# Can anyone explain me this?

## Homework Statement

Let m be the number of numbers fromantic the set {1,2,3,...,2014} which can be expressed as difference of squares of two non negative integers. The sum of the digits of m is ....

## The Attempt at a Solution

I got a solution from a magazine but I didn't under stand how it came
Can anyone explain me how it came.
2n+1=(n+1)^2 -n^2
n^3=[n (n+1)/2]^2 - [n (n-1)/2]^2
Therefore m contains all odd numbers and the even numbers 2^3,4^3,8^3,10^3,12^3.
Therefore m=1007+7=1013 with digit sum 5.

RUber
Homework Helper
Containing all odd numbers is a consequence of the first statement, 2n+1=(n+1)^2 -n^2. This clearly shows that all odd numbers can be created as a difference of squares.
For even numbers, a different identity is used, n^3=[n (n+1)/2]^2 - [n (n-1)/2]^2. This is not necessarily even, but is definitely a difference of square integers.
So it can be concluded that any cubed integer is also in the set.
Looking at the bounds [1,2014], the largest even cube is 12^3 = 1728.
Odd numbers are 2n+1 for n = 0 to 1006, or 1007 in the set.
Even numbers are the cubes of 2, 4, 6, 8, 10, and 12, or 6 in the set.
This gives 1007+6 = 1013.

The truth of the identities used can be shown through simple expansion.

My question is how can one be sure that other even numbers should be excluded from the set?

Oops... think about RUber's question first :-)
##(n+1)^2-n^2=2n+1##
##(n+2)^2-n^2=4n+4=4(n+1)##
Therefore any odd number and any multiple of 4 can be expressed as the difference of two squares.

So m is wrong, and so is the digit sum.

haruspex