The following is based on the fact that a perfect square can be generated by adding odd numbers in a series starting from 1. 1+3+5+7+9+11 = 36. Since there are 6 summands the number is 62. Since sigmas are hard to make in html I will use the following notation to indicate the sum of a series:
sumk=1[2k-1]n = n2
Partial sums of arithmetic series adhere to the formula: (n/2)[2a1 + (n-1)d], where
| n = the number of terms |
| a1 = the first term |
| d = the difference between terms |
Example: 9+11+13+15+17+19+21, n=7, a1=9
72+(9-1)(7) = 49+56 = 105
Now we can demonstrate that 105 = (7)(15), there are 7 terms and the middle term is 15
and 105 = 112 - 42 = 121-16 = 105
| 42 | = | 1 | + | 3 | + | 5 | + | 7 | ||||||||||||||
| 112 | = | 1 | + | 3 | + | 5 | + | 7 | + | 9 | + | 11 | + | 13 | + | 15 | + | 17 | + | 19 | + | 21 |
| 112-42 | = | 9 | + | 11 | + | 13 | + | 15 | + | 17 | + | 19 | + | 21 |
105 is also equal to (5)(21) or 132 - 82 = 169 - 64
17+19+21+23+25 = 105
To generate the difference series when the squares are known:
example: 562 - 332, since 2k-1, k=56 will determine the last term, 111
the first term is 2 plus the last term of the small square, k=33, 2k-1 + 2 = 67.
67+69+71+73+75+77+79+81+83+85+87+89+91+93+95+97+99+101+103+105+107+109+111
there are 2 ways to determine n, 56 - 33 = 23 = n or [(111 - 67)/2] + 1 = 23 = n
plugging into our sum formula we get sumk=34[2k-1]56 = 232 + (67 - 1)23 = 529 + 1518 = 2047 = (23)(89)