Integer as Difference between Two Squares/Formulation 2

Theorem

Any integer can be expressed as the difference of two squares if and only if that integer is NOT $n \equiv 2 \pmod 4$

$n \equiv 0 \pmod 4$

$n \equiv 1 \pmod 4$

$n \equiv 2 \pmod 4$

$n \equiv 3 \pmod 4$

$\ds 4x$

$=$

$\ds \paren{x + 1 }^2 - \paren{x - 1 }^2$

$\ds 4x + 1$

$=$

$\ds \paren{2x + 1 }^2 - 4x^2$

$\ds 4x + 3$

$=$

$\ds \paren{2x + 2 }^2 - \paren{2x + 1 }^2$

For $n \equiv 2 \pmod 4$, it is impossible to represent such an integer as the difference of two squares.

Taking the squares of each of the residue classes, we have:

$0^2 \equiv 0 \pmod 4$

$1^2 \equiv 1 \pmod 4$

$2^2 \equiv 0 \pmod 4$

$3^2 \equiv 1 \pmod 4$

Therefore, when taking the difference of two squares, $n \equiv 2 \pmod 4$ is never a result.

$0 - 0 \equiv 0 \pmod 4$

$1 - 1 \equiv 0 \pmod 4$

$0 - 1 \equiv 3 \pmod 4$

$1 - 0 \equiv 1 \pmod 4$

$\blacksquare$