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...