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Question Number 97746 by bemath last updated on 09/Jun/20

$$\mathrm{Determine}\mathrm{all}\mathrm{pairs}(\mathrm{x},\mathrm{y})$$$$\mathrm{of}\mathrm{integers}\mathrm{satisfying}1+2^{\mathrm{x}}+2^{2\mathrm{x}+1}={\mathrm{y}}^2$$

Commented by john santu last updated on 09/Jun/20

$$2^x(1+2^{x+1})=(y-1)(y+1)$$$$showthatthefactorsy-1andy+1$$$$areeven,\mathrm{exactly}\mathrm{one}\mathrm{of}\mathrm{them}$$$$\mathrm{divisible}\mathrm{by}4.\mathrm{Hence}\mathrm{x}⩾3\mathrm{and}\mathrm{one}$$$$\mathrm{of}\mathrm{these}\mathrm{factors}\mathrm{is}\mathrm{divisible}\mathrm{by}{2}^{\mathrm{x}-1}$$$$\mathrm{but}\mathrm{not}\mathrm{by}{2}^{\mathrm{x}}.\mathrm{so}\mathrm{y}=2^{\mathrm{x}-1}\mathrm{m}+ϵ,\mathrm{m}\mathrm{odd},ϵ=\pm 1$$$$\mathrm{plugging}\mathrm{this}\mathrm{into}\mathrm{the}\mathrm{original}$$$$\mathrm{equation}\mathrm{we}\mathrm{obtain}{2}^{\mathrm{x}}(1+2^{\mathrm{x}+1})={(2^{\mathrm{x}-1}\mathrm{m}+ϵ)}^2-1$$$$=2^{2\mathrm{x}-2}{\mathrm{m}}^2+2^{\mathrm{x}}\mathrm{m}ϵ$$$$\mathrm{or}\mathrm{equivalently}1+2^{\mathrm{x}+1}=2^{\mathrm{x}-2}{\mathrm{m}}^2+\mathrm{m}ϵ$$$$\iff \mathrm{thus}\mathrm{we}\mathrm{have}\mathrm{the}\mathrm{complete}$$$$\mathrm{list}\mathrm{solutions}(\mathrm{x},\mathrm{y}):(0,2),(0,-2),$$$$(4,23),(4,-23).$$

Answered by Rasheed.Sindhi last updated on 18/Jun/20

$$1+2^{\mathrm{x}}+2^{2\mathrm{x}+1}={\mathrm{y}}^2$$$$2{\left(2^x\right)}^2+2^x+1-y^2=0$$$$2^x=\frac{-1\pm \sqrt{1-8(1-y^2)}}{4}$$$$=\frac{-1\pm \sqrt{-7+8y^2)}}{4}$$$$8y^2-7=u^2$$$$y^2=\frac{u^2+7}{8}$$$$u^2=0,1,4,9,16,....$$$$u^2=1\Rightarrow y=\pm 1$$$$u^2=25\Rightarrow y=\pm 2$$$$u^2=121\Rightarrow y=\pm 4$$$$$$$$Continue$$

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