Can each odd composite integer be represented as $\sqrt{a^2-b^2}$

Question

Each odd composite integer can be represented as $a^2-b^2$, it's the basics for Fermat's factorization method.

I got an idea of exploring an integer factorization algorithm(see my other question) that is based on the assumption that each odd composite integer can be represented as $\sqrt{a^2-b^2}$.

Can you prove this assumption, or show where it fails?

Answer 1

Yes. This is equivalent to solving $n^2=a^2-b^2.$

But if $n$ is odd then so is $n^2.$ So your original result applies to $n^2.$

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(You don’t need $n$ composite for your first result, just $n$ odd. If $n=2k-1$ then $k^2-(k-1)^2=2k-1=n.$)