Let \(N=2^k\times4^m\times8^n\)where \(k,m,n\) are positive whole numbers. Then \(N\) will definitely be a square number whenever

\(k\) is even;

\(k+n\) is odd;

\(k\) is odd but \(m+n\) is even;

\(k+n\) is even.

We’ll start by rewriting all the terms as powers of \(2\).

By doing this, we will get \[N=2^k\times2^{2m}\times 2^{3n}\Longrightarrow N=2^{k+2m+3n}.\]

Since square numbers are all of the form \(x^{y}\), where \(y\) is an even number, \(N\) will be a square number when \(k+2m+3n\) is even.

That is, as \(2m+2n\) is always even, when \(k+n\) is even so the answer is (d).

Alternatively, we could have noticed that \(4^m\) is already a square, as \(4^m=\left(2^m\right)^2\), and rewrite \(8^n\) as \(4^n\times2^n\).

Since \(4^n\) is also a square, for \(N\) to be a square we will need only \(2^k\times 2^n=2^{k+n}\) to be a square, and, therefore, \(k+n\) to be even.