Question Number 165480 by naka3546 last updated on 02/Feb/22
$${Find}\:\:{the}\:\:{value}\:\:{of}\:\:{integers}\:\:{x},\:{y}\:\:{which}\:\:{satisfy}\: \\ $$$$\:\:\:\:\mathrm{9}\left({x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:+\:\mathrm{1}\right)\:+\:\mathrm{6}{xy}\:\:=\:\mathrm{2001} \\ $$
Answered by mindispower last updated on 02/Feb/22
![(3x+y)^2 +8y^2 =1992 a^2 +8b^2 =1992 a^2 =0[8]⇒4∣a a=4v ⇔2v^2 +b^2 =249=3.83 ⇒∣v∣≤11 2v^2 =248−4k−4k^2 =4(62−k−k^2 )⇒2∣v v=2u ⇔8u^2 +b^2 =249 ⇒∣u∣∈{0,1,2,3,4,5} ∣u∣=5⇒b^2 =49⇒b∈{7,−7} ∣u∣=4⇒b^2 =121⇒∣b∣∈{11,−11} ∣u∣=3⇒b^2 =177 not possibl ∣u∣∣=2⇒b^2 =217 not possibl ∣u∣=1⇒b^2 =248 not possible ∣u∣=0⇒b^2 =249 not possible a=4v=8u ∣a∣=32,∣b∣=11;a=∣40∣,∣b∣=7 ∣3x+y∣=32 ∣y∣=11 ∣3x+11∣=32,x=7 ∣3x−11∣=32,3x−11=−32=x=−7 ∣3x+y∣=40 ∣y∣=7,x=11,y=7 x=−11,y=−7 (x,y)={(11,7);(7,11);(−7,−11);(−11,−7)}](Q165484.png)
$$\left(\mathrm{3}{x}+{y}\right)^{\mathrm{2}} +\mathrm{8}{y}^{\mathrm{2}} =\mathrm{1992} \\ $$$${a}^{\mathrm{2}} +\mathrm{8}{b}^{\mathrm{2}} =\mathrm{1992} \\ $$$${a}^{\mathrm{2}} =\mathrm{0}\left[\mathrm{8}\right]\Rightarrow\mathrm{4}\mid{a} \\ $$$${a}=\mathrm{4}{v} \\ $$$$\Leftrightarrow\mathrm{2}{v}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{249}=\mathrm{3}.\mathrm{83} \\ $$$$\Rightarrow\mid{v}\mid\leqslant\mathrm{11}\: \\ $$$$\mathrm{2}{v}^{\mathrm{2}} =\mathrm{248}−\mathrm{4}{k}−\mathrm{4}{k}^{\mathrm{2}} =\mathrm{4}\left(\mathrm{62}−{k}−{k}^{\mathrm{2}} \right)\Rightarrow\mathrm{2}\mid{v} \\ $$$${v}=\mathrm{2}{u} \\ $$$$\Leftrightarrow\mathrm{8}{u}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{249} \\ $$$$\Rightarrow\mid{u}\mid\in\left\{\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5}\right\} \\ $$$$\mid{u}\mid=\mathrm{5}\Rightarrow{b}^{\mathrm{2}} =\mathrm{49}\Rightarrow{b}\in\left\{\mathrm{7},−\mathrm{7}\right\} \\ $$$$\mid{u}\mid=\mathrm{4}\Rightarrow{b}^{\mathrm{2}} =\mathrm{121}\Rightarrow\mid{b}\mid\in\left\{\mathrm{11},−\mathrm{11}\right\} \\ $$$$\mid{u}\mid=\mathrm{3}\Rightarrow{b}^{\mathrm{2}} =\mathrm{177}\:{not}\:{possibl} \\ $$$$\mid{u}\mid\mid=\mathrm{2}\Rightarrow{b}^{\mathrm{2}} =\mathrm{217}\:{not}\:{possibl} \\ $$$$\mid{u}\mid=\mathrm{1}\Rightarrow{b}^{\mathrm{2}} =\mathrm{248}\:{not}\:{possible} \\ $$$$\mid{u}\mid=\mathrm{0}\Rightarrow{b}^{\mathrm{2}} =\mathrm{249}\:{not}\:{possible} \\ $$$${a}=\mathrm{4}{v}=\mathrm{8}{u} \\ $$$$\mid{a}\mid=\mathrm{32},\mid{b}\mid=\mathrm{11};{a}=\mid\mathrm{40}\mid,\mid{b}\mid=\mathrm{7} \\ $$$$\mid\mathrm{3}{x}+{y}\mid=\mathrm{32} \\ $$$$\mid{y}\mid=\mathrm{11} \\ $$$$\mid\mathrm{3}{x}+\mathrm{11}\mid=\mathrm{32},{x}=\mathrm{7} \\ $$$$\mid\mathrm{3}{x}−\mathrm{11}\mid=\mathrm{32},\mathrm{3}{x}−\mathrm{11}=−\mathrm{32}={x}=−\mathrm{7} \\ $$$$\mid\mathrm{3}{x}+{y}\mid=\mathrm{40} \\ $$$$\mid{y}\mid=\mathrm{7},{x}=\mathrm{11},{y}=\mathrm{7} \\ $$$${x}=−\mathrm{11},{y}=−\mathrm{7} \\ $$$$\left({x},{y}\right)=\left\{\left(\mathrm{11},\mathrm{7}\right);\left(\mathrm{7},\mathrm{11}\right);\left(−\mathrm{7},−\mathrm{11}\right);\left(−\mathrm{11},−\mathrm{7}\right)\right\} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by naka3546 last updated on 02/Feb/22
$$\mathrm{Thank}\:\:\mathrm{you},\:\:\mathrm{sir}. \\ $$
Commented by mindispower last updated on 04/Mar/22
$${withe}\:{pleasur} \\ $$