Formally proving that if $x^2 + 1$ is even, then $x$ is odd. 
Theorem:
If $x^2 + 1$ is even, then $x$ is odd.


I have to mention, that I am a beginner at this. So, sorry if it is very wrong.
Suppose that $x^2+1$ is even, such that there exists an integer $k$ such that
$$
\begin{align}
x^2+1 &= 2k \\
x^2 &= 2k-1
\end{align}
$$
since $k$ is an integer and $L+1$ is also an integer $k := L+1$
$$
\begin{align}
x^2 &= 2(L+1)-1 \\
x^2 &= 2L+2-1 \\
x^2 &= 2L+1 \\
\end{align}
$$
Is this right? Did I miss something? It doesn't look quite right to me… What type of proof is this? Thanks.
 A: I agree with your logic, although there may be clearer ways of expressing it. Moreover, you've only shown that $x^2$ is odd. You still need to show that $x$ is odd as well.
To clean up your argument so far, I would keep 

Suppose that $x^2+1$ is even, such that there exists an integer $k$ such that $$x^2+1=2k$$ or equivalently [my edit] $$x^2=2k-1.$$

Next, I would phrase the introduction of $L$ in a slightly different way:

Since $k-1$ is an integer, let $L=k-1$. Then $L+1=k$, and it follows that
  \begin{align}
x^2&=2(L+1)-1 \\
&= (2L+2)-1 \\
&= 2L + 1,
\end{align}
  implying that $x^2$ is odd.

Finally, you have to show that $x^2$ being odd implies $x$ is also odd. My hint to you is to use the contrapositive statement as opposed to any direct methods. Logically, the statement
$$ \text{If } x^2 \text{ is odd then } x \text{ is odd}$$
is equivalent to the statement
$$ \text{If } x \text{ is not odd then } x^2 \text{ is not odd.}$$
Since an integer that is not odd is even (prove this!), you simply have to show that if $x$ is even, it follows that $x^2$ is even. From here, following your nose and using definitions will finish up the proof.
A: This is a classic exercise in contraposition. 
Suppose that $x$ is not odd (i.e. even). Then, we can write $x = 2k$ for $k \in \mathbb N$. Then:
$x^2 + 1 = (2k)^2 + 1 = 2(2k^2) + 1=$ odd.
Then, if $x^2+1$ is not odd, that is even, then $x$ is odd.
A: I recommend proving the contrapositive. The implication "if $A$ then $B$" is logically equivalent to the so-called contrapositive implication "if not $B$ then not $A$."
The original implication is "if $x^2+1$ is even then $x$ is odd." Do you know what the opposite of even is? Do you know what the opposite of odd is? So then what is the contrapositive implication, and can you figure out how to prove it?
A: What you have shown is that $x^2$ is odd. That is not quite the same thing as showing that $x$ is odd. Is it true that $x^2$ is odd only if $x$ is odd? 
We can shorten your proof to the pair of lines
$$\begin{align}
x^2 + 1 = 2k \implies
x^2 = 2k - 1,
\end{align}$$
so that $x^2$ is odd. You have more to say to prove the desired claim.
A: You asked

What type of proof is this?

This is called a direct proof, and the idea is that you set out from the get-go to prove the desired property. 
There are answers that are suggesting proving the contrapositive of the result (proof by contraposition). This works fine. You can also prove that a number is odd iff its square is odd, and even iff its square is even. However, if you want to do a direct proof, I suggest the following.
Let $x = 2j + r$, where $r \in \{ 0, 1\}$. You can write $x$ this way by the division theorem. Substitute $2j + r$ for $x$ into your equations. If you conclude $r = 0$, then $x$ is even. If you can conclude $r = 1$, then $x$ is odd.
A: contrapositive easiest but can also use Proof by Contradiction(still trivial):
$i)x^2+1$ even
$1)$assume $x$ even
$2)x=2k,k\in Z^+$             $\space\space\space\space1)$*
$3)x^2+1=4k^2+1=2(2k^2)+1$  $\space\space\space\space2)$
$4)x^2+1$ odd$\space\space\space\space3)$
$5)F_0$ $\space\space\space\space i)$ and $4)$
$6)x$ even $\rightarrow F_0$ $\space\space\space\space5)$
$7)\neg(x$ even$)\vee F_0$$\space\space\space\space6)$
$\therefore x$ odd$\space\space\space\space7)$
