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.

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TopNewest\(a|b, b \gt 3a\), 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 · 11 months, 2 weeks ago

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– Benjamin Ononogbu · 11 months, 2 weeks ago

nice one sir. i did mine by using compendo et divendo rule.Log in to reply