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Question Number 111503 by Aina Samuel Temidayo last updated on 04/Sep/20

$$\mathrm{Find}\mathrm{four}\mathrm{values}\mathrm{of}\mathrm{n}\mathrm{satisfying}$$$$1⩽\mathrm{n}⩽2000\mathrm{and}{2}^{\mathrm{n}}={\mathrm{n}}^2\left(\mathrm{mod}1024\right)$$

Answered by 1549442205PVT last updated on 04/Sep/20

$$2^{\mathrm{n}}={\mathrm{n}}^2\left(\mathrm{mod}1024\right)\iff 2^{\mathrm{n}}-{\mathrm{n}}^2=\mathrm{k}.2^{10}$$$$\iff {\mathrm{n}}^2=2^{\mathrm{n}}-\mathrm{k}.2^{10}=2^{10}(2^{\mathrm{n}-10}-\mathrm{k})(\mathrm{k}\in {\mathbb{N}}^{\ast })$$$$\iff {\left(\frac{\mathrm{n}}{32}\right)}^2=2^{\mathrm{n}-10}-\mathrm{k}.\Rightarrow \mathrm{n}=32\mathrm{m}(\mathrm{m}\in {\mathbb{N}}^{\ast })$$$$1<\mathrm{n}⩽2000\Rightarrow 32\mathrm{m}<2000\Rightarrow 1⩽\mathrm{m}⩽62$$$$\Rightarrow {\mathrm{m}}^2=2^{32\mathrm{m}-10}-\mathrm{k}\left(1\right)$$$$\mathrm{m}=1\Rightarrow \mathrm{n}=32,\mathrm{k}=2^{22}-1$$$$\mathrm{m}=2\Rightarrow \mathrm{n}=64,\mathrm{k}=2^{54}-4$$$$\mathrm{m}=3\Rightarrow \mathrm{n}=96,\mathrm{k}=2^{86}-9$$$$\mathrm{m}=4\Rightarrow \mathrm{n}=128,\mathrm{k}=2^{118}-16$$$$\mathrm{Thus},\mathrm{we}\mathrm{found}\mathrm{four}\mathrm{values}\mathrm{of}\mathrm{n}$$$$\mathrm{satisfying}{2}^{\mathrm{n}}={\mathrm{n}}^2\left(\mathrm{mod}1024\right)\mathrm{are}$$$$\mathrm{n}\in \{32,64,96,128\}$$

Commented by Aina Samuel Temidayo last updated on 04/Sep/20

$$\mathrm{Thanks}.$$

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