If $p^2$ is even then $p$ is even. Let $p^2=6$. Is $\sqrt{6}$ even?
I believe $\sqrt{6}$ is not even an integer. Could you help me find the flaw in this statement. Thanks.
 A: "Even" and "odd" are properties of integers, not of arbitrary real numbers.  The correct statement is "if $p$ is an integer and $p^2$ is even, then $p$ is even."  
Someone might omit the "if $p$ is an integer" part if they think it is obvious from the context.  But apparently it was not obvious to you.
A: As Robert noted, this is only true if $p$ is an integer.
Here is a proof:
Let $p$ be an integer and $p^2$ be even, so $p^2 = 2a$ for some integer $a$. Consider $p^2 + p = p(p+1)$. We know that $p(p+1)$ must be even because regardless of if $p$ is even, we will have an $\text{even} \cdot \text{odd}$, which must be even. Thus $p^2 + p = 2b$ for some integer $b$. Now note that $p = (p^2 + p) - p^2 = 2b - 2a = 2(b-a)$, where $b, a$ are integers and $b>a$. So $p$ must be even.
A: For integral $p$ we can also obtain this result by noting that the square of an odd integer must be odd:  if
$p = 2k + 1, \tag 1$
then
$p^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1, \tag 2$
also odd.  Therefore if $p^2$ is even, $p$ cannot be odd, hence is also even.
