Integer as Difference between Two Squares/Formulation 1

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $n$ be a positive integer.

Then $n$ can be expressed as:

$n = a^2 - b^2$

if and only if $n$ has at least two distinct divisors of the same parity that multiply to $n$.


Proof

\(\ds n\) \(=\) \(\ds a^2 - b^2\)
\(\ds \) \(=\) \(\ds \paren {a + b} \paren {a - b}\) Difference of Two Squares


Thus $n = p q$ where:

\(\text {(1)}: \quad\) \(\ds p\) \(=\) \(\ds \paren {a + b}\)
\(\text {(2)}: \quad\) \(\ds q\) \(=\) \(\ds \paren {a - b}\)
\(\ds \leadsto \ \ \) \(\ds p + q\) \(=\) \(\ds 2 a\) $(1) + (2)$
\(\ds p - q\) \(=\) \(\ds 2 b\) $(1) - (2)$
\(\ds \leadsto \ \ \) \(\ds a\) \(=\) \(\ds \dfrac {p + q} 2\)
\(\ds b\) \(=\) \(\ds \dfrac {p - q} 2\)


Thus for $a$ and $b$ to be integers, both $p$ and $q$ must be:

distinct, otherwise $p = q$ and so $b = 0$
either both even or both odd, otherwise both $p + q$ and $p - q$ will be odd, and so neither $\dfrac {p + q} 2$ nor $\dfrac {p - q} 2$ are defined in $\Z$.


Hence the result.

$\blacksquare$