# 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.

Note by Benjamin Ononogbu
2 years, 3 months ago

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$$a|b, b \gt 3a$$, 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 $$(b - a)|(b + a)$$.

- 2 years, 3 months ago

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

- 2 years, 3 months ago

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