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

$$\mathrm{If}\:{x}\:\mathrm{and}\:{y}\:\mathrm{are}\:\mathrm{integers}\:,\:\mathrm{prove} \\ $$$$\mathrm{that}\:{x}^{\mathrm{3}} −\mathrm{7}{x}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{3}\: \\ $$

Commented by bobhans last updated on 10/Jun/20

let x is an integer. we can write x^3 −7x = 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 (3k+1)^2 = 3[ k(3k+2)] +1 , k∈Z

$$\mathrm{let}\:{x}\:\mathrm{is}\:\mathrm{an}\:\mathrm{integer}.\:\mathrm{we}\:\mathrm{can}\:\mathrm{write} \\ $$$${x}^{\mathrm{3}} −\mathrm{7}{x}\:=\:{x}\left({x}^{\mathrm{2}} −\mathrm{7}\right)\: \\ $$$$\mathrm{if}\:{x}\:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{3}\:,\:\mathrm{then}\:\mathrm{we}'\mathrm{re}\:\mathrm{done}. \\ $$$$\mathrm{otherwise}\:,{x}\:\mathrm{must}\:\mathrm{have}\:\mathrm{remainder}\:\mathrm{1}\:\mathrm{or}\:\mathrm{2} \\ $$$$\mathrm{in}\:\mathrm{the}\:\mathrm{divisibility}\:\mathrm{by}\:\mathrm{3}.\:\mathrm{thus}\:{x}^{\mathrm{2}} \:\mathrm{must} \\ $$$$\mathrm{has}\:\mathrm{remainder}\:\mathrm{1} \\ $$$$\left(\mathrm{3}{k}+\mathrm{1}\right)^{\mathrm{2}} \:=\:\mathrm{3}\left[\:{k}\left(\mathrm{3}{k}+\mathrm{2}\right)\right]\:+\mathrm{1}\:,\:\mathrm{k}\in\mathbb{Z} \\ $$

Commented by john santu last updated on 10/Jun/20

$$\mathrm{great}\:\mathrm{all}\:\mathrm{answer}\: \\ $$

Answered by Rio Michael last updated on 10/Jun/20

let f(x) = x^3 −7x ⇒ f(x + 1) = (x + 1)^3 −7(x + 1) −[x^3 −7x] f(x + 1)−f(x) = x^3 + 3x^(2 ) + 3x + 1−7x−7 −x^3 + 7x = 3x^2 + 3x −6 = 3(x^2 + x −2) , (x^2 + x −2) ∈ Z by mathematical induction.

$$\mathrm{let}\:{f}\left({x}\right)\:=\:{x}^{\mathrm{3}} −\mathrm{7}{x}\:\Rightarrow\:{f}\left({x}\:+\:\mathrm{1}\right)\:=\:\left({x}\:+\:\mathrm{1}\right)^{\mathrm{3}} −\mathrm{7}\left({x}\:+\:\mathrm{1}\right)\:−\left[{x}^{\mathrm{3}} −\mathrm{7}{x}\right] \\ $$$${f}\left({x}\:+\:\mathrm{1}\right)−{f}\left({x}\right)\:=\:{x}^{\mathrm{3}} \:+\:\mathrm{3}{x}^{\mathrm{2}\:} \:+\:\mathrm{3}{x}\:+\:\mathrm{1}−\mathrm{7}{x}−\mathrm{7}\:−{x}^{\mathrm{3}} \:+\:\mathrm{7}{x} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\:\mathrm{3}{x}^{\mathrm{2}} \:+\:\mathrm{3}{x}\:−\mathrm{6} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{3}\left({x}^{\mathrm{2}} \:+\:{x}\:−\mathrm{2}\right)\:\:\:,\:\left({x}^{\mathrm{2}} \:+\:{x}\:−\mathrm{2}\right)\:\in\:\mathbb{Z} \\ $$$$\mathrm{by}\:\mathrm{mathematical}\:\mathrm{induction}. \\ $$

Answered by MJS last updated on 10/Jun/20

(1) x=3n x^3 −7x=3n(9n^2 −7) (2) x=3n+1 x^3 −7x=3(3n+1)(3n^2 +2n−2) (3) x=3n+2 x^3 −7x=3(3n+2)(3n^2 +4n−1)

$$\left(\mathrm{1}\right)\:{x}=\mathrm{3}{n} \\ $$$${x}^{\mathrm{3}} −\mathrm{7}{x}=\mathrm{3}{n}\left(\mathrm{9}{n}^{\mathrm{2}} −\mathrm{7}\right) \\ $$$$\left(\mathrm{2}\right)\:{x}=\mathrm{3}{n}+\mathrm{1} \\ $$$${x}^{\mathrm{3}} −\mathrm{7}{x}=\mathrm{3}\left(\mathrm{3}{n}+\mathrm{1}\right)\left(\mathrm{3}{n}^{\mathrm{2}} +\mathrm{2}{n}−\mathrm{2}\right) \\ $$$$\left(\mathrm{3}\right)\:{x}=\mathrm{3}{n}+\mathrm{2} \\ $$$${x}^{\mathrm{3}} −\mathrm{7}{x}=\mathrm{3}\left(\mathrm{3}{n}+\mathrm{2}\right)\left(\mathrm{3}{n}^{\mathrm{2}} +\mathrm{4}{n}−\mathrm{1}\right) \\ $$

Answered by floor(10²Eta[1]) last updated on 10/Jun/20

x^3 −7x≡x^3 −x(mod 3) x^3 −x=x(x^2 −1)=x(x−1)(x+1) but x−1, x and x+1 are 3 consecutive numbers so at least one of them have to be a multiple of 3 ⇒x^3 −7x≡0(mod 3)

$${x}^{\mathrm{3}} −\mathrm{7}{x}\equiv{x}^{\mathrm{3}} −{x}\left({mod}\:\mathrm{3}\right) \\ $$$${x}^{\mathrm{3}} −{x}={x}\left({x}^{\mathrm{2}} −\mathrm{1}\right)={x}\left({x}−\mathrm{1}\right)\left({x}+\mathrm{1}\right) \\ $$$${but}\:{x}−\mathrm{1},\:{x}\:{and}\:{x}+\mathrm{1}\:{are}\:\mathrm{3}\:{consecutive}\:{numbers} \\ $$$${so}\:{at}\:{least}\:{one}\:{of}\:{them}\:{have}\:{to}\:{be}\:{a}\:{multiple}\:{of}\:\mathrm{3} \\ $$$$\Rightarrow{x}^{\mathrm{3}} −\mathrm{7}{x}\equiv\mathrm{0}\left({mod}\:\mathrm{3}\right) \\ $$

Answered by 1549442205 last updated on 10/Jun/20

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

$$\mathrm{x}^{\mathrm{3}} −\mathrm{7x}=\left[\left(\mathrm{x}−\mathrm{1}\right)\mathrm{x}\left(\mathrm{x}+\mathrm{1}\right)−\mathrm{6x}\right]\vdots\mathrm{3}\:\mathrm{since} \\ $$$$\left(\mathrm{x}−\mathrm{1}\right)\mathrm{x}\left(\mathrm{x}+\mathrm{1}\right)\vdots\mathrm{3}\left(\mathrm{among}\:\mathrm{three}\:\mathrm{consecutive}\right. \\ $$$$\left.\mathrm{always}\:\mathrm{have}\:\mathrm{at}\:\mathrm{least}\:\mathrm{one}\:\mathrm{number}\:\mathrm{divide}\:\mathrm{by}\:\mathrm{3}\right)\:\mathrm{and}\:\mathrm{6x}\vdots\mathrm{3} \\ $$

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