Question Number 196576 by sonukgindia last updated on 27/Aug/23 | ||
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Commented by Frix last updated on 27/Aug/23 | ||
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$$\mathrm{Infinite}\:\mathrm{solutions}\:\mathrm{for}\:{x},\:{y}\:\in\mathbb{R} \\ $$$$\mathrm{No}\:\mathrm{solution}\:\mathrm{for}\:{x},\:{y}\:\in\mathbb{Z} \\ $$ | ||
Answered by sniper237 last updated on 27/Aug/23 | ||
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$${Such}\:{numbers}\:{can}'{t}\:{exist}!!! \\ $$$$\mathrm{4}^{{x}} \equiv\mathrm{1}\left[\mathrm{3}\right]\:;\:\mathrm{48}^{\mathrm{12}} \equiv\mathrm{0}\left[\mathrm{3}\right]\Rightarrow\:\mathrm{1}−\mathrm{6}^{{y}} \equiv\mathrm{0}\left[\mathrm{3}\right] \\ $$$${So}\:\:{y}=\mathrm{0}.\:{Then}\:\:\mathrm{4}^{{x}} =\mathrm{1}+\mathrm{48}^{\mathrm{12}} \\ $$$$\mathrm{48}^{\mathrm{12}} \equiv\mathrm{0}\left[\mathrm{2}\right]\Rightarrow\:\mathrm{4}^{{x}} \equiv\mathrm{1}\left[\mathrm{2}\right]\Rightarrow{x}=\mathrm{0} \\ $$$${But}\:\:\mathrm{4}^{\mathrm{0}} −\mathrm{6}^{\mathrm{0}} \neq\mathrm{48}^{\mathrm{12}} .\: \\ $$ | ||
Commented by deleteduser1 last updated on 27/Aug/23 | ||
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$${That}'{s}\:{if}\:{x},{y}\in\mathbb{Z} \\ $$ | ||