If $x$ squared and $x$ cubed are both even, why is $x$ necessarily even? Can anyone help me with an analytical proof for this? This was a question on a Data Sufficiency question; and while I can empirically see the answer, proving it analytically is beyond my abilities.
 A: If $x^2$ is even and $x$ is a whole number, then $x$ is even. This is easiest to prove by proving that if $x$ is odd, then $x^2$ is odd.
A: I don't know if this is analytical enough for you, but here goes: let's say $x = 2n$. Then $x^2 = 2^2 n^2$ and $x^3 = 2^3 n^3$. But if $x = 2n + 1$, then $x^2 = 4n^2 + 4n + 1$ and $x^3 = 8n^3 + 12n^2 + 6n + 1$.
A: Since $x^2$ and $x^3$ are even by assumption, there are natural numbers $n,m\in \mathbb N$ such that
$$
x^2 = 2n, x^3 = 2m.
$$
Trying to produce $x$ by division gives
$$\begin{split}
x &= \frac{x^3}{x^2} =\frac{2m}{2n} = \frac mn \\
&= \frac{x^4}{x^3} = \frac{4n^2}{2m} = \frac{2n^2}{m},
\end{split}$$
which implies
$$
m^2 = 2n^3.
$$
Hence $m$ must be divisible by $n$, so $x$ is a natural number. Since $x^2$ is even, $x$ cannot be odd.
A: If $x^2\in2\mathbb{Z}$ and $x^3\in2\mathbb{Z}$, then $x=x^3/x^2\in\mathbb{Q}$. 
Since $x^2-2k=0$ for some $k\in\mathbb{Z}$, we have that $x$ is an algebraic integer.
As shown in this answer, the intersection of $\mathbb{Q}$ and the algebraic integers is $\mathbb{Z}$. Thus, $x\in\mathbb{Z}$. Now we can use the fact that $x$ must be even, for if it were odd, $x^2$ would be odd.
A: The question might be better phrased as

If $x^2$ and $x^3$ are both even integers, is $x$ necessarily an even integer?

I don't think the question is a priori assuming that $x$ is even an integer.
I would begin by showing that since $x^2$ and $x^3$ are both integers, $x=x^3/x^2$ must be a rational number.
A slightly challenging (but very standard) thing to prove here is that if the square root of an integer is rational, then it is also an integer.  Hence $x=\pm\sqrt{x^2}$ is an integer (since it is the rational square root of an integer.)
Since $x$ is an integer, it is either even or odd.  If it is odd, then $x^2$ and $x^3$ are both odd.  Thus $x$ must be even, since $x^2$ and $x^3$ are certainly not both odd.
A: First $x=\dfrac{x^3}{x^2}$ is a rational number. Write $x=\dfrac mn$, $\,m\wedge n=1$. Then $x^2=\dfrac{m^2}{n^2}$ is an (even) integer. As $m^2$ and $n^2$  are coprime, this implies $n^2=1$, hence $x$ in an integer. 
Now, if $x$ is an integer and $x^2$ is even, $x$ is even (by contraposition: if $x$ were odd, $x^2$ would be too).
