What is the significance of Fermat's Last Theorem in modern mathematics?

[PFanalog57]

Why not put the Fermat equation in terms of one variable, where the polynomial must have positive integer roots, for degree n = prime ?

f[x] + g[x] = h[x]

f[x] = x^p

g[x] = [x+y]^p

h[x] = [x+z]^p

where y and z are arbitrary integer constants > 0.


f[x]+g[x] = h[x]


1 + g[x]/f[x] = h[x]/f[x]


{f[x]g'[x] - g[x]f '[x] }/{f[x]}^2 =

{f[x]h'[x] - h[x]f '[x] }/{f[x]}^2


divide out the {f [x]}^2

rearrange terms

f[x]{g'[x] - h'[x]} = f '[x]{g[x] - h[x]}

f[x]/f '[x] = {g[x] - h[x]}/{g'[x] - h'[x]}

f[x]/f '[x] also equals:

{h[x] - g[x]}/{h'[x] - g'[x]}


[1.] x^p + y^p = z^p

differentiate [1.]

p*x^[p-1] * x' + p*y^[p-1] * y' = p*z^[p-1] * z'

divide both sides by p

[2.] x^[p-1] * x' + y^[p-1] * y' = z^[p-1] * z'

[1.] - [2.] & factor

x^[p-1] *[xy'-yx'] = z^[p-1] *[zy'- yz']

[xy' - yx'] = z^[p-1] *[zy'-yz']/x^[p-1]

Since x,y,z are relatively prime, z^[p-1] does not divide x^[p-1]

interesting...

 

[matt grime]

non-formally differentiating functions of integers is it now?

 

[PFanalog57]

Originally posted by matt grime
non-formally differentiating functions of integers is it now?

No, in the last paragraph, x,y,z are polynomial functions of x.

 

[PFanalog57]

Types are entities built up in a finite number of steps, beginning with an arbitrarily chosen object, 0, also identifiable with the number 0, in accordance to certain rules.

[1.]

0 is a type

[2.]

If n is a positive integer and t_1,...,t_n are types then the sequence [t_1,...,t_n] is a type. Types can be assigned to all individuals, sets, and relations, as needed, with regards to the universe under consideration.

[3.]

Let the set of all types be T.

If, for every type t, B_t, contains all relations of A_t, then the higher order structure, M, which is any set, B_t, that is indexed in T( or basically function defined on T ), such, that for any non-empty set of individuals A_t, where B_t is a subset of A_t, with every element of A_t appearing as a function satisfying certain necessary and sufficient conditions, then the higher order structure M, is full.

If no B_t contains any relation repeatedly, such that the value a function, f, takes on each value at most "once", then the higher order structure is normal. Loosely speaking, a structure is full and normal if B_t = A_t for all t.

At the very rock bottom of the set theory universe, is the singular entity called the "empty set" At first there is nothing at all, with possibly an undefined, unknown, realm in the reverse direction of the empty set, such that the imagination runs amok contemplating the possibilities.

The whole universe of set theory is constructed from the empty set, and since it is the raw idea of set formation, the set theory universe in ever increasing levels of complexity.

The empty set is the simplest set, which is the set that has no elements, represented as { }

The next most elementary set is the set containing the empty set, i.e. { {} }

etc...etc...etc...


A topological group G is a topological space, and also a group, such, that group multiplication ab = c is a continuous function from G x G into G and the inverse operation a^-1 = b is also a continuous function in a topological space. Yes, it is much to the bewilderment and chagrin of certain friends at physics forums. So one of the defining characteristics of a topological space, is, that if ab = c and W is an open neighborhood of c, then there exist open neighborhoods U and V of a and b respectively, such that UV is a subset of W.

 

[PFanalog57]

Pythagorean triples:

http://grail.cba.csuohio.edu/~somos/rtritab.txt

The Fibonacci numbers are also within Pythagorean triples

a^2 + b^2 = c^2 .

15^2 + 8^2 = 17^2

35^2 +12^2 = 37^2

99^2 + 20^2 = 101^2

255^2 + 32^2 = 257^2

675^2 + 52^2 = 677^2

1763^2 + 84^2 = 1765^2

4623^2 + 136^2 = 4625^2

etc.

At closer inspection:

(4^2-1)^2 + (2*4)^2 = (4^2+1)^2

(6^2-1)^2 + (2*6)^2 = (6^2+1)^2

(10^2-1)^2 + (2*10)^2 = (10^2+1)^2

(16^2-1)^2 + (2*16)^2 = (16^2+1)^2

(26^2-1)^2 + (2*26)^2 = (26^2+1)^2

(42^2-1)^2 + (2*42)^2 = (42^2+1)^2

(68^2-1)^2 + (2*68)^2 = (68^2+1)^2

etc...

