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Question Number 195342 by Matica last updated on 31/Jul/23

$$1.\mathrm{Prove}\mathrm{that}\mathrm{\forall }n\in {\mathbb{N}}^{\ast },4^n{(n!)}^3<{(n+1)}^{3n}.$$$$2.\mathrm{Solve}\mathrm{the}\mathrm{equations}\mathrm{in}{\mathbb{Z}}^2:$$$$a./2x^3+xy-7=0,$$$$b./x(x+1)(x+7)(x+8)=y^2.$$

Answered by Rasheed.Sindhi last updated on 31/Jul/23

$$a./2x^3+xy-7=0$$$$x(2x^2+y)=7$$$$\{\begin{array}{l}x=1\wedge 2x^2+y=7\Rightarrow y=5\\ x=-1\wedge 2x^2+y=-7\Rightarrow y=-9\\ x=7\wedge 2x^2+y=1\Rightarrow y=-97\\ x=-7\wedge 2x^2+y=-1\Rightarrow y=-99\end{array}$$

Commented by Matica last updated on 31/Jul/23

$$Thankyou.Sothetrickis‘‘{don}^{\prime }tforget$$$$x,y\in \mathbb{Z}.$$

Answered by Rasheed.Sindhi last updated on 31/Jul/23

$$b./x(x+1)(x+7)(x+8)=y^2$$$$SOMEcases:$$$$\bullet x(x+1)=(x+7)(x+8)$$$$x^2+x=x^2+15x+56$$$$14x+56=0\Rightarrow x=-4✓$$$$\bullet x(x+7)=(x+1)(x+8)$$$$x^2+7x=x^2+9x+8$$$$2x+8=0\Rightarrow x=-4✓$$$$\bullet x(x+8)=(x+1)(x+7)$$$$x^2+8x=x^2+8x+7\left(Noresult\right)$$$$\bullet x=x(x+1)(x+7)(x+8)(x\notin \mathbb{Z})$$$$\bullet x+1=x(x+7)(x+8)(x\notin \mathbb{Z})$$$$\bullet x+7=x(x+1)(x+8)(x\notin \mathbb{Z})$$$$\bullet x(x+1)(x+7)(x+8)=0$$$$x=0,-1,-7,-8✓$$$$\bullet x(x+1)(x+7)(x+8)=144$$$$x=1,-4,-9✓$$

Commented by Matica last updated on 01/Aug/23

$$Thankyou.$$

Answered by witcher3 last updated on 31/Jul/23

$$\mathrm{k}(\mathrm{n}-\mathrm{k})⩽\frac{{\mathrm{n}}^2}{4}$$$$\mathrm{n}!⩽\frac{{(\mathrm{n}+1)}^{\mathrm{n}}}{4^{\frac{\mathrm{n}}{3}}}$$$$\mathrm{if}\mathrm{n}=2\mathrm{m}$$$$\mathrm{n}!=\underset{\mathrm{k}=1}{\prod ^{\mathrm{m}}}\mathrm{k}(2\mathrm{m}+1-\mathrm{k})⩽\underset{\mathrm{k}=1}{\prod ^{\mathrm{m}}}{\left(\frac{\mathrm{k}+2\mathrm{m}+1-\mathrm{k})}{2}\right)}^2$$$$\mathrm{Am}\mathrm{GM},\mathrm{xy}⩽{\left(\frac{\mathrm{x}+\mathrm{y}}{2}\right)}^{2...}$$$$⩽\frac{{(2\mathrm{m}+1)}^{2\mathrm{m}}}{4^{\mathrm{m}}}=\frac{{(\mathrm{n}+1)}^{\mathrm{n}}}{4^{\frac{\mathrm{n}}{2}}}⩽\frac{{(\mathrm{n}+1)}^{\mathrm{n}}}{4^{\frac{\mathrm{n}}{3}}},\frac{\mathrm{n}}{2}>\frac{\mathrm{n}}{3}$$$$\mathrm{if}\mathrm{n}=2\mathrm{m}+1$$$$\mathrm{n}!=(\mathrm{m}+1)\underset{\mathrm{k}=1}{\prod ^{\mathrm{m}}}(2\mathrm{m}+2-\mathrm{k})\mathrm{k}$$$$⩽(\mathrm{m}+1)\underset{\mathrm{k}=1}{\prod ^{\mathrm{m}}}.{\left(\frac{2\mathrm{m}+2}{2}\right)}^2=(\mathrm{m}+1)\left(\frac{{(2\mathrm{m}+2)}^{2\mathrm{m}}}{4^{\mathrm{m}}}\right)$$$$=\frac{{(2\mathrm{m}+2)}^{2\mathrm{m}+1}}{4^{\mathrm{m}+\frac{1}{2}}}=\frac{{(\mathrm{n}+1)}^{\mathrm{n}}}{4^{\frac{\mathrm{n}}{2}}}<\frac{{(\mathrm{n}+1)}^{\mathrm{n}}}{4^{\frac{\mathrm{n}}{3}}}$$$$\Rightarrow \mathrm{\forall }\mathrm{n}\in \mathbb{N}{4}^{\mathrm{n}}{(\mathrm{n}!)}^3⩽{(\mathrm{n}+1)}^{3\mathrm{n}}$$

Commented by Matica last updated on 01/Aug/23

$$Thankyoubrother.$$

Commented by witcher3 last updated on 01/Aug/23

$$\mathrm{withe}\mathrm{Pleasur}\mathrm{god}\mathrm{bless}\mathrm{You}$$

Answered by Rasheed.Sindhi last updated on 31/Jul/23

$$b./x(x+1)(x+7)(x+8)=y^2$$$$Letx+4=a$$$$\Rightarrow (a-4)(a-3)(a+3)(a+4)=y^2$$$$(a^2-16)(a^2-9)=y^2$$$$a^4-25a^2+144=y^2$$$$a^4-25a^2+144-y^2=0$$$$a^2=\frac{25\pm \sqrt{625-4(144-y^2)}}{2}$$$$=\frac{25\pm \sqrt{49+4y^2}}{2}\in {\mathbb{Z}}^{+}$$$$\sqrt{49+4y^2}⩽25\wedge 4y^2+49isperfectsquare.$$$$4y^2+49⩽625$$$$y^2⩽144\Rightarrow -12⩽y⩽12$$$$Thevalueofyforwhich4y^2+49$$$$isperfectsquare:$$$$y=0,\pm 12$$$$y=0:$$$$a^2=\frac{25\pm \sqrt{49+4{\left(0\right)}^2}}{2}=16,9$$$$a=\pm 4,\pm 3$$$$x=\pm 4-4,\pm 3-4$$$$=0,-8,-1,-7✓$$$$$$$$y=\pm 12:$$$$a^2=\frac{25\pm \sqrt{49+4{(\pm 12)}^2}}{2}$$$$=\frac{25\pm 25}{2}=25,0$$$$a=\pm 5,0$$$$x=\pm 5-4,0-4$$$$x=1,-9,-4✓$$$$$$$$(x,y)=(0,0),(-1,0),(-7,0),(-8,0),$$$$(1,\pm 12),(-4,\pm 12),(-9,\pm 12)$$$$\left(Total10solutions\right)$$

Commented by Matica last updated on 01/Aug/23

$$Tahnkyoualot.$$

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