Factoring of integers using the product of complex conjugates

Question

I saw this technique somewhere a while ago, and I wanted to ask about a generalization I noticed.

Let $x \in Z$. There is some $a \in Z | x =(a+i)(a-i)$

What I found when plugging in values for ‘a’ and multiplying out was that $a=x-1$.

Letting this substitution hold:

$x = (x - 1 + i)(x-1+i)$

$x=(x-1)^2 -i ^2$

$x=x-1+1$

$x=x$

This seems lie it would work for all $x,a \in Z$. Any thoughts/comments would be appreciated.

Thank you, Jordan W. Effinger.

Answer 1

The substitution you have thought of does not work because the expansion of $(x-1)^2$ is $x^2-2x+1$ and not the $x-1$ which you have used.