Notice:

4 = 2*2

6 = 2*3

10 = 2*5

16 = 2*8

26 = 2*13

42 = 2*21

68 = 2*34

2*Fibonacci

etc...

Interesting...




While researching "squares" of numbers I find that ratios of certain types of squares always involve something like a + (2b)^(1/2).

I found many squares that have successive ratios that tend towards the number
3 + (8)^(1/2)

This is of course one solution to the polynomial x^2 - 6x + 1 = 0.

The challenge is to answer the question "why"
are these squares ratios tending towards
3 + (8)^(1/2) as the numbers are getting larger?

I will give some "lists" of these squares!

3^2 + 4^2 = 5^2

20^2 + 21^2 = 29^2

119^2 + 120^2 = 169^2

696^2 + 697^2 = 985^2

4059^2 + 4060^2 = 5741^2

23660^2 + 23661^2 = 33461^2

137903^2 + 137904^2 = 195025^2

Now the ratios

29/5 = 5.8

169/29 = 5.827586207...

985/169 = 5.828402367...

5741/985 = 5.828426396...

33461/5741 = 5.828427103...

195025/33461 = 5.828427124...

3 + (8)^(1/2) = 5.828427125...

This number also appears for the successive ratios of any of the three numbers
(a, b, c), a2/a1, b2/b1, or c2/c1 of the Pythagorean triples on the above list.

Here are some more...

3 = 2^2 - 1

17 = 4^2 + 1

99 = 10^2 - 1

577 = 24^2 + 1

3363 = 58^2 - 1

19601 = 140^2 + 1

114243 = 338^2 - 1

665857 = 816^2 + 1

The ratios

17/3 = 5.666666667...

99/17 = 5.823529412...

577/99 = 5.828282828...

3363/577 = 5.828422877...

19601/3363 = 5.828427...

114243/19601 = 5.828427121...

665857/114243 = 5.828427125...

Also approaching 3 + (8)^(1/2)




The mathematicians tell me "no-problemo" an easy problem!

This is what one mathematician explained to me...

Hi, this is easy to explain.

Notice that your equation is basically of the form

a^2 + (a+1)^2 + 1 = b^2

Never mind, for the moment, that you are considering consecutive ratios of b. First start by trying to understand which a's and b's can be solutions to this equation.

Well, this equation is the same as:

2a^2 + 2a + 1 - b^2 = 0.

Which is the same as

4a^2 + 4a + 2 - 2b^2 = 0

which is

(2a+1)^2 -2b^2 = -1

This is a special case of Pell's Equation

X^2 - 2Y^2 = -1 (i.e. where X=2a+1 and Y=b).

Now observe that (X,Y) = (1,1) is a solution.

It turns out that all the solutions
(X_k, Y_k) to this equation are given by

X_k + sqrt(2) Y_k = (1+sqrt(2))^k where k is odd.

Now your question relates to those X_k which are odd. Now compute some examples with k = 1,3,5,7,9,11,13... you will see a pattern. This explains your observations.

1+sqrt2)^2 = (3 + sqrt8)

(1+sqrt2)^3 = (7 + sqrt50)

(1+sqrt2)^4 = (17 + sqrt288)

(1+sqrt2)^5 = (41 + sqrt1682)

(1+sqrt2)^6 = (99 + sqrt9800)

(1+sqrt2)^7 = (239 + sqrt57122)

(1+sqrt2)^8 = (577 + sqrt332928)


The Fibonacci sequence:

1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144,
233...

1+0 = 1

1+1 = 2

1+2 = 3

2+3 = 5

3+5 = 8

5+8 = 13

8+13 = 21

etc...

Fibonacci squares... also generate Fibonacci numbers...

1^2 - 0^2 = 1

1^2 + 1^2 = 2

2^2 - 1^2 = 3

2^2 + 1^2 = 5

3^2 - 1^2 = 8

3^2 + 2^2 = 13

5^2 - 2^2 = 21

5^2 + 3^2 = 34

8^2 - 3^2 = 55

8^2 + 5^2 = 89

13^2 - 5^2 = 144

13^2 + 8^2 = 233

21^2 -8^2 = 377

21^2 + 13^2 = 610

etc...

The consecutive ratios of these numbers... are of course approaching (1 + sqrt(5))/2

The "golden" ratio...

Sum the digits of 2^n and get a repeating pattern:

2^1 = 2

2^2 = 4

2^3 = 8

2^4 = 16, 1+6 = 7

2^5 = 32, 3+2 = 5

2^6 = 64, 6+4 = 10, 1+0 = 1

2^7 ---> 2

2^8 ---> 4

2^9 ---> 8

2^10 ---> 7

2^11 ---> 5

2^12 ---> 1

2^13 ---> 2

2^14 ---> 4

etc...

etc...

etc...