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...