Question: 2^x +2^(2x+1) +1=y^2 solve this equation if x,y𝛆Z


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Question Number 118452 by 2004 last updated on 17/Oct/20

$Q u e s t i o n :$ $2^{x} + 2^{2 x + 1} + 1 = y^{2} s o l v e t h i s e q u a t i o n i f$ $x, y ϵ Z$
Commented by prakash jain last updated on 18/Oct/20

$2^{x} + 2^{2 x + 1} + 1 = y^{2}$ $x = 0, y = \pm 2 is obvious solution$ $x > 0 LHS is odd$ $x = 1 has no solution .$ $x ⩾ 2$ $y = 2 k + 1$ $2^{x} + 2^{2 x + 1} = 4 k^{2} + 4 k$ $k (k + 1) = 2^{2 x - 1} + 2^{x - 2} = 2^{x - 2} (2^{x + 1} + 1)$ $w i l l c o n t i n u e$
Commented by 2004 last updated on 18/Oct/20

$Can you continue sir (I could come to this step)$
Commented by 2004 last updated on 18/Oct/20

$T h a n k s$
	Answered by floor(10²Eta[1]) last updated on 18/Oct/20
	$if x<0⇒x≤−1 ⇒2^x ≤(1/2), 2^(2x+1) ≤(1/2) 1<1+2^x +2^(2x+1) ≤1+(1/2)+(1/2)=2 ⇒1<y^2 ≤2⇒no solution for x<0. if x=0: ⇒1+2^x +2^(2x+1) =4=y^2 ⇒y=±2 (x, y)={(0, −2), (0, 2)} if x=1: ⇒1+2+2^3 =11=y^2 ⇒no sol. x=2: 1+2^2 +2^5 =37=y^2 ⇒no sol. Suppose x≥3: 1+2^x +2^(2x+1) =y^2 2^x +2^(2x+1) =y^2 −1=(y+1)(y+1) 2^x (1+2^(x+1) )=(y+1)(y−1) LHS is even⇒RHS is also even but y+1 and y−1 have the same parity ⇒y+1 and y−1 are even ⇒gcd(y+1,y−1)∣2 ⇒gcd(y+1,y−1)=2 ⇒y+1 or y−1 is divisible by 2 but not a larger power of 2 like 4 (y+1)(y−1)=2.2^(x−1) (1+2^(x+1) ) so y+1 or y−1 is a multiple of 2^(x−1) Case 1: y+1=2^(x−1) .m (m≥1, odd) (If m was even then that would imply that y−1 was odd because y+1 would take all of the even part of the RHS, but we know that they are both even. And m≥1 because if (x,y) is a sol. so is (x, −y), so we only need to look for y≥0) ⇒y=2^(x−1) .m−1 1+2^x +2^(2x+1) =(2^(x−1) .m−1)^2 =2^(2x−2) .m^2 −2^x .m+1 ⇒1+2^(x+1) =2^(x−2) .m^2 −m m+1=2^(x−2) .m^2 −2^(x+1) =2^(x−2) (m^2 −8) m+1=2^(x−2) (m^2 −8)≥2(m^2 −8) 2m^2 −m−17≤0 ((1−(√(137)))/2)≤m≤((1+(√(137)))/2) but since m≥1: 1≤m≤((1+(√(137)))/4)<4⇒m=1 or m=3 m=1⇒2=2^(x−2) (−7) (no sol.) m=3⇒4=2^(x−2) ⇒x=4 1+2^4 +2^9 =y^2 ⇒y=±23 Case 2: y−1=2^(x−1) .m (m≥1, odd) y=2^(x−1) .m+1 1+2^x +2^(2x+1) =(2^(x−1) .m+1)^2 =2^(2x−2) .m^2 +2^x .m+1 1+2^(x+1) =2^(x−2) .m^2 +m 1−m=2^(x−2) .m^2 −2^(x+1) =2^(x−2) (m^2 −8) 0≥1−m=2^(x−1) (m^2 −8) ⇒m^2 −8≤0⇒m=1 (doesn′t work) all the solutions are: (x, y)=(0, −2), (0, 2), (4, −23), (4, 23)$
