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Question Number 155344 by mathdanisur last updated on 29/Sep/21

Solve for integers:    x^2  - 3x(y^2  + y - 1) + 4y^2  + 4y - 6 = 0

$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{integers}: \\ $$$$ \\ $$$$\mathrm{x}^{\mathrm{2}} \:-\:\mathrm{3x}\left(\mathrm{y}^{\mathrm{2}} \:+\:\mathrm{y}\:-\:\mathrm{1}\right)\:+\:\mathrm{4y}^{\mathrm{2}} \:+\:\mathrm{4y}\:-\:\mathrm{6}\:=\:\mathrm{0} \\ $$

Answered by ghimisi last updated on 29/Sep/21

y^2 +y−1=t≥−1  x^2 −3xt+4t−2=0  if t=−1⇒x^2 +3x−6=0⇒x∉Z  △=9t^2 −16t+8=k^2    if t≥0⇒(3t−3)^2 <△<(3t−1)^2 ⇒△=(3t−2)^2   ⇒9t^2 −16t+8=9t^2 −12t+4⇒t=1⇒  y^2 −y−1=1  x^2 −3x+2=0⇒  (1,2);(1,−1);(2,2);(2,−1)

$${y}^{\mathrm{2}} +{y}−\mathrm{1}={t}\geqslant−\mathrm{1} \\ $$$${x}^{\mathrm{2}} −\mathrm{3}{xt}+\mathrm{4}{t}−\mathrm{2}=\mathrm{0} \\ $$$${if}\:{t}=−\mathrm{1}\Rightarrow{x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{6}=\mathrm{0}\Rightarrow{x}\notin{Z} \\ $$$$\bigtriangleup=\mathrm{9}{t}^{\mathrm{2}} −\mathrm{16}{t}+\mathrm{8}={k}^{\mathrm{2}} \\ $$$$\:{if}\:{t}\geqslant\mathrm{0}\Rightarrow\left(\mathrm{3}{t}−\mathrm{3}\right)^{\mathrm{2}} <\bigtriangleup<\left(\mathrm{3}{t}−\mathrm{1}\right)^{\mathrm{2}} \Rightarrow\bigtriangleup=\left(\mathrm{3}{t}−\mathrm{2}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{9}{t}^{\mathrm{2}} −\mathrm{16}{t}+\mathrm{8}=\mathrm{9}{t}^{\mathrm{2}} −\mathrm{12}{t}+\mathrm{4}\Rightarrow{t}=\mathrm{1}\Rightarrow \\ $$$${y}^{\mathrm{2}} −{y}−\mathrm{1}=\mathrm{1} \\ $$$${x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{2}=\mathrm{0}\Rightarrow \\ $$$$\left(\mathrm{1},\mathrm{2}\right);\left(\mathrm{1},−\mathrm{1}\right);\left(\mathrm{2},\mathrm{2}\right);\left(\mathrm{2},−\mathrm{1}\right) \\ $$

Commented by mathdanisur last updated on 29/Sep/21

Very nice solution Ser thankyou

$$\mathrm{Very}\:\mathrm{nice}\:\mathrm{solution}\:\boldsymbol{\mathrm{S}}\mathrm{er}\:\mathrm{thankyou} \\ $$

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