Odd numbers expressed as : $x^2-y^2$ How to prove following statement :

Conjecture:
An odd number $n$ , $(n>1)$ can be uniquely expressed as : $n= x^2-y^2$ ; $x,y \in \mathbb{Z}^{*}$
if and only if $n$ is a prime number .

If $x-y=m$ , where $m>1$ then $m \mid n$
Proof :
$n=x^2-y^2=(y+m)^2-y^2=y^2+2\cdot y\cdot m +m^2-y^2 \Rightarrow$
$\Rightarrow n=m\cdot (2\cdot y+m) \Rightarrow m \mid n$
Therefore , if $m \neq 1$ it follows that $n$ is a composite number , but how to prove that every odd
composite number ,other than $1$ , has representation : $x^2-y^2$ , where $x-y>1$ ?

HINT $\ $ (Non) uniqueness follows from the composition law for differences of squares
$\rm\qquad\qquad\ (a^2-b^2)\ (A^2-B^2)\ =\ (a\:A+b\:B)^2-(a\:B+A\:b)^2$
$\rm\qquad\qquad\ \phantom{(a^2-b^2)\ (A^2-B^2)}\ =\ (a\:A-b\:B)^2-(a\:B-A\:b)^2$
E.g. composing $\rm\ 7 = 4^2 - 3^2\ $ with $\ 11 = 6^2 - 5^2\ $ yields for $\rm\: 7\cdot 11\:$ the following $\,2\,$ rep's
$\rm\qquad\qquad\ (4^2-3^2)\ (6^2-5^2)\ =\ (4\cdot 6+3\cdot 5)^2-(4\cdot 5+6\cdot 3)^2\ =\ 39^2 - 38^2$
$\rm\qquad\qquad\ \phantom{(4^2-3^2)\ (6^2-5^2)}\ =\ (4\cdot 6-3\cdot 5)^2-(4\cdot 5-6\cdot 3)^2\ =\ 9^2 - 2^2$
Remark $ $ Just like the e Brahmagupta–Fibonacci identity for composition of sums of squares can be viewed as arising from the multiplicativity of the norms of Gaussian integers, the above can be viewed as arising from norm multiplicativity of split-complex numbers (aka double or perplex numbers), $\,a+b\:\!j\,$ where $\,j^2 = 1.\ \ $
