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Question Number 126950 by deleteduser1 last updated on 05/Dec/22

Answered by floor(10²Eta[1]) last updated on 25/Dec/20

$$\mathrm{A}.1+2^{2^{\mathrm{n}}}+2^{2^{\mathrm{n}+1}}\equiv 1+{(-1)}^{2^{\mathrm{n}}}+{(-1)}^{2^{\mathrm{n}+1}}\equiv 1+1+1\equiv 0\left(\mathrm{mod}3\right)$$$$\mathrm{A}\mathrm{is}\mathrm{true}\mathrm{for}\mathrm{all}\mathrm{n}\in \mathbb{N}\left({\mathbb{Z}}_{+}^{\ast }\right)$$$$$$$$\mathrm{B}.1+2^{2^{\mathrm{n}}}+2^{2^{\mathrm{n}+1}}\left(\mathrm{mod}7\right)$$$$2^{\mathrm{n}}={6\mathrm{q}}_1+{\mathrm{r}}_1$$$$2^{\mathrm{n}+1}=6.{2\mathrm{q}}_1+{2\mathrm{r}}_1\Rightarrow 0⩽{2\mathrm{r}}_1<5\Rightarrow 0⩽{\mathrm{r}}_1⩽2$$$$\left(\mathrm{I}\right)2^{\mathrm{n}}={6\mathrm{q}}_1,\left(\mathrm{II}\right)2^{\mathrm{n}}={6\mathrm{q}}_1+1,\left(\mathrm{III}\right)2^{\mathrm{n}}={6\mathrm{q}}_1+2$$$$\left(\mathrm{I}\right):1+{\left(2^{{\mathrm{q}}_1}\right)}^6+{\left(2^{{2\mathrm{q}}_1}\right)}^6\equiv 3\equiv 0\left(\mathrm{mod}7\right)$$$$\left(\mathrm{II}\right):1+{\left(2^{{\mathrm{q}}_1}\right)}^6.2^1+{\left(2^{{2\mathrm{q}}_1}\right)}^6.2^2\equiv 1+2+4\equiv 0\left(\mathrm{mod}7\right)$$$$\left(\mathrm{III}\right):1+{\left(2^{{\mathrm{q}}_1}\right)}^6.2^2+{\left(2^{{2\mathrm{q}}_1}\right)}^6.2^4\equiv 1+4+16\equiv 0\left(\mathrm{mod}7\right)$$$$\mathrm{B}\mathrm{is}\mathrm{true}\mathrm{for}\mathrm{all}\mathrm{n}\in \mathbb{N}$$$$$$$$\mathrm{C}.1+2^{2^{\mathrm{n}}}+2^{2^{\mathrm{n}+1}}\mid 1+2^{2^{\mathrm{n}+1}}+2^{2^{\mathrm{n}+2}}$$$$(....)$$$$$$$$\mathrm{D}.{\mathrm{s}}_{\mathrm{n}}^2>{\mathrm{s}}_{\mathrm{n}+1}\Rightarrow$$$${(1+2^{2^{\mathrm{n}}}+2^{2^{\mathrm{n}+1}})}^2>1+2^{2^{\mathrm{n}+1}}+2^{2^{\mathrm{n}+2}}$$$$1+2^{2^{\mathrm{n}+1}}+2^{2^{\mathrm{n}+2}}+2^{2^{\mathrm{n}}+1}+2^{2^{\mathrm{n}+1}+1}+2^{2^{\mathrm{n}}.3+1}>1+2^{2^{\mathrm{n}+1}}+2^{2^{\mathrm{n}+2}}$$$$\mathrm{D}\mathrm{is}\mathrm{true}\mathrm{for}\mathrm{all}\mathrm{n}\in \mathbb{N}$$$$$$$$\mathrm{E}.{12\mathrm{s}}_1{\mathrm{s}}_2...{\mathrm{s}}_{\mathrm{n}}<{\mathrm{s}}_{\mathrm{n}+1}$$$$(....)$$

Commented by deleteduser1 last updated on 26/Sep/22

$$C\Rightarrow 1+2^{2^n}+2^{2^{n+1}}\mid 1+2^{2^{n+1}}+2^{2^{n+2}}$$$$Let{2}^{2^n}=p$$$$\therefore C\Rightarrow 1+p+p^2\mid 1+p^2+p^4$$$$1+p^2+p^4={(p^2+1)}^2-p^2=(p^2+1-p)(p^2+1+p)$$$$\Rightarrow Cistrue$$

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