Z = an even number
(Z + 1) = odd
n = integer
(Z+1)^2 = z^2+2z+1
so we're trying to prove that
z^2 + 2z + 1 = 8n + 1
so
z^2 + 2z = 8n
since z is even, it must be 2w where w is an integer
(2w)^2 + 2(2w) = 8n
4w^2 + 4w = 8n
4*(w^2 + w) = 8n
4*[w*(w+1)] = 8n
8*[w*(w+1)/2] = 8n //notice that I turned the 4 into 8/2
[w*(w+1)]/2 = n
If you multiply two integers together, and at least one of them is even, the product will be even. Therefore, w*(w+1) is even, since one of them has to be even, and one of them has to be odd. So if [w*(w+1)] is even, then [w*(w+1)]/2 is an integer, equal to the n that we started with (8n+1)
It just so happens that [w*(w+1)]/2 = the sum of all the non-negative integers up to and including w (0 + 1 + 2 + 3 + ...... + w). So that is the pattern for n (0, 1, 3, 6, 10, 15).
1^2 = 8(0) + 1
3^2 = 8(1) + 1
5^2 = 8(3) + 1
7^2 = 8(6) + 1
9^2 = 8(10) + 1
11^2 = 8(15) + 1
.
.
. |
