Multiple Choice Question (MCQ)

If m = 2006^2+2006^2+2007^2+2007^2, then m

  1. is a perfect square and odd number

  2. ×

    is a perfect square and even number

  3. ×

    is a non-perfect square but an odd number

  4. ×

    is a non-perfect square but an even number

Collected in the board: Number Theory

Steven Zheng posted 38 minutes ago

Answer

  1. Let n = 2006

    then

    n^2+n^2(n+1)^2+(n+1)^2

    =n^4+2n^3+3n^2+2n+1

    =(n^4+n^3+n^2)+(n^3+n^2+n)+(n^2+n+1)

    =(n^2+n+1)^2

    Therefore

    (n^2+n+1)^2 is a perfect square number

    On the other hand,

    n^2+n+1

    =n(n+1)+1

    n(n+1) is the product of two consecutive number which is even.

    then

    n(n+1)+1 is odd.

    Square of odd number results in odd number.

    Therefore

    m = 2006^2+2006^2+2007^2+2007^2 is a perfect square and even number.

    A is the correct choice

Steven Zheng posted 29 minutes ago

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