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Question Number 87533 by mr W last updated on 04/Apr/20

Commented by MJS last updated on 05/Apr/20

$$a=\frac{49}{76}\pm \frac{457\sqrt{87}}{6612}$$$$b=\frac{49}{76}\mp \frac{457\sqrt{87}}{6612}$$$$x=-7\pm \sqrt{87}$$$$y=-7\mp \sqrt{87}$$$$\mathrm{answer}\mathrm{is}20$$

Commented by mr W last updated on 05/Apr/20

$$thankssir!$$

Answered by ajfour last updated on 05/Apr/20

$$let\mathit{a}\mathit{x}=\mathit{s},\mathit{b}\mathit{y}=\mathit{t}$$$$\Rightarrow s+t=3....\left(i\right)$$$$sx+ty=7....\left(ii\right)$$$${sx}^2+{ty}^2=16...\left(iii\right)$$$${sx}^3+{ty}^3=42...\left(iv\right)$$$$\mathit{Q}={sx}^4+{ty}^4=?$$$$\left(ii\right)-y\times \left(i\right)$$$$s(x-y)=7-3y.....\left(A\right)$$$$\left(iii\right)-y\times \left(ii\right)$$$$sx(x-y)=16-7y....\left(B\right)$$$$\Rightarrow \mathit{x}=\frac{16-7\mathit{y}}{7-3\mathit{y}}.....\left(\mathit{I}\right)$$$$\left(iii\right)-y^2\times \left(i\right)$$$$s(x^2-y^2)=16-3y^2....\left(C\right)$$$$\left(iv\right)-y^2\times \left(ii\right)$$$$sx(x^2-y^2)=42-7y^2....\left(D\right)$$$$\Rightarrow \mathit{x}=\frac{42-7{\mathit{y}}^2}{16-3{\mathit{y}}^2}.....\left(\mathit{I}\mathit{I}\right)$$$$equating\left(I\right)\mathrm{\&}\left(II\right)$$$$\frac{16-7\mathit{y}}{7-3\mathit{y}}=\frac{42-7{\mathit{y}}^2}{16-3{\mathit{y}}^2}$$$$256-48y^2-112y+21y^3$$$$=294-49y^2-126y+21y^3$$$$$$$$\Rightarrow {\mathit{y}}^2+14\mathit{y}-38=0$$$${\mathit{y}}_1=-7-\sqrt{87},{\mathit{y}}_2=-7+\sqrt{87}$$$${\mathit{x}}_1=\frac{16-7{\mathit{y}}_1}{7-3{\mathit{y}}_1}=\frac{16+7(\sqrt{87}+7)}{7+3(\sqrt{87}+7)}$$$$=\frac{65+7\sqrt{87}}{28+3\sqrt{87}}=-7+\sqrt{87}$$$$letsproceedwithjust{x}_1,y_1.$$$$\mathit{s}+\mathit{t}=3$$$$\mathit{s}\mathit{x}+\mathit{t}\mathit{y}=7\Rightarrow sx+(3-s)y=7$$$$\Rightarrow \mathit{s}=\frac{7-3y}{x-y}=\frac{7+3(7+\sqrt{87})}{2\sqrt{87}}$$$$=\frac{28+3\sqrt{87}}{2\sqrt{87}}=\frac{261+28\sqrt{87}}{174}$$$$\mathit{t}=3-\mathit{s}=\frac{261-28\sqrt{87}}{174}$$$$\mathit{Q}={sx}^4+{ty}^4$$$$=\frac{261+28\sqrt{87}}{174}\times {\{\sqrt{87}-7\}}^4$$$$+\frac{261-28\sqrt{87}}{174}\times {\{-\sqrt{87}-7\}}^4$$$$=2\times {10}^1.$$

Commented by mr W last updated on 05/Apr/20

$$thankssir!$$

Commented by ajfour last updated on 05/Apr/20

$$Haveyoudevisedsome$$$$stillshorterbetterwaysir,$$$$ifso,pleaseshare..$$

Commented by mr W last updated on 05/Apr/20

$$ihavetriedthesamewayassantusir,$$$$ihavenootherbetterway.$$$$i^{\prime }mconsideringhowtoget{\mathit{a}\mathit{x}}^{\mathit{n}}+{\mathit{b}\mathit{y}}^{\mathit{n}}=?$$$$forn⩾5,seeanattemptbelow.$$

Answered by john santu last updated on 05/Apr/20

$$\left(\mathrm{i}\right)({\mathrm{ax}}^4+{\mathrm{by}}^4)(\mathrm{x}+\mathrm{y})=42(\mathrm{x}+\mathrm{y})$$$${\mathrm{ax}}^5+{\mathrm{b}}^{\mathrm{y}}+{\mathrm{ax}}^4\mathrm{y}+{\mathrm{bxy}}^4=42(\mathrm{x}+\mathrm{y})$$$${\mathrm{ax}}^5+{\mathrm{by}}^5+\mathrm{xy}({\mathrm{ax}}^3+{\mathrm{by}}^3)=42(\mathrm{x}+\mathrm{y})$$$$\left(\mathrm{ii}\right)({\mathrm{ax}}^2+{\mathrm{by}}^2)(\mathrm{x}+\mathrm{y})=7(\mathrm{x}+\mathrm{y})$$$${\mathrm{ax}}^3+{\mathrm{by}}^3+\mathrm{xy}(\mathrm{ax}+\mathrm{by})$$$$7(\mathrm{x}+\mathrm{y})=16+\mathrm{xy}.3$$$$16(\mathrm{x}+\mathrm{y})=42+\mathrm{xy}.7$$$$\mathrm{x}+\mathrm{y}=-14\wedge \mathrm{xy}=-38$$$$\Rightarrow {\mathrm{ax}}^5+{\mathrm{by}}^5=42(-14)+\left(38\right)\left(16\right)$$$$=-588+608=20$$

