Bunuel wrote:
Official Solution:
(1) \(n^2\) is an integer. Now, \(n^2\) to be an integer \(n\) must be either an integer or an irrational number (for example: \(\sqrt{3}\)), (note that \(n\) cannot be a reduced fraction, for example \(\frac{2}{3}\) or \(\frac{11}{3}\) as in this case \(n^2\) won't be an integer). But as \(n\) can be expressed as the ratio of 2 integers, \(n=\frac{p}{q}\), then it cannot be an irrational number (definition of irrational number: an irrational number is any real number which cannot be expressed as a fraction \(\frac{a}{b}\), where \(a\) and \(b\) are integers), so only one option is left: \(n\) is an integer. Sufficient.
(2) \(\frac{2n+4}{2}\) is an integer. Given: \(\frac{2n+4}{2}=n+2=\text{integer}\). So, \(n=\text{integer}\). Sufficient.
Answer: D
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