Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 199170 by tri26112004 last updated on 28/Oct/23

$$n^4+2n^3+2n^2+n+7=a^2(a\in N)$$$$\to n=¿(n\in N)$$

Commented by Rasheed.Sindhi last updated on 31/Oct/23

$$\mathbb{G}\mathbf{reat}!$$$$\mathcal{T}han\mathcal{X}\mathcal{S}ir$$

Commented by Rasheed.Sindhi last updated on 30/Oct/23

$$\bullet a^2⩾{\left(1\right)}^4+2{\left(1\right)}^3+2{\left(1\right)}^2+\left(1\right)+7$$$$a^2>13\Rightarrow a⩾4^{★}$$$$\bullet n^3+2n^2+2n+1=\frac{a^2-7}{n}\in \mathbb{N}$$$$\bullet n^4+2n^3+2n^2+n+7=a^2$$$$\Rightarrow 2n^3+2n^2+n+7=a^2-n^4$$$$=(a-n^2)(a+n^2)\in \mathbb{N}$$$$\Rightarrow n^2<a\Rightarrow n<\sqrt{a}$$

Commented by Rasheed.Sindhi last updated on 30/Oct/23

$${}^{★}{What}^{\prime }s{a}^{\prime }supperlimit?$$

Commented by Frix last updated on 30/Oct/23

$$\mathbb{SOLUTION}\mathrm{Part}1$$$$\mathrm{We}\mathrm{can}\mathrm{factorize}\mathrm{it}.$$$$n^4+2n^3+2n^3+n-a^2+7=0$$$$n=t-\frac{1}{2}$$$$t^4+\frac{t^2}{2}-a^2+\frac{109}{16}=0$$$${\mathrm{It}}^{\prime }\mathrm{s}\mathrm{easy}\mathrm{to}\mathrm{get}$$$$n=-\frac{1}{2}\pm \frac{\sqrt{-1\pm 2\sqrt{4a^2-27}}}{2}\left[{}_{\mathrm{of}\mathrm{the}\pm \mathrm{signs}}^{\mathrm{all}4\mathrm{combinations}}\right]$$$$n\in {\mathbb{R}}^{+}\Rightarrow n=-\frac{1}{2}+\frac{\sqrt{-1+2\sqrt{4a^2-27}}}{2}$$$$\Rightarrow \mathrm{for}a\in \mathbb{R}$$$$4a^2-27⩾0\wedge -1+2\sqrt{4a^2-27}⩾0$$$$\Rightarrow a⩽-\frac{\sqrt{109}}{4}\vee a⩾\frac{\sqrt{109}}{4}$$$$a\in \mathbb{N}\Rightarrow a⩾3$$$$\mathrm{No}\mathrm{upper}\mathrm{limit}.$$

Commented by Rasheed.Sindhi last updated on 30/Oct/23

$$\mathcal{T}\mathit{h}\mathit{a}\mathit{n}\mathit{k}\mathit{s}\mathit{F}\mathit{r}\mathit{i}\mathit{x}\mathit{s}\mathit{i}\mathit{r}!$$

Commented by Frix last updated on 30/Oct/23

$$\mathbb{SOLUTION}\mathrm{Part}2$$$$4a^2-27=b^2$$$$4a^2-b^2=27$$$$(2a-b)(2a+b)=27$$$$2a-b=1\wedge 2a+b=27\Rightarrow a=7\wedge b=13\Rightarrow n=2★$$$$2a-b=3\wedge 2a+b=9\Rightarrow a=3\wedge b=3\Rightarrow n\notin \mathbb{N}$$$$\mathrm{No}\mathrm{other}\mathrm{possibilities}.$$

Commented by Frix last updated on 30/Oct/23

$$\mathrm{I}\mathrm{was}\mathrm{in}\mathrm{a}\mathrm{hurry}\mathrm{before},{\mathrm{didn}}^{\prime }\mathrm{t}\mathrm{see}\mathrm{this}\mathrm{but}$$$${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{obvious}...$$