Commented by ajfour last updated on 05/Apr/20

$$\mathcal{S}uperb\mathcal{S}ir!Congratulations,$$$$ingeniousway.$$

Commented by jagoll last updated on 05/Apr/20

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

Commented by mr W last updated on 05/Apr/20

$$thankssir!$$

Commented by mr W last updated on 05/Apr/20

$$ithinkthisisthebestway.butcanwe$$$$applyittoget{\mathit{a}\mathit{x}}^{\mathit{n}}+{\mathit{b}\mathit{y}}^{\mathit{n}}generally?$$$$ithinkthisispossible,seebelow.$$

Answered by mr W last updated on 26/Oct/21

$$S_n={ax}^n+{by}^n$$$$S_1=3$$$$S_2=7$$$$S_3=16$$$$S_4=42$$$$S_{n+1}={ax}^{n+1}+{by}^{n+1}$$$$(x+y)S_n=(x+y)({ax}^n+{by}^n)={ax}^{n+1}+{by}^{n+1}+xy({ax}^{n-1}+{by}^{n-1})$$$$(x+y)S_n=S_{n+1}+{xyS}_{n-1}$$$$\Rightarrow S_{n+1}=(x+y)S_n-{xyS}_{n-1}$$$$letx+y=p,xy=q$$$$\Rightarrow S_{n+1}={pS}_n-{qS}_{n-1}...\left(i\right)$$$$S_3={pS}_2-{qS}_1$$$$S_4={pS}_3-{qS}_2$$$$\Rightarrow p=\frac{S_1{S}_4-S_2{S}_3}{S_1{S}_3-S_2^2}=\frac{3\times 42-7\times 16}{3\times 16-7^2}=-14$$$$\Rightarrow q=\frac{S_2{S}_4-S_3^2}{S_1{S}_3-S_2^2}=\frac{7\times 42-{16}^2}{3\times 16-7^2}=-38$$$$$$$$a=\frac{S_2-S_{1}y}{x(x-y)}$$$$b=\frac{S_2-S_{1}x}{y(y-x)}$$$$$$$$S_2={pS}_1-{qS}_0$$$$\Rightarrow S_0=a+b=\frac{{pS}_1-S_2}{q}$$$$$$$$from\left(i\right)wehavethecharacteristiceqn.$$$$t^2-pt+q=0$$$$t=\frac{p\pm \sqrt{p^2-4q}}{2}$$$$let\alpha =\frac{p+\sqrt{p^2-4q}}{2},\beta =\frac{p-\sqrt{p^2-4q}}{2}$$$$\Rightarrow S_n=A{\alpha }^n+B{\beta }^n...\left(ii\right)$$$$$$$$S_1=A\alpha +B\beta ...\left(iii\right)$$$$S_2=A{\alpha }^2+B{\beta }^2...\left(iv\right)$$$$\Rightarrow A=\frac{\beta S_1-S_2}{\alpha (\beta -\alpha )}$$$$\Rightarrow B=\frac{\alpha S_1-S_2}{\beta (\alpha -\beta )}$$$$$$$$withourexample:$$$$\alpha =\frac{-14+\sqrt{{14}^2+4\times 38}}{2}=-7+\sqrt{87}$$$$\beta =\frac{-14-\sqrt{{14}^2+4\times 38}}{2}=-7-\sqrt{87}$$$$A=\frac{1}{76}(49+\frac{457\sqrt{87}}{87})$$$$B=\frac{1}{76}(49-\frac{457\sqrt{87}}{87})$$$$$$$$S_n=\frac{1}{76}\{(49+\frac{457\sqrt{87}}{87}){(-7+\sqrt{87})}^n+(49+\frac{457\sqrt{87}}{87}){(-7-\sqrt{87})}^n\}$$$$weget:$$$$S_5=20$$$$S_6=1316$$$$S_7=-17446$$$$S_8=297304$$$$S_9=-4833488$$$$S_{10}=79966384$$

Commented by john santu last updated on 05/Apr/20

$$\mathrm{great}\mathrm{sir}$$

Commented by mr W last updated on 05/Apr/20

$$recursivelywecanget$$$$S_5=-14S_4+38S_3=-14\times 42+38\times 16=20$$$$S_6=-14S_5+38S_4=-14\times 20+38\times 42=1316$$$$S_7=-14S_6+38S_5=-14\times 1316+38\times 20=-17664$$$$...$$

Commented by TawaTawa1 last updated on 05/Apr/20

$$\mathrm{Wow}\mathrm{great}\mathrm{sir}.$$$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

Commented by TawaTawa1 last updated on 05/Apr/20

$$\mathrm{Sir},\mathrm{what}\mathrm{of}\mathrm{if}\mathrm{x}\pm \frac{1}{\mathrm{x}}=\mathrm{a}$$$$\mathrm{Then}{\mathrm{x}}^{\mathrm{n}}\pm \frac{1}{{\mathrm{x}}^{\mathrm{n}}}=???$$$$\mathrm{What}\mathrm{is}\mathrm{the}\mathrm{general}\mathrm{form}\mathrm{sir}?$$

Commented by mr W last updated on 05/Apr/20

$$thisquestionisalreadyansweredn$$$$times,pleasetrytofindanoldpost$$$$concerningthisquestion.thesolution$$$$isalsoeasy:getxfromx\pm \frac{1}{x}=a,this$$$$shouldbesolvable.thenyoucanget$$$$x^n\pm \frac{1}{x^n}.$$

Commented by TawaTawa1 last updated on 05/Apr/20

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

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