## Thursday, November 4, 2010

### An Interesting Property of the Odd Numbers

The first odd numbers ...
1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, ...
Their squares ....
1, 9, 25, 49, 81, 121, 169, 225, 289, 361, 441, 529, 625, 729, 841, 961, 1089, ...

Notice that
1^2 = 1 = 8*0 + 1
3^2 = 9 = 8*1 + 1
5^2 = 25 = 8*3 + 1
7^2 = 49 = 8*6 + 1
9^2 = 81 = 8*10 + 1
11^2 = 121 = 8*15 + 1
13^2 = 169 = 8*21 + 1
15^2 = 225 = 8*28 + 1
17^2 = 289 = 8*36 + 1
19^2 = 361 = 8*45 + 1
21^2 = 441 = 8*55 + 1
23^2 = 529 = 8*66 + 1
25^2 = 625 = 8*78 + 1
27^2 = 729 = 8*91 + 1
29^2 = 841 = 8*105 + 1
31^2 = 961 = 8*120 + 1
33^2 = 1089 = 8*136 + 1

So, if we take any odd number and square it, it will be a multiple of 8 plus 1.

x is an even number, then (x + 1) is odd
n = integer

(x + 1)^2 = x^2 + 2x + 1
So, x^2 + 2x + 1 = 8n + 1
Or, x^2 + 2x = 8n
x is an even, so x = 2p, where p is an integer.

(2p)^2 + 2*(2p) = 8n
4p^2 + 4p = 8n
4*(p^2 + p) = 8n
4*[p*(p + 1)] = 8n
Or,
8*[p*(p + 1)/2] = 8n
[p*(p + 1)/2] = n

p, (p + 1) are 2 consecutive numbers, so one of them is even.
Hence, their product is even.
And p*(p + 1)/2 is an integer which is equal to n.

p*(p + 1)/2 is that the sum of consecutive integers (non-negative) from 1 to p.
(0 + 1 + 2 + 3 + ... + p)
So the pattern for n is (0, 1, 3, 6, 10, 15, ...)

The pattern for n is the triangular number series.

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What if we use higher powers, can we still find interesting properties?

The first odd numbers ...
1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, ...

Their cubes ...
1, 27, 125, 343, 729, 1331, 2197, 3375, 4913, 6859, 9261, 12167, 15625, 19683, 24389, 29791, 35937, ...

Raised to the 4-th power ....
1, 81, 625, 2401, 6561, 14641, 28561, 50625, 83521, 130321, ...

Raised to the 5-th power ....
1, 243, 3125, 16807, 59049, 161051, 371293, 759375, 1419857, 2476099, ...