If x and y are integers , prove that x^3 −7x divisible by 3


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Question Number 97823 by john santu last updated on 10/Jun/20

$If x and y are integers, prove$ $that x^{3} - 7 x divisible by 3$
Commented by bobhans last updated on 10/Jun/20

$let x is an integer . we can write$ $x^{3} - 7 x = x (x^{2} - 7)$ $if x is divisible by 3, then {we}^{'} re done .$ $otherwise, x must have remainder 1 or 2$ $in the divisibility by 3 . thus x^{2} must$ $has remainder 1$ ${(3 k + 1)}^{2} = 3 [k (3 k + 2)] + 1, k \in Z$
Commented by john santu last updated on 10/Jun/20

$great all answer$
	Answered by Rio Michael last updated on 10/Jun/20

	$let f (x) = x^{3} - 7 x \Rightarrow f (x + 1) = {(x + 1)}^{3} - 7 (x + 1) - [x^{3} - 7 x]$ $f (x + 1) - f (x) = x^{3} + 3 x^{2} + 3 x + 1 - 7 x - 7 - x^{3} + 7 x$ $= 3 x^{2} + 3 x - 6$ $= 3 (x^{2} + x - 2), (x^{2} + x - 2) \in Z$ $by mathematical induction .$
	Answered by MJS last updated on 10/Jun/20

	$(1) x = 3 n$ $x^{3} - 7 x = 3 n (9 n^{2} - 7)$ $(2) x = 3 n + 1$ $x^{3} - 7 x = 3 (3 n + 1) (3 n^{2} + 2 n - 2)$ $(3) x = 3 n + 2$ $x^{3} - 7 x = 3 (3 n + 2) (3 n^{2} + 4 n - 1)$
	Answered by floor(10²Eta[1]) last updated on 10/Jun/20

	$x^{3} - 7 x \equiv x^{3} - x (m o d 3)$ $x^{3} - x = x (x^{2} - 1) = x (x - 1) (x + 1)$ $b u t x - 1, x a n d x + 1 a r e 3 c o n s e c u t i v e n u m b e r s$ $s o a t l e a s t o n e o f t h e m h a v e t o b e a m u l t i p l e o f 3$ $\Rightarrow x^{3} - 7 x \equiv 0 (m o d 3)$
	Answered by 1549442205 last updated on 10/Jun/20

	$x^{3} - 7 x = [(x - 1) x (x + 1) - 6 x] ⋮ 3 since$ $(x - 1) x (x + 1) ⋮ 3 (among three consecutive$ $always have at least one number divide by 3) and 6 x ⋮ 3$
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