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Math Help - algebra squares problem

  1. #1
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    algebra squares problem

    (i) If n is a positive integer, such that 2n+1 is a perfect square, show that n+1 is the sum of two successive perfect squares.

    (ii) If 3n+1 is a perfect square, show that n+1 is the sum of three perfect square.
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  2. #2
    Senior Member vincisonfire's Avatar
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    (i)A square of an odd number
     (2k+1)^2=4k^2+4k+1
    If  n = 2k^2+2k then  (2k+1)^2=2n+1 and that shows that  2n+1 is the square of an odd number.
    Now consider  n+1 = 2k^2 + 2k + 1 = k^2 + (k^2 + 2k + 1) =  k^2 + (k+1)^2 that is the sum of two successive squares.
    (ii) A number can be of the form  3q+r, r\in \{0,1,2\}
    Square can thus be of the form :
     (3q)^2 = 9 q^2
     (3q+1)^2 = 9 q^2 +6q +1 = 3(3q^2+2q)+1
     (3q+2)^2 = 9 q^2 +12q +4 = 3(3q^2+4q+1)+1
    So if  3n +1 a square then  n =3q^2+2q \text{ or } n=3q^2+4q+1
    Then n +1 is form (when 3n+1 is odd)  3q^2 + 4q + 2 = q^2 + 2q+1 +q^2 +2q+1 + q^2 = 2\cdot (q+1)^2 + q^2 or is form (when 3n+1 is even)  3q^2 + 2q + 1 = q^2 + 2q+1 +q^2+ q^2=(q+1)^2 + 2 \cdot q^2
    Last edited by vincisonfire; November 27th 2008 at 07:37 AM.
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