If-x-and-y-are-integers-prove-that-x-3-7x-divisible-by-3-

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