Integer as Difference between Two Squares/Formulation 1
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.
$\blacksquare$