Find the value of integers x, y which satisfy 9(x^2 + y^2 + 1) + 6xy = 2001


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Question Number 165480 by naka3546 last updated on 02/Feb/22

$F i n d t h e v a l u e o f i n t e g e r s x, y w h i c h s a t i s f y$ $9 (x^{2} + y^{2} + 1) + 6 x y = 2001$
	Answered by mindispower last updated on 02/Feb/22
	$(3x+y)^2 +8y^2 =1992 a^2 +8b^2 =1992 a^2 =0[8]⇒4∣a a=4v ⇔2v^2 +b^2 =249=3.83 ⇒∣v∣≤11 2v^2 =248−4k−4k^2 =4(62−k−k^2 )⇒2∣v v=2u ⇔8u^2 +b^2 =249 ⇒∣u∣∈{0,1,2,3,4,5} ∣u∣=5⇒b^2 =49⇒b∈{7,−7} ∣u∣=4⇒b^2 =121⇒∣b∣∈{11,−11} ∣u∣=3⇒b^2 =177 not possibl ∣u∣∣=2⇒b^2 =217 not possibl ∣u∣=1⇒b^2 =248 not possible ∣u∣=0⇒b^2 =249 not possible a=4v=8u ∣a∣=32,∣b∣=11;a=∣40∣,∣b∣=7 ∣3x+y∣=32 ∣y∣=11 ∣3x+11∣=32,x=7 ∣3x−11∣=32,3x−11=−32=x=−7 ∣3x+y∣=40 ∣y∣=7,x=11,y=7 x=−11,y=−7 (x,y)={(11,7);(7,11);(−7,−11);(−11,−7)}$
	${(3 x + y)}^{2} + 8 y^{2} = 1992$ $a^{2} + 8 b^{2} = 1992$ $a^{2} = 0 [8] \Rightarrow 4 ∣ a$ $a = 4 v$ $\Leftrightarrow 2 v^{2} + b^{2} = 249 = 3.83$ $\Rightarrow∣ v ∣⩽ 11$ $2 v^{2} = 248 - 4 k - 4 k^{2} = 4 (62 - k - k^{2}) \Rightarrow 2 ∣ v$ $v = 2 u$ $\Leftrightarrow 8 u^{2} + b^{2} = 249$ $\Rightarrow∣ u ∣\in {0, 1, 2, 3, 4, 5}$ $∣ u ∣= 5 \Rightarrow b^{2} = 49 \Rightarrow b \in {7, - 7}$ $∣ u ∣= 4 \Rightarrow b^{2} = 121 \Rightarrow∣ b ∣\in {11, - 11}$ $∣ u ∣= 3 \Rightarrow b^{2} = 177 n o t p o s s i b l$ $∣ u ∣∣= 2 \Rightarrow b^{2} = 217 n o t p o s s i b l$ $∣ u ∣= 1 \Rightarrow b^{2} = 248 n o t p o s s i b l e$ $∣ u ∣= 0 \Rightarrow b^{2} = 249 n o t p o s s i b l e$ $a = 4 v = 8 u$ $∣ a ∣= 32, ∣ b ∣= 11; a =∣ 40 ∣, ∣ b ∣= 7$ $∣ 3 x + y ∣= 32$ $∣ y ∣= 11$ $∣ 3 x + 11 ∣= 32, x = 7$ $∣ 3 x - 11 ∣= 32, 3 x - 11 = - 32 = x = - 7$ $∣ 3 x + y ∣= 40$ $∣ y ∣= 7, x = 11, y = 7$ $x = - 11, y = - 7$ $(x, y) = {(11, 7); (7, 11); (- 7, - 11); (- 11, - 7)}$
	Commented by naka3546 last updated on 02/Feb/22

	$Thank you, sir .$
	Commented by mindispower last updated on 04/Mar/22

	$w i t h e p l e a s u r$
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