Is the square for a number $n\in \mathbb{Z}$ even, then is also $n$ even - prove this via contraposition? 
Is the square for a number $n \in Z$ even, then is also $n$ even -
  prove this via contraposition?

Ok my prove is like that:
if $ a \implies b$ then (contraposition) $\mathrm{not}(b) \implies \mathrm{not}(a)$.
for all $k$: 
$$n=2 \times k + 1 $$
$$n^2=(2 \times k + 1)^2 = 4k^2 + 4k + 1 = 2k \times (2k +2) + 1 $$
when I put in:
$$k=2 $$
then I get $$25$$ for the equation
however I follow now that this has to be incorrect that if the square for a number $n \in Z$ even, then is also n even, because of my prove.
Do I have a small chance to be right, against a hundret year old law?
 A: By letting $k = 2$, you tested $n^2 = (2 \times k + 1) = 5^2 = 25$, so for $k = 2$, we have $n = 5$ is odd. It is the odd $n$, and its associated square in which we are interested.
Filling in the details of your argument:
(We use the following:


*

*For every integer $n$, either $n$ is even or $n$ is odd but not both.

*$n$ is an even positive integer means that $2$ divides $n$ (i.e., $n$ even means there exists an integer $k$ such that $n = 2k$, or equivalently, that n is a multiple of 2.)

*$n$ is an odd integer if and only if  there exists an integer $k$ such that $n = 2k + 1$ ($n$ is odd if and only if $2$ does NOT divide $n$).)


to prove the following:

$(1)$ If the square of an integer $n$ is an even, then $n$ is even.

$\quad\quad$ by proving the contrapositive of $(1)$ which is:

$(2)$ For $n\in \mathbb{Z}$, if $n$ is not an even integer, then $n^2$ is not even.

$\quad\quad$Using the first bullet above, $(2)$ is equivalent to $(3)$:

$(3)$ If $n$ is an odd integer, then $n^2$ is an odd integer.


Assume $n$ is odd: 
Then for all odd $n$, there is an integer $k$ such that,
$$n=2 \cdot k + 1 $$
(it doesn't matter if $k$ is odd or even. We need only know that $n$ is odd.)
Then, 
$$n^2=(2k + 1)^2 = 4k^2 + 4k + 1 = 4(k^2 + k) + 1$$
Since $2$ divides $4$, but does not divide $1$, $n^2$ must be odd.
Or, as you have written: $2$ divides $2k(2k + 2)$ but does not divide $1$,
Hence, for integer $n$, $n$ is not an even integer $\implies$ $n^2$ not even...
So having proven its contrapositive, we conclude that if $n^2$ is even, then $n$ must be even.
A: First, we have to prove that the contrapositive is true, and then claim that our assertion is true since the contrapositive is always true for a given statement.
The original statement is:

(For $n \in \mathbb{Z})$  If $n^2$ is even, then $n$ is even.

Now, let's hunt for the contrapositive, viz.

(For $n \in \mathbb{Z}) $  If $n$ is not even, then $n^2$ is not even.

Recall that the above statement is the same as:

(For $n \in \mathbb{Z}$) If $n$ is odd, then $n^2$ is odd.

We have a direct proof for that:

Knowing that an odd number is in some form $2k + 1, k \in \mathbb{Z}$, let us represent $n = 2k + 1$. Thus, $n^2 = (2k + 1)^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1$.
Knowing that a number in the form $2x + 1, x \in \mathbb{Z}$ is an odd number, we may say that $2(2k^2 + 2k) + 1$ is odd (a special case of $x = 2k^2 + 2k$).
$\boxed{}$

We have proved that the contrapositive is true, and as told, we may claim that the assertion is true because given $\neg q \Rightarrow \neg p$, $\ \ p \Rightarrow q$ is always true.
A: The contrapositive of what you would like to show is the square of an odd numbers is again an odd number, that is, can be written as $2x+1$ for some integer $x$.
Suppose $m$ and $n$ are odd integers. Now, since $m$ and $n$ are odd, we know $m=2a+1$ and $n=2b+1$. Thus, $$ mn=(2a+1)(2b+1)=4ab+2a+2b+1=2(2ab+a+b)+1.$$ Hence, the product of odd numbers is odd. As a special case, we see that any odd number squared will again be odd. Hence, if a number squared is even, the original number must have been even.
A: What you need are the following theorems:


*

*For every positive integer n, either n is even or n is odd but not both.

*n is an even positive integer if an only if there is a positive integer m such that n = 2m.

*n is an odd positive integer if and only if either n = 1 or there is a positive integer m such that n = 2m+1.
I find it an interesting exercise in using the Peano axioms for the integers to define even and odd and then prove these theorems.
Once you have these theorems, many problems involving evenness and oddness become straightforward.
The next step is to go from even and not even to multiple of k and not a multiple of k for k an integer greater than 2.
