<s>[INST] 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?
 [/INST] Since $n$ is composite $a,b>1$ (WLOG $a\geq b$)
$$x-y=\dfrac{a+b}{2}-\dfrac{a-b}{2}=b>1$$
$$x>y+1$$
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