Brent´s solution is perfect and full of important details.

I would like to add some comments, related to dangerous situations usually explored in traps for the uncautious students...

\[{x^2}\,\,{\text{even}}\,\,\,\,{\text{does}}\,\,{\text{NOT}}\,\,{\text{imply}}\,\,\,\,\,x\,\,{\text{even}}\,\,\,\,\,\,\,{\text{ }}\left( {x = \sqrt 2 \,\,\,{\text{for}}\,\,{\text{example}}} \right)\]

\[{y^2}\,\,{\text{odd}}\,\,\,\,{\text{does}}\,\,{\text{NOT}}\,\,{\text{imply}}\,\,\,\,\,y\,\,{\text{odd}}\,\,\,\,\,\,\,{\text{ }}\left( {y = \sqrt 3 \,\,\,{\text{for}}\,\,{\text{example}}} \right)\]

On the other hand, as it is the case in the problem proposed,

\[x\,\,\,\operatorname{int} \,,\,\,\,{x^2}\,\,{\text{even}}\,\,\,\,{\text{imply}}\,\,\,\,\,\,x\,\,{\text{even}}\,\,\,\,\,\,\,{\text{ }}\left( {x\,\,{\text{is}}\,\,{\text{odd}}\,\,{\text{or}}\,\,{\text{even,}}\,\,{\text{but}}\,\,\,{\text{it}}\,\,{\text{is}}\,\,{\text{not}}\,\,{\text{odd}}...} \right)\]

\[y\,\,\,\operatorname{int} \,,\,\,\,{y^2}\,\,{\text{odd}}\,\,\,\,{\text{imply}}\,\,\,\,\,\,y\,\,{\text{odd}}\,\,\,\,\,\,\,{\text{ }}\left( {y\,\,{\text{is}}\,\,{\text{odd}}\,\,{\text{or}}\,\,{\text{even,}}\,\,{\text{but}}\,\,\,{\text{it}}\,\,{\text{is}}\,\,{\text{not}}\,\,{\text{even}}...} \right)\]


This kind of observation follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

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Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.

(n+1)^2 is an even integer
=> n+1 is an even integer
=> n is an odd integer

Condition 1)
Since "n-1 is an even integer" is equivalent to "n is an odd integer", condition 1) is sufficient.

Condition 2)
"(n-1)^2 is an even integer" is equivalent to "n-1 is an even integer", which is condition 1).
Condition 2) is also sufficient.

Therefore, D is the answer.

Answer: D

Note: Tip 1) of the VA method states that D is most likely to be the answer if condition 1) gives the same information as condition 2).

Theorem: If n is an odd integer, then n2 is an odd integer. Proof: Since n is an odd integer, there exists an integer k such that n=2k+1. Therefore, n2 = (2k+1)2 = 4k2+4k+1 = 2(2k2+2k)+1

mobdro