Commented by Frix last updated on 31/Oct/23

��

Commented by tri26112004 last updated on 01/Nov/23

$$a⩾4andn<\sqrt{a}$$$${It}^{\prime }snotOK$$

Answered by Frix last updated on 28/Oct/23

$$\mathrm{Just}\mathrm{try}\mathrm{a}\mathrm{few}\mathrm{values}\mathrm{for}n$$$$\Rightarrow$$$$n=2\wedge a=7$$

Commented by tri26112004 last updated on 29/Oct/23

$$Exactly,onlyn=2satyfied$$

Answered by Rasheed.Sindhi last updated on 29/Oct/23

$$n^4+2n^3+2n^2+n+7=A^2(A\in N)$$$$\to n=¿(n\in N)$$$$f\left(n\right)=n^4+2n^3+2n^2+n+7-A^2=0(A\in N)$$$$Leta,b,c,dareroots,atleastone$$$$ofwhichisnaturalnumber.say$$$$a\in \mathbb{N}$$$$a+b+c+d=-2$$$$ab+ac+ad+bc+bd+cd=2$$$$abc+abd+acd+bcd=-1$$$$abcd=7-A^2\Rightarrow a=\frac{7-A^2}{bcd}$$$$A=1,2,...$$$$A=1\Rightarrow a=\frac{6}{bcd}\Rightarrow bcd=1,2,3,6\Rightarrow a=6,3,2,1$$$$f\left(n\right){isn}^{\prime }tsatisfied$$$$A=2\Rightarrow abcd=7-2^2=3\Rightarrow a=\frac{3}{bcd}\Rightarrow bcd=1,3\Rightarrow a=3,1$$$$f\left(n\right){isn}^{\prime }tsatisfied$$$$....$$$$A=7\Rightarrow abcd=7-7^2\Rightarrow a=\frac{-42}{bcd}$$$$\Rightarrow bcd=-1,-2,-3,-6,-7,-14,-21,-42$$$$\Rightarrow a=42,21,14,7,6,3,2,1$$$$(A,n)=(7,2)satisfyf\left(n\right)$$$$n=2:$$$$\bullet 2+b+c+d=-2$$$$\overline{)\begin{array}{c}b+c+d=-4\end{array}}$$$$\bullet ab+ac+ad+bc+bd+cd=2$$$$2(b+c+d)+bc+bd+cd=2$$$$2(-4)+bc+bd+cd=2$$$$\overline{)\begin{array}{c}bc+bd+cd=10\end{array}}$$$$\bullet abc+abd+acd+bcd=-1$$$$2bc+2bd+2cd+bcd=-1$$$$2(bc+bd+cd)+bcd=-1$$$$2\left(10\right)+bcd=-1$$$$\overline{)\begin{array}{c}bcd=-21\end{array}}$$$$f\left(n\right)=(n-2)(n^3+4n^2+10n+21)$$$$=(n-2)(n+3)(n^2+n+7)$$$$n=2isonlynaturalroot.$$$$Notagoodapproach!$$

Commented by Rasheed.Sindhi last updated on 29/Oct/23

$$Oncewefoundn=2\wedge A=7$$$${We}^{\prime }vetoprovethat{there}^{\prime }sno$$$$othernaturalrootoff\left(n\right).$$$$ForthisputA=7inf\left(n\right)and$$$$thentryingtosolvef\left(n\right)whose$$$$onefactorisalreadyknown:n-2$$$$$$$$n^4+2n^3+2n^2+n+7-A^2=0(A\in N)$$$$n^4+2n^3+2n^2+n+7-7^2=0$$$$n^4+2n^3+2n^2+n-42=0$$$$(n-2)(n+3)(n^2+n+7)=0$$$$\therefore n=2isonlynaturalroot.$$$$$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com