Understanding the Solution for Finding the Sum of Digits of m

[Murtuza Tipu]

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 ...

Homework Equations

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.
Answer is as follows:
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]

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?

 

[Joffan]

Oops... think about RUber's question first :-)

Spoiler

$(n+1)^2-n^2=2n+1$
$(n+2)^2-n^2=4n+4=4(n+1)$
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]

Hint: What are the quadratic residues modulo 8?