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

Question

Uniteasy Admin · Accepted Answer

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

Multiple Choice Question (MCQ)

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