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Can anyone explain me this?

  1. Jan 5, 2015 #1
    1. The problem statement, all variables and given/known data
    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 ....

    2. Relevant equations


    3. 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.
     
  2. jcsd
  3. Jan 5, 2015 #2

    RUber

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    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?
     
  4. Jan 5, 2015 #3
    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.
     
  5. Jan 5, 2015 #4

    haruspex

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    Hint: What are the quadratic residues modulo 8?
     
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