1
2 3
4 5 6
7 8 9 10
11 12 13 14 15
16 17 18 19 20 21
22 23 24 25 26 27 28
etc...
Show that adding all the numbers that are not crossed off from 1 to the end of a line, the result is a 4^{th} power.
Example: 1+4+5+6=16=2^{4}
The sum of n consecutive odd numbers is = n^2 (beginning with 1).
So the sum of a square number of consecutive odd numbers is a power of 4. (beginning with 1). [If n=a^2: n^2=a^4).
I can dispose a square number of terms in rows with increasingly number of odd terms (1, 3, 5, 7... terms). At the end of each row the cumulative number of terms will be an square, as:
1
3, 5, 7
9, 11, 13, 15, 17
19, 21, 23, 25, 27, 29, 31
And the cumulative sum at the end of each row will be a power of 4.
It is now possible to change the terms of each row somehow but leaving unchanged the sum in each row. Always the global sum will be a 4th power.
Note now the equivalence between my above rows and those in the puzzle.
Edited on June 28, 2016, 6:40 pm

Posted by armando
on 20160628 17:43:52 