solve for x, y ∈N (√x)+(√y)=(√(2023))


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Question Number 198103 by mr W last updated on 10/Oct/23

$s o l v e f o r x, y \in N$ $\sqrt{x} + \sqrt{y} = \sqrt{2023}$
	Answered by Safojon last updated on 10/Oct/23

	$\sqrt{7} + 16 \sqrt{7} = \sqrt{2023}$ $2 \sqrt{7} + 15 \sqrt{7} = \sqrt{2023}$ $3 \sqrt{7} + 14 \sqrt{7} = \sqrt{2023}$ $4 \sqrt{7} + 13 \sqrt{7} = \sqrt{2023}$ $5 \sqrt{7} + 12 \sqrt{7} = \sqrt{2023}$ $6 \sqrt{7} + 11 \sqrt{7} = \sqrt{2023}$ $7 \sqrt{7} + 10 \sqrt{7} = \sqrt{2023}$ $8 \sqrt{7} + 9 \sqrt{7} = \sqrt