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

Determine all pairs (x,y) of integers satisfying 1+2^x +2^(2x+1) =y^2

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

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

2^x (1+2^(x+1) ) = (y−1)(y+1) show that the factors y−1 and y+1 are even, exactly one of them divisible by 4. Hence x≥3 and one of these factors is divisible by 2^(x−1) but not by 2^x . so y = 2^(x−1) m+ε , m odd , ε = ±1 plugging this into the original equation we obtain 2^x (1+2^(x+1) )=(2^(x−1) m+ε)^2 −1 = 2^(2x−2) m^2 +2^x mε or equivalently 1+2^(x+1) =2^(x−2) m^2 +mε ⇔ thus we have the complete list solutions (x,y) : (0,2),(0,−2), (4,23),(4,−23).

$$\mathrm{2}^{{x}} \left(\mathrm{1}+\mathrm{2}^{{x}+\mathrm{1}} \right)\:=\:\left({y}−\mathrm{1}\right)\left({y}+\mathrm{1}\right) \\ $$$${show}\:{that}\:{the}\:{factors}\:{y}−\mathrm{1}\:{and}\:{y}+\mathrm{1} \\ $$$${are}\:{even},\:\mathrm{exactly}\:\mathrm{one}\:\mathrm{of}\:\mathrm{them}\: \\ $$$$\mathrm{divisible}\:\mathrm{by}\:\mathrm{4}.\:\mathrm{Hence}\:\mathrm{x}\geqslant\mathrm{3}\:\mathrm{and}\:\mathrm{one} \\ $$$$\mathrm{of}\:\mathrm{these}\:\mathrm{factors}\:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{2}^{\mathrm{x}−\mathrm{1}} \\ $$$$\mathrm{but}\:\mathrm{not}\:\mathrm{by}\:\mathrm{2}^{\mathrm{x}} .\:\mathrm{so}\:\mathrm{y}\:=\:\mathrm{2}^{\mathrm{x}−\mathrm{1}} \mathrm{m}+\epsilon\:,\:\mathrm{m}\:\mathrm{odd}\:,\:\epsilon\:=\:\pm\mathrm{1} \\ $$$$\mathrm{plugging}\:\mathrm{this}\:\mathrm{into}\:\mathrm{the}\:\mathrm{original}\: \\ $$$$\mathrm{equation}\:\mathrm{we}\:\mathrm{obtain}\:\mathrm{2}^{\mathrm{x}} \left(\mathrm{1}+\mathrm{2}^{\mathrm{x}+\mathrm{1}} \right)=\left(\mathrm{2}^{\mathrm{x}−\mathrm{1}} \mathrm{m}+\epsilon\right)^{\mathrm{2}} −\mathrm{1} \\ $$$$=\:\mathrm{2}^{\mathrm{2x}−\mathrm{2}} \mathrm{m}^{\mathrm{2}} +\mathrm{2}^{\mathrm{x}} \mathrm{m}\epsilon \\ $$$$\mathrm{or}\:\mathrm{equivalently}\:\mathrm{1}+\mathrm{2}^{\mathrm{x}+\mathrm{1}} =\mathrm{2}^{\mathrm{x}−\mathrm{2}} \mathrm{m}^{\mathrm{2}} +\mathrm{m}\epsilon \\ $$$$\Leftrightarrow\:\mathrm{thus}\:\mathrm{we}\:\mathrm{have}\:\mathrm{the}\:\mathrm{complete} \\ $$$$\mathrm{list}\:\mathrm{solutions}\:\left(\mathrm{x},\mathrm{y}\right)\::\:\left(\mathrm{0},\mathrm{2}\right),\left(\mathrm{0},−\mathrm{2}\right), \\ $$$$\left(\mathrm{4},\mathrm{23}\right),\left(\mathrm{4},−\mathrm{23}\right).\: \\ $$

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

1+2^x +2^(2x+1) =y^2 2(2^x )^2 +2^x +1−y^2 =0 2^x =((−1±(√(1−8(1−y^2 ))))/4) =((−1±(√(−7+8y^2 ))))/4) 8y^2 −7=u^2 y^2 =((u^2 +7)/8) u^2 =0,1,4,9,16,.... u^2 =1⇒y=±1 u^2 =25⇒y=±2 u^2 =121⇒y=±4 Continue

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

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