Correctness of Fermat Factorization Proof I have asked similar questions regarding this proof. But now I would like to know if my reformulation (after perseverance and different thinking) is correct.
Prove: An odd integer $n \in \mathbb{N}$ is composite $\iff$ $n$ can be written as $n = x^2 - y^2 s.t. y+1 < x$
Proof: $\leftarrow$
Let $n$ be odd and consider $n= x^2-y^2 s.t. y+1 < x$
We can factor the difference of two squares as: $(x+y)(x-y)$ and we note for $n$ to be composite (not prime) $(x-y) > 1$. Thus we have shown this direction.
$\rightarrow$
Let $n$ be odd an composite. By definition of composite we have $n=ab$ for some odd integers $a$ and $b$. Now, working backwards:
$\dfrac{4ab}{4} = \dfrac{(a^2 + 2b + b^2)}{4} - \dfrac{(a^2-2b +b^2)}{4} = (\dfrac{a+b}{2})^2 - (\dfrac{a-b}{2})^2$. Thus, we shall take $x = \dfrac{a+b}{2}$ and $y =\dfrac{a-b}{2}$. And we have $n = x^2 - y^2$. 
How do I get $y+1 < x$ in this direction?

You can't. However, you can show that $y+1<x$, which is what you should have been asking. Here it is, with $y=(a-b)/2, x=(a+b)/2$:
$$
\begin{align}
y+1 = \frac{a-b}{2}+1&=\frac{a-b+2}{2}<\frac{a+b}{2}\\
 &\Longleftrightarrow a-b+2<a+b\\
 &\Longleftrightarrow 2<2b \\
 &\Longleftrightarrow 1<b \\
\end{align}
$$
which you know is true, since if $n=ab$ is composite with $b\le a$ we must have $b>1$. (Note: in your statement we can't just have $a,b$ odd integers.) Other than that, your proof is just fine. 
