what is the number of ordered pairs of positif integers (x,y) that satisfy x^2 +y^2 −xy=37


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Question Number 100178 by bobhans last updated on 25/Jun/20

$what is the number of ordered pairs of positif$ $integers (x, y) that satisfy x^{2} + y^{2} - xy = 37$
	Answered by 1549442205 last updated on 25/Jun/20
	$x^2 −xy+y^2 −37=0.We look at it like a quadratic equation respect with x.Then Δ=y^2 −4y^2 +148=−3y^2 +148.The above eqs.always has roots ,so Δ=−3y^2 +148≥0 ⇒y^2 ≤((148)/3).Since y∈N^(∗ ) ,y∈{1,2,...,7}. i)If y∈{1,2,5,6}then Δ∈{145,136,73,40} aren′t perfect numbers,so are rejected ii)for y=3⇒Δ=121=11^2 ⇒x=((3+11)/2)=7(we reject negative root) we get the root (x,y)=(7;3) iii)for y=4 ⇒Δ=100=10^2 ⇒x=((4+10)/2)=7 we get the root (x;y)=(7;4) iv)for y=7 ⇒Δ=1⇒x=((7±1)/2)⇒x∈{4;3} we get two roots (x;y)∈{(4;7);(3;7)} Thus ,the given equation has four roots (x;y)∈{(7;3);(7;4);(3;7);(4;7)}$
	$x^{2} - xy + y^{2} - 37 = 0 . We look at it like a$ $quadratic equation respect with x . Then$ $Δ = y^{2} - {4 y}^{2} + 148 = - {3 y}^{2} + 148 . The above$ $eqs . always has roots, so Δ = - {3 y}^{2} + 148 ⩾ 0$ $\Rightarrow y^{2} ⩽ \frac{148}{3} . Since y \in N^{*}, y \in {1, 2, . . ., 7} .$ $i) If y \in {1, 2, 5, 6} then Δ \in {145, 136, 73, 40}$ ${aren}^{'} t perfect numbers, so are rejected$ $ii) for y = 3 \Rightarrow Δ = 121 = 11^{2} \Rightarrow x = \frac{3 + 11}{2} = 7 (we reject negative root)$ $we get the root (x, y) = (7; 3)$ $iii) for y = 4 \Rightarrow Δ = 100 = 10^{2} \Rightarrow x = \frac{4 + 10}{2} = 7$ $we get the root (x; y) = (7; 4)$ $iv) for y = 7 \Rightarrow Δ = 1 \Rightarrow x = \frac{7 \pm 1}{2} \Rightarrow x \in {4; 3}$ $we get two roots (x; y) \in {(4; 7); (3; 7)}$ $Thus, the given equation has four roots$ $(x; y) \in {(7; 3); (7; 4); (3; 7); (4; 7)}$
	Commented by bemath last updated on 25/Jun/20

	$how with (4, - 3), (- 4, 3), (3, - 4), (- 3, 4) ?$
	Commented by 1549442205 last updated on 25/Jun/20

	$your problem require to find the pairs of$ $positive integers$
	Answered by Rasheed.Sindhi last updated on 25/Jun/20

	$\underline{A Novel Method}$ $x^{2} + y^{2} - xy = 37$ ${(x - y)}^{2} = 37 - xy . . . . . . . . . . . . A$ ${(x + y)}^{2} = 37 + 3 xy . . . . . . . . . . . . B$ $x, y \in Z^{+} \land 37 - xy is perfect square$ $37 - xy = 36, 25, 16, 9, 4, 1 (possible values)$ $xy = 1, 12, 21, 28, 33, 36$ $From above values of xy acceptable$ $values are those for which 37 + 3 xy$ $is also perfect square are 21 & 28$ $When xy = 21$ $A \Rightarrow x - y = 4 (assume x ⩾ y)$ $B \Rightarrow x + y = 10$ $x = 7, y = 3$ $(or x = 3, y = 7 due to symmetry)$ $When xy = 28$ $A \Rightarrow x - y = 3$ $B \Rightarrow x + y = 11$ $x = 7, y = 4$ $(or x = 4, y = 7 due to symmetry)$ $(7, 3), (3, 7), (7, 4), (4, 7)$
	Commented by mr W last updated on 25/Jun/20

	$n i c e s o l u t i o n s i r!$
	Commented by bobhans last updated on 26/Jun/20

	$thank you sir$
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