## Monday, October 10, 2005

### Basic Lemmas Needed for FLT: n = 5

Here are some basic lemmas that are used by the proof for Fermat's Last Theorem: n=5.

Lemma 1: (p+q)5 + (p-q)5 = 2p(p4 + 10p2q2 + 5q4)

(1) Using the Binomial Theorem:
(p + q)5 = p5 + 5p4q + 10p3q2 + 10p2q3 + 5pq4 + q5
(p - q)5 = p5 - 5p4q + 10p3q2 - 10p2q3 + 5pq4 - q5

(2) Adding these two values together gives us:
2p5 + 20p3q2 + 10pq4 = 2p(p4 + 10p2q2 + 5q4)

QED

Lemma 2:

If:

t = q4 + 50q2r2 + 125r4
u = q2 + 25r2
v = 10r2

Then:

t = u2 - 5v2

(1) u2 = (q2 + 25r2)2 = q4 + 50q2 r2 + 625r4

(2) -5v2 = -5(10r2)2 = -500r4

(3) (q2 + 25r2)2 + -5(10r2)2 = q4 + 50q2 r2 + 625r4 + -500r4 = q4 + 50q2 r2 + 125r4

QED

## Sunday, October 09, 2005

### Some Simple Division Lemmas

In this blog, I want to outline some simple implications of the division which are very useful in basic mathematical reasoning:

Lemma 1:

If:

(a) c = a + b
(b) d divides a
(c) d divides b

Then:

d divides c

(1) By assumption (b) and (c), we know that there exists a',b' such that:

a = a'd
b = b'd

(2) So,

c = (a'd) + (b'd) = d(a' + b')

Lemma 2:

If

(a) c = a + b
(b) d divides c
(c) d divides a

Then

d divides b

(1) c = a + b → b = c -a = c + (-a)

(2) This results follows from Lemma 1 above.