**Parity**

http://mathworld.wolfram.com/Parity.html

Let be a positive integer and be any integer with the same parity as

with

These odd integers are all positive.

for example,

If n = 2, m = 4

8 = 3 + 5

the product is equal to the sum of 2 (n = 2) consecutive odd integers

If n = 3, m = 5

15 = 3 + 5 + 7

the product is equal to the sum of 3 (n = 3) consecutive odd integers

If n = 5, m = 7

35 = 3 + 5 + 7 + 9 + 11

the product is equal to the sum of 5 (n = 5) consecutive odd integers

Can you prove the product will always equal to the sum of consecutive odd integers?

——————————————

Proof –

The sum of the first odd integers is :

From this, we have that the sum of any sequence of consecutive odd integers is :

So an integer can be written as the sum of consecutive odd integers if it can be expressed as

where

and

or

or

is odd, is odd :

If

The first odd integer in the sequence is: 2(1) + 1 = 3

The last odd integer in the sequence is 2(4) – 1 = 7

We have the solution

is even, is even :

E.g.

a = 12, b = 6

n = (a+b)/2 n = (12+6)/2 = 9

m = (a-b)/2 m = (12-6)/2 = 3

1st odd integer is : 2(3+1) – 1 = 7

last odd integer is : 2(9) – 1 = 17

We have the solution 7 + 9 + 11 + 13 + 15 + 17 = 72