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Question Number 196576 by sonukgindia last updated on 27/Aug/23

Commented by Frix last updated on 27/Aug/23

$$\mathrm{Infinite}\mathrm{solutions}\mathrm{for}x,y\in \mathbb{R}$$$$\mathrm{No}\mathrm{solution}\mathrm{for}x,y\in \mathbb{Z}$$

Answered by sniper237 last updated on 27/Aug/23

$$Suchnumbers{can}^{\prime }texist!!!$$$$4^x\equiv 1\left[3\right];{48}^{12}\equiv 0\left[3\right]\Rightarrow 1-6^y\equiv 0\left[3\right]$$$$Soy=0.Then{4}^x=1+{48}^{12}$$$${48}^{12}\equiv 0\left[2\right]\Rightarrow 4^x\equiv 1\left[2\right]\Rightarrow x=0$$$$But{4}^0-6^0\ne {48}^{12}.$$

Commented by deleteduser1 last updated on 27/Aug/23

$${That}^{\prime }sifx,y\in \mathbb{Z}$$

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