	$if x < 0 \Rightarrow x ⩽ - 1$ $\Rightarrow 2^{x} ⩽ \frac{1}{2}, 2^{2 x + 1} ⩽ \frac{1}{2}$ $1 < 1 + 2^{x} + 2^{2 x + 1} ⩽ 1 + \frac{1}{2} + \frac{1}{2} = 2$ $\Rightarrow 1 < y^{2} ⩽ 2 \Rightarrow no solution for x < 0 .$ $if x = 0 :$ $\Rightarrow 1 + 2^{x} + 2^{2 x + 1} = 4 = y^{2} \Rightarrow y = \pm 2$ $(x, y) = {(0, - 2), (0, 2)}$ $if x = 1 :$ $\Rightarrow 1 + 2 + 2^{3} = 11 = y^{2} \Rightarrow no sol .$ $x = 2 :$ $1 + 2^{2} + 2^{5} = 37 = y^{2} \Rightarrow no sol .$ $Suppose x ⩾ 3 :$ $1 + 2^{x} + 2^{2 x + 1} = y^{2}$ $2^{x} + 2^{2 x + 1} = y^{2} - 1 = (y + 1) (y + 1)$ $2^{x} (1 + 2^{x + 1}) = (y + 1) (y - 1)$ $LHS is even \Rightarrow RHS is also even but$ $y + 1 and y - 1 have the same parity$ $\Rightarrow y + 1 and y - 1 are even$ $\Rightarrow \gcd (y + 1, y - 1) ∣ 2$ $\Rightarrow \gcd (y + 1, y - 1) = 2$ $\Rightarrow y + 1 or y - 1 is divisible by 2 but not a$ $larger power of 2 like 4$ $(y + 1) (y - 1) = 2 . 2^{x - 1} (1 + 2^{x + 1})$ $so y + 1 or y - 1 is a multiple of 2^{x - 1}$ $Case 1 : y + 1 = 2^{x - 1} . m (m ⩾ 1, odd)$ $(If m was even then that would imply$ $that y - 1 was odd because y + 1 would take$ $all of the even part of the RHS, but$ $we know that they are both even .$ $And m ⩾ 1 because if (x, y) is a sol . so is$ $(x, - y), so we only need to look for y ⩾ 0)$ $\Rightarrow y = 2^{x - 1} . m - 1$ $1 + 2^{x} + 2^{2 x + 1} = {(2^{x - 1} . m - 1)}^{2} = 2^{2 x - 2} . m^{2} - 2^{x} . m + 1$ $\Rightarrow 1 + 2^{x + 1} = 2^{x - 2} . m^{2} - m$ $m + 1 = 2^{x - 2} . m^{2} - 2^{x + 1} = 2^{x - 2} (m^{2} - 8)$ $m + 1 = 2^{x - 2} (m^{2} - 8) ⩾ 2 (m^{2} - 8)$ ${2 m}^{2} - m - 17 ⩽ 0$ $\frac{1 - \sqrt{137}}{2} ⩽ m ⩽ \frac{1 + \sqrt{137}}{2}$ $but since m ⩾ 1 :$ $1 ⩽ m ⩽ \frac{1 + \sqrt{137}}{4} < 4 \Rightarrow m = 1 or m = 3$ $m = 1 \Rightarrow 2 = 2^{x - 2} (- 7) (no sol .)$ $m = 3 \Rightarrow 4 = 2^{x - 2} \Rightarrow x = 4$ $1 + 2^{4} + 2^{9} = y^{2} \Rightarrow y = \pm 23$ $Case 2 : y - 1 = 2^{x - 1} . m (m ⩾ 1, odd)$ $y = 2^{x - 1} . m + 1$ $1 + 2^{x} + 2^{2 x + 1} = {(2^{x - 1} . m + 1)}^{2} = 2^{2 x - 2} . m^{2} + 2^{x} . m + 1$ $1 + 2^{x + 1} = 2^{x - 2} . m^{2} + m$ $1 - m = 2^{x - 2} . m^{2} - 2^{x + 1} = 2^{x - 2} (m^{2} - 8)$ $0 ⩾ 1 - m = 2^{x - 1} (m^{2} - 8)$ $\Rightarrow m^{2} - 8 ⩽ 0 \Rightarrow m = 1 ({doesn}^{'} t work)$ $all the solutions are :$ $(x, y) = (0, - 2), (0, 2), (4, - 23), (4, 23)$
	Commented by 1549442205PVT last updated on 18/Oct/20
	$If so such then 1+2^x =2^(x−3) .m^2 −m your argument later still true.You can go by path base on system { ((2^(x−1) .m=y+1)),((((2.(2^(x+1) +1))/m)=y−1)) :} Thank you$
	$If so such then 1 + 2^{x} = 2^{x - 3} . m^{2} - m$ $your argument later still true . You$ $can go by path base on system$ ${\begin{cases} 2^{x - 1} . m = y + 1 \\ \frac{2 . (2^{x + 1} + 1)}{m} = y - 1 \end{cases}$ $Thank you$
	Answered by TANMAY PANACEA last updated on 18/Oct/20

	$b y i n s p e c t i o n$ $x = 0 y = 2$
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