This document proves that if n is an odd integer, then n^3 - n is divisible by 8. It first proves in a lemma that the product of two consecutive even integers is divisible by 8. It then shows that if n is odd, n-1 and n+1 are consecutive even integers. Their product is divisible by 8, and since (n-1)(n+1)n = n^3 - n, it follows that n^3 - n is also divisible by 8. However, it notes that this is not true for all integers n, providing a counterexample of n = 6 where 6^3 - 6 is not divisible by 8.

1. PROVE THAT IF N IS ODD,
THEN N3 – N IS DIVISIBLE BY 8
N ∈ ℤ
BY MATTHEW SARAGUSA

2. METHOD OF PROOF
• I will be using the fact that the product of two consecutive even integers is divisible by 8.
• I will also have to prove that the product of two consecutive integers is divisible by 8 in a lemma.
• I will also prove that n3-n is not divisible by 8 for all n ∈ ℤ by subsitution

3. LEMMA
THE PRODUCT OF TWO CONSECUTIVE EVEN INTEGERS IS DIVISIBLE BY 8
Two consecutive integers can be written as:
x = 2k
y = 2k + 2
for some k ∈ ℤ
Case 1: Assume that k is even, then k = 2j
x = 2(2j) = 4j
y = 2(2j) + 2 = 4j + 2
So xy = 4j (4j + 2) = 16j2 + 8j = 8(2j2 + j) Which is divisible by 8
Case 2: Assume that k is odd, then k = 2j + 1
x = 2(2j + 1) = 4j + 2
y = 2(2j + 1) + 2 = 4j + 4
So xy = (4j + 2)(4j + 4) = 16j2 + 8j + 16j + 8 = 16j2 + 24j + 8 = 8(2j2 + 3j + 1). Which is divisible by 8.
So the product of two consecutive integers is divisible by 8.

4. PROOF
IF N IS ODD, THEN N3 – N IS DIVISIBLE BY 8
We will assume that n is odd.
We have shown that the product of any two consecutive integers is divisible by 8. We must show that n3-n
is the product of two consecutive integers.
n3-n = n(n2-1)
= n(n+1)(n-1)
= (n-1) . n . (n+1)
Now since n is odd, then clearly, (n-1) and (n+1) is even and (n-1) and (n+1) are consecutive even integers.
Then, the product of (n-1) and (n+1) is divisible by 8.
Since (n-1).(n+1) is divisible by 8, then (n-1).(n+1).n is also divisible by 8.
Therefore, if n is odd, then then n3 – n is divisible by 8 □

5. BUT, N3-N IS NOT DIVISIBLE BY 8 FOR ALL N ∈ ℤ
Let n = 6
Then n3 – n = 210 which is not divisible by 8.
Therefore we have found an n ∈ ℤ where n3 – n is not divisible by 8.