f(a+1) to f(b). d = b-a+1
  • then finishing off with space aⁿ again
  So the initial and final spaces a are the same, but the territory they occupy in the sequence differs.
• Approaching the problem from the search perspective, we can start the search for aⁿ exactly in the middle of the volume cⁿ: v_m = cⁿ/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.

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}

row	c2	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, 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

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

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

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

e.g. c=5,c² = 25, 2*4² - 2*3.5 = 25

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

A few more notes for next evening ...