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Question Number 19413 by Tinkutara last updated on 10/Aug/17

How many integer pairs (x, y) satisfy  x^2  + 4y^2  − 2xy − 2x − 4y − 8 = 0?

$$\mathrm{How}\:\mathrm{many}\:\mathrm{integer}\:\mathrm{pairs}\:\left({x},\:{y}\right)\:\mathrm{satisfy} \\ $$$${x}^{\mathrm{2}} \:+\:\mathrm{4}{y}^{\mathrm{2}} \:−\:\mathrm{2}{xy}\:−\:\mathrm{2}{x}\:−\:\mathrm{4}{y}\:−\:\mathrm{8}\:=\:\mathrm{0}? \\ $$

Commented by mrW1 last updated on 11/Aug/17

(0,−1)  (−2,0)  (4,0)  (0,2)  (6,2)  (4,3)  is this correct?

$$\left(\mathrm{0},−\mathrm{1}\right) \\ $$$$\left(−\mathrm{2},\mathrm{0}\right) \\ $$$$\left(\mathrm{4},\mathrm{0}\right) \\ $$$$\left(\mathrm{0},\mathrm{2}\right) \\ $$$$\left(\mathrm{6},\mathrm{2}\right) \\ $$$$\left(\mathrm{4},\mathrm{3}\right) \\ $$$$\mathrm{is}\:\mathrm{this}\:\mathrm{correct}? \\ $$

Answered by mrW1 last updated on 11/Aug/17

x^2   − 2(y+1)x − 4(2+y−y^2 ) = 0  ⇒x=((2(y+1)±(√(4(y+1)^2 +16(2+y−y^2 ))))/2)  x=y+1±(√(y^2 +2y+1+8+4y−4y^2 ))  x=y+1±(√(3(3+2y−y^2 )))  x=y+1±(√(3[4−(y−1)^2 ]))  4−(y−1)^2 ≥0  (y−1)^2 ≤4  since x is integer, 4−(y−1)^2 =0 or 3  ⇒(y−1)^2 =4 or 1  ⇒y−1=±2 or ±1  ⇒y=1±2 or 1±1  ⇒y=3,−1,2,0  ⇒x=4,0,3±3,1±3  ⇒(x,y)=(4,3)/(0,−1)/(6,2)/(0,2)/(4,0)/(−2,0)

$${x}^{\mathrm{2}} \:\:−\:\mathrm{2}\left({y}+\mathrm{1}\right){x}\:−\:\mathrm{4}\left(\mathrm{2}+{y}−\mathrm{y}^{\mathrm{2}} \right)\:=\:\mathrm{0} \\ $$$$\Rightarrow\mathrm{x}=\frac{\mathrm{2}\left(\mathrm{y}+\mathrm{1}\right)\pm\sqrt{\mathrm{4}\left(\mathrm{y}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{16}\left(\mathrm{2}+\mathrm{y}−\mathrm{y}^{\mathrm{2}} \right)}}{\mathrm{2}} \\ $$$$\mathrm{x}=\mathrm{y}+\mathrm{1}\pm\sqrt{\mathrm{y}^{\mathrm{2}} +\mathrm{2y}+\mathrm{1}+\mathrm{8}+\mathrm{4y}−\mathrm{4y}^{\mathrm{2}} } \\ $$$$\mathrm{x}=\mathrm{y}+\mathrm{1}\pm\sqrt{\mathrm{3}\left(\mathrm{3}+\mathrm{2y}−\mathrm{y}^{\mathrm{2}} \right)} \\ $$$$\mathrm{x}=\mathrm{y}+\mathrm{1}\pm\sqrt{\mathrm{3}\left[\mathrm{4}−\left(\mathrm{y}−\mathrm{1}\right)^{\mathrm{2}} \right]} \\ $$$$\mathrm{4}−\left(\mathrm{y}−\mathrm{1}\right)^{\mathrm{2}} \geqslant\mathrm{0} \\ $$$$\left(\mathrm{y}−\mathrm{1}\right)^{\mathrm{2}} \leqslant\mathrm{4} \\ $$$$\mathrm{since}\:\mathrm{x}\:\mathrm{is}\:\mathrm{integer},\:\mathrm{4}−\left(\mathrm{y}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0}\:\mathrm{or}\:\mathrm{3} \\ $$$$\Rightarrow\left(\mathrm{y}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{4}\:\mathrm{or}\:\mathrm{1} \\ $$$$\Rightarrow\mathrm{y}−\mathrm{1}=\pm\mathrm{2}\:\mathrm{or}\:\pm\mathrm{1} \\ $$$$\Rightarrow\mathrm{y}=\mathrm{1}\pm\mathrm{2}\:\mathrm{or}\:\mathrm{1}\pm\mathrm{1} \\ $$$$\Rightarrow\mathrm{y}=\mathrm{3},−\mathrm{1},\mathrm{2},\mathrm{0} \\ $$$$\Rightarrow\mathrm{x}=\mathrm{4},\mathrm{0},\mathrm{3}\pm\mathrm{3},\mathrm{1}\pm\mathrm{3} \\ $$$$\Rightarrow\left(\mathrm{x},\mathrm{y}\right)=\left(\mathrm{4},\mathrm{3}\right)/\left(\mathrm{0},−\mathrm{1}\right)/\left(\mathrm{6},\mathrm{2}\right)/\left(\mathrm{0},\mathrm{2}\right)/\left(\mathrm{4},\mathrm{0}\right)/\left(−\mathrm{2},\mathrm{0}\right) \\ $$

Commented by Tinkutara last updated on 11/Aug/17

Thank you very much Sir!

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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