Since

**a** is even,

**a**'s

prime factorization must contain at least one factor of 2. Suppose

**a** is not a multiple of 4, so

**a** has only one factor of 2. Thus

**a**^2 has exactly two factors of 2. However,
the equality requires

**a**^2 to be expressible as

**b**^3 with

**b** an integer. This would require

**a**^2 to have all entries in the prime decomposition to be to the power of a multiple of three. The number of factors of 2 in

**a**'s prime decomposition is two, which is not a multiple of 3. Thus our supposition that

**a** could fail to be a multiple of 4 is false.

QED.

There is an implied other side to this: showing that there is at least one case where the equality holds. I put forward 8^2 = 4^3 = 64.

But more can be said! In fact, if **a**^2 = **b**^3, then **a** is a cube and **b** is a square (unless they're both 1 or 0 ). But that is for another node...