Prove if 2 divides $a^2$, then 2 divides $a$. If 2 divides $a^2$, then 2 divides a.
I know that 2 divides $a^2$ means there is some integer $n$ such that $a^2 = 2n$, 
and similarly, 2 divides $a$ means there is some integer $m$ such that $a = 2m$
I thought I could rewrite $a^2 = 2n$ into this $=  a = 2(n/a)$ but I don't think that helps, because I'm not sure $n/a$ is an integer.  
Thank you for any help!
 A: $$RTP: 2|a^2\implies 2|a$$
Or equivalently using the fact that $A\implies B$ is equivalent to $B^c\implies A^c:$
$$RTP:2\not| a\implies 2\not| a^2$$
Suppose $2\not|a$. Then we can write $a=2k+1$ for some integer $k$.
$\implies a^2=(2k+1)^2 =4k^2+4k+1=2(2k^2+2k)+1\equiv 1\bmod 2\implies2\not|a^2\quad\text{as required}$
A: Your proof doesn't work, because as you said you have no justification for $a$ dividing $n$. For an alternative idea, see below. As an exercise:
(i) Prove that $n(n+1)$ must always be even.
(ii) Prove that the difference of two even numbers is also even.

Note that if $n^2$ is even then $n(n+1) - n^2 = n$ is the difference of two even numbers, so it must be even itself. 
A: By division algorithm, $a=2q+r$ where $r=0\ \text{or}\ 1$ and $q\in\mathbb{Z}$. Now $a^2=4q^2+4qr+r^2$. Since $2|a^2$ it follows that $2|r^2$, whence $r=0$. 
OR use the fact that if $p$ is a prime such that $p$ divides $ab$ then $p$ divides $a \ \text{or}\ b$.
A: $2|a(a-1)  \Rightarrow 2|(a^2-a) \Rightarrow 2|a$. But be cautioned this proof doesnt generalize as well as the answers suggested by others, for instance by @Kaj Hansen
A: Okay, because I think people should think for themselves and work things out, I will continue your process:
$a = 2\frac an$.  If $\frac an$ is an integer we are done.
If $\frac an$ is not an integer, the $2\frac an$ is an integer so $\frac an = k + \frac 12$ for some integer $k$.
If so then $a^2 = (2\frac an)^2 =$
$ [2(k + \frac 12)]^2 =$
$ 4(k^2 + k + \frac 14) = $
$4k^2 + 4k + 1$ which is an odd number.  So $2$ does not divide $a^2$ so that is a contradiction.
So $\frac an$ is an integer and we are done.
So if $2$ divides $a^2$ then $2$ divides $a$.
That is not the most straightforward way to do it, but it does work.
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The most straightforward way to do it in my opinion is:
Let $a = \prod p_i^{b_i}$ be the prime factorization of $a$.
Then $a^2 = \prod p_i^{2b_i}$.
If a prime $q$ divides $a^2$ then $q$ must be one of the $p_i$.  And if $q$ is one of the $p_i$ then $q$ divides $a=\prod p_i^{b_i}$.
$2$ is prime.
A: We can write,
$$a^2=a×a$$$$$$
Now if $a=2^n k$,
$$a^2=2^{2n} k^2$$$$$$
If $a^2$ is divisible by 2,
$$2n\gt 0$$
$$n\gt 0$$$$$$
So , As
$$a=2^nk\text{ and } n\gt 0$$
a is divisible by 2.
A: You can assume that $a$ is non-negative and you can represent it in Base Two. So with $b$ equal to two,
$a = x_sb^s+\dots+x_1b^1+x_0$ where the $x's$ are equal to $1$ or $0$.
So
(1) $a^2 = q b + x_0^2$ for some non-negative $q$.
But since $x_0^2$ can only be 0 or 1, equation (1) is in fact Euclidean Division, which spits out a unique quotient and remainder. But if $a^2$ is divisible by $b$, $\; x_0$ must be 0. But then $a$ is also divisible by two.
A: This problem can be solved by a simple contrapositive proof. 
By the definition of contrapositive
$2|a^2\Rightarrow 2|a\equiv 2\nmid a\Rightarrow 2\nmid a^2$
Thus we assume that $2\nmid a\Rightarrow a=2n+1$ for some $n\in\mathbb{Z}$
$\Rightarrow a^2=4n^2+4n+1\Rightarrow a$ is odd by definition of odd numbers $\Rightarrow 2\nmid a^2$
Therefore the contrapositive is proven to be true.
Thus $(2\nmid a\Rightarrow 2\nmid a^2)\iff (2|a^2\Rightarrow 2|a)$
Since we have proved the contrapositive to be true, the biimplication means that $2|a^2\Rightarrow 2|a$ is also true
A: You can think about it like this:
If $2|a^{2}$ then $a^{2}= 2 \cdots$, for $a=p_{1} \cdot p_{2} \cdots p_{n}$ you will have $a^{2}= 2 \cdots ={(p_{1} \cdot p_{2} \cdots p_{n})}^{2} $, then  if 2 appears in $a^{2}$ it must appear in $a$ due to the unicity of factorization (Fundamental theorem of arithmetic). In other words, taking the decomposition of $a^{2}$, is also taking the decomposition of $a$ and square it, it means that you woudn't find any new prime, you will just have new exponents.
