rxs0005
If sqrt( 4a ) = integer is sqrt ( a ) an integer
(1) a is a positive integer
(2) a = n^6 where n is an integer
Given: \(\sqrt{4a}=integer\) --> \(2\sqrt{a}=integer\): so either \(\sqrt{a}=integer\) (note that in this case
\(a\) will be an integer, more it'll be a perfect square: 1, 4, 9, ...) or \(\sqrt{a}=\frac{odd \ integer}{2}\) (1/2, 3/2, 5/2, ... note that in this case
\(a\) won't be an integer it'll equal to: 1/4, 9/4, 25/4, ...). Question asks whether we have the first case: is \(\sqrt{a}=integer\) (or as we can have only two cases question basically asks whether \(a\) is an integer)?
(1) a is a positive integer --> we have the first case. Sufficient.
Or: square root of an integer (\(\sqrt{a}\)) is either an integer or an irrational number, so we can not have the second case (if \(a=integer\) then \(\sqrt{a}\) can not equal to \(\frac{odd \ integer}{2}\)). Sufficient.
(2) a = n^6 where n is an integer --> \(\sqrt{a}=n^3\) and as \(n=integer\) then \(\sqrt{a}=n^3=integer\). Sufficient.
Answer: D.