Statement begs for proof

Hey guys, this is something little I observed:

If $a$ and $b$ are positive integers, then the following holds. if $a$ divides $b$ , then $(b-a)$ can never divide $(b+a)$ provided $b>3a$ . Is this statement obvious? if no please prove it.

#NumberTheory

Easy Math Editor

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Comments

$a|b, b \gt 3a$ , implies that $b = na$ for some positive integer $n \gt 3$ . So

$b - a = na - a = (n - 1)a$ and $b + a = (n + 1)a$ .

Then $\dfrac{b + a}{b - a} = \dfrac{(n + 1)a}{(n - 1)a} = \dfrac{n + 1}{n - 1} = 1 + \dfrac{2}{n - 1}$ ,

which can only be an integer when $(n - 1)|2$ . This only occurs when either $n - 1 = 1 \to n = 2$ or $n - 1 = 2 \to n = 3$ , so with the condition that $n \gt 3$ we can never have $(b - a)|(b + a)$ .

Brian Charlesworth - 5 years, 2 months ago

nice one sir. i did mine by using compendo et divendo rule.

Benjamin ononogbu - 5 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$