An even number can be expressed as the square of an integer as well as a cube of another integer. Then the number has to be necessarily divisible by:

To determine the properties of an even number that can be expressed as both the square of an integer and the cube of another integer, let's break down the problem.

1. Break Down the Problem

• Let the even number be $N$.
• Since $N$ is even, we can write it as $N = 2k$ for some integer $k$.
• We know $N = m^2$ (square of an integer) and $N = n^3$ (cube of another integer) for integers $m$ and $n$.

2. Relevant Concepts

• For $N$ to be both a perfect square and a perfect cube, it must be a perfect sixth power: $N = p^6$ where $p$ is an integer. This is because the least common multiple of 2 (for square) and 3 (for cube) is 6.

3. Analysis and Detail

• Since $N$ is an even number, $p$ must contain at least one factor of 2 to ensure $N$ is even.
• Thus, we can express $p$ as $p = 2^a \cdot q$, where $a \geq 1$ and $q$ is an odd integer.

Now we compute $N$: $N = (2^a \cdot q)^6 = 2^{6a} \cdot q^6$ To ensure $N$ is even, and since $a \geq 1$, we can simplify this to: $N = 2^{6} \cdot (2^{a-1} \cdot q)^6$ Which indicates that $N$ is at least divisible by $2^6$.

4. Verify and Summarize

• Thus, if $N$ can be expressed as both a square and a cube and must be even, it must be divisible by $2^6$.
• The smallest even number that can satisfy these conditions is $64$, since $64 = 8^2 = 4^3$.

Final Answer

The number has to be necessarily divisible by $64$.