Question and Answers Forum

All Questions      Topic List

Arithmetic Questions

Previous in All Question      Next in All Question      

Previous in Arithmetic      Next in Arithmetic      

Question Number 21067 by youssoufab last updated on 11/Sep/17

$$\mathrm{\forall }n\in \mathbb{N},prove9\mid [n^3+{(n+1)}^3+{(n+2)}^3]$$

Answered by dioph last updated on 12/Sep/17

$$n^3+{(n+1)}^3+{(n+2)}^3=$$$$=n^3+n^3+3n^2+3n+1+n^3+6n^2+12n+8=$$$$=3n^3+9n^2+15n+9$$$$=3n(n^2+5)+9(n^2+1)=x$$$$\mathrm{Case}1:3\mid n$$$$\Rightarrow n=3k$$$$\Rightarrow x=9k(9k^2+5)+9(9k^2+1)$$$$\Rightarrow 9\mid x$$$$\mathrm{Case}2:3\nmid n$$$$\Rightarrow n^2+5\equiv n^2-1\equiv (n+1)(n-1)$$$$\equiv 0\left(\mathrm{mod}3\right)$$$$\Rightarrow n^2+5=3h$$$$\Rightarrow x=9nh+9(n^2+1)$$$$\Rightarrow 9\mid x$$$$◼$$

Commented by youssoufab last updated on 14/Sep/17

$$thanksforhelp$$

Answered by mrW1 last updated on 12/Sep/17

$$\mathrm{let}\mathrm{m}=\mathrm{n}+1$$$$\Rightarrow {(\mathrm{m}-1)}^3+{\mathrm{m}}^3+{(\mathrm{m}+1)}^3$$$$={\mathrm{m}}^3-{3\mathrm{m}}^2+3\mathrm{m}-1+{\mathrm{m}}^3+{\mathrm{m}}^3+{3\mathrm{m}}^2+3\mathrm{m}+1$$$$={3\mathrm{m}}^3+6\mathrm{m}$$$$=3\mathrm{m}({\mathrm{m}}^2+2)$$$$$$$$\mathrm{if}\mathrm{m}=3\mathrm{k}:$$$$3\mathrm{m}({\mathrm{m}}^2+2)=9\mathrm{k}({9\mathrm{k}}^2+2)\equiv 0\left(\mathrm{mod}9\right)$$$$$$$$\mathrm{if}\mathrm{m}=3\mathrm{k}\pm 1:$$$$3\mathrm{m}({\mathrm{m}}^2+2)=3(3\mathrm{k}\pm 1)({9\mathrm{k}}^2\pm 6\mathrm{k}+1+2)$$$$=9(3\mathrm{k}\pm 1)({3\mathrm{k}}^2\pm 2\mathrm{k}+1)\equiv 0\left(\mathrm{mod}9\right)$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com