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Question Number 199133 by hardmath last updated on 28/Oct/23

$${\mathrm{a}}^2\mathrm{b}-1=1999$$$$\mathrm{how}\mathrm{many}\mathrm{natural}\mathrm{solutions}\mathrm{of}\mathrm{the}$$$$\mathrm{equation}(\mathrm{a},\mathrm{b})\mathrm{have}?$$

Answered by Rasheed.Sindhi last updated on 28/Oct/23

$${\mathrm{a}}^2\mathrm{b}-1=1999$$$${\mathrm{a}}^2\mathrm{b}=2000=2^4.5^3$$$$Possibilitiesfor{\mathrm{a}}^2$$$${\mathrm{a}}^2=1^{2k},2^2,2^4,5^2,2^2.5^2,2^4.5^2$$$$Sixnaturalsoutions.$$

Commented by hardmath last updated on 28/Oct/23

$$\mathrm{thank}\mathrm{you}\mathrm{professor}$$

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