If a is a positive integer, is a^2 a multiple of 8 ?

Is \(a^2 = 8k\)? \(k\) is an constant
This can be rephrased as, Is \(a^2 = 2^3k\)?
As it is given that \(a \) is an integer. Thus, constant \(k \) will have one \(2 \) in it. Thus, \(k = 2m\) (\(m \) another constant).
Thus, the question becomes, Is \(a^2 = 2^4m\) ?
Take square root of the above equation: so the question becomes, Is \(a = 2^2 m\) => Is \(a = 4m\) ? Or is \(a \) an multiple of \(4\)?

1) \(a^3 = 16c.\)
=> \(a^3 = 2^4c\) (\(c \) is any constant).
Again, since \(a \) is an integer. \(c \) should have \(2^2\) for \(a \) to be an integer. Thus, \(a^3 = 2^6c\) =>Take cube root on both sides, \(a = 2^2n\) (\(n \) is a constant) => \(a = 4n\). So, yes, \(a \) is a multiple of \(4\). Sufficient.

2) \((a + 4)^2\) is a multiple of \(8\).
\(a^2 + 16 + 8a = 8p\\
\)
\(a^2 = 8p - 16 - 8a\\
\)
\(a^2 = 8 (p - 16 - a). \\
\)

Thus, \(a^2\) is a multiple of \(8\). Sufficient. Option (D).