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Whole-number right-angled triangles
Consider a right angled triangle with whole-number sides a, b and c where c is the hypotenuse.
Pythagoras’s equation is a2 + b2 = c2.

Deriving the equations for ‘b’ and ‘c’

a2 + b2 = c2

Let the difference in length, between c and another side be d.

For example, d = c - b

So, a2 = (b + d)2 - b2
b = (a2 - d2) / 2d ......................................................... (1)

Similarly, c = (a2 + d2) / 2d (or c = b + d) .................. (2)


Equations (1) and (2) enable us to easily produce a series of instances of the triplet a, b, c for each whole-number value of d, where a2 + b2 = c2 and a, b and c are positive integers.

Here are the first few triplets of each of the first five series.

d = 1
a b c
1 0 1
3 4 5
5 12 13
7 24 25
9 40 41
11 60 61
13 84 85
15 112 113
17 144 145
19 180 181
etc.
d = 2
a b c
2 0 2
4 3 5
6 8 10
8 15 17
10 24 26
12 35 37
14 48 50
16 63 65
18 80 82
20 99 101
etc.
d = 3
a b c
3 0 3
9 12 15
15 36 39
21 72 75
27 120 123
33 180 183
39 252 255
45 336 339
51 432 435
57 540 543
etc.
d = 4
a b c
4 0 4
8 6 10
12 16 20
16 30 34
20 48 52
24 70 74
28 96 100
32 126 130
36 160 164
40 198 202
etc.
d = 5
a b c
5 0 5
15 20 25
25 60 65
35 120 125
45 200 205
55 300 305
65 420 425
75 560 565
85 720 725
95 900 905
etc.

The limiting conditions 1, 0, 1 etc. have been included for completeness.

Classification
The triplets can be classified in an infinite number of series; each series derived from a different value of d.

A beautiful equation
I have since found that the triplets can best be found using the equation (u2 - v2)2 + 4u2v2 = (u2 + v2)2,where u and v are whole numbers. Not my own work, but I feel that I should have been able to devise this equation myself. If you replace u and v with any two whole numbers, the three terms in the equation are a set of whole number triplets.

I call this equation beautiful because, in spite of its simplicity, it can generate all possible triplets.

I found the 'beautiful equation' in a general solution for generating Pythagorean Triplets in Ian Stewart's book 'The Problems of Mathematics'.

Fermat's Last Theorem

The conjecture that became known as Fermat's Last Theorem was first put forward by Pierre de Fermat in 1637. It states that no three positive integers a, b, and c can satisfy the equation an + bn = cn for any integer value of n greater than two.

 Fermat claimed to have proved it but did not leave details of his proof. It was finally proved by Andrew Wiles in 1995 using mathematical techniques not available to Fermat.

 Did Fermat find a proof using the techniques available to him? What kind of approach would he have used?

When n = 0
When n = 0 the equation does not balance as it gives 1 + 1 = 1.

When n = 1
This is the trivial case. Any integer values of  a and b give an integer value for c.

When n = 2
Using the 'beautiful  equation'  I have derived a partial solution. I cannot show why it is true for any integer value of n, greater than two, but I can show why it is true for n = 2.

(u2 - v2)2 + 4u2v2 = (u2 + v2)2,  where u and v are any positive integers.
 
It can also be easily derived from:
(a + b)2 - (a - b)2 = 4ab,  by making a = u2, and b = v2.

Fermat could have known of this relationship. 

What are the features of (a + b)2 - (a - b)2 = 4ab  that enables whole number solutions to be generated?
Obviously, (a + b)2 and (a - b)2 are squares but what about 4ab?

1. ab can be guaranteed to be a square by making a = u2 and b = v2, where u and v are integers.

 2. 4 is a square.
This comes from 2ab + 2ab, because 2 + 2 = 2 x 2.

 Incidentally, 2 is the only number that has this property:
                                                                                     n x n = n + n
                                                                                    n2 = 2n
                                                                                    n = 2

 Also, 4ab is a single term which makes things easy compared with the equations for n = 3  and above which result in multiple terms.

Only the equation for n = 2 generates a single term.

The third power
(a + b)3 - (a - b)3 = 2(3a2b + b3) = 2b(3a2 + b2)

2b(3a2 + b2) is the critical term. Can this ever be a cube, for integer values of a and b?

The argument might go something like this:
If b(3a2 + b2) were to be an integer, the cube root of 2b(3a2 + b2) would be a multiple of 21/3 and therefore not an integer.

Or this:
In order for the cube root of 2b(3a2 + b2) to be an integer, b(3a2 + b2) would have to be a multiple of 22/3. This is not possible if a and b are integers.


The fourth power
(a + b)4 - (a - b)4 = 2(4a3b + 4ab3) = 8ab(a2 + b2)

8ab(a2 + b2) is the critical term. Can this ever be a fourth power, for integer values of a and b?

If ab(a2 + b2) were to be an integer, the fourth root of 8ab(a2 + b2) would be a multiple of 81/4 and therefore not an integer.
In order for the fourth root of 8ab(a2 + b2) to be an integer, ab(a2 + b2) would have to be a multiple of 83/4. This is not possible if a and b are integers.

The  nth power
For even values of n:
(a + b)n - (a - b)n = 2(nan-1b/1 + n(n - 1)(n - 2)an-3b3/1.2.3 ... + n(n - 1)(n - 2)an-3b3/1.2.3 + nan-1b/1)

For odd values of n:
(a + b)n - (a - b)n = 2(nan-1b/1 + n(n - 1)(n - 2)an-3b3/1.2.3 ... + n(n - 1)a2bn-2/1.2 + bn)


It may be possible to use the argument employed for n = 3 and n = 4, for all higher values of n.

With the mathematics of his time Fermat could easily have followed this line of reasoning and assumed the result for n = 3 and above.

Mike Holden - Sep 2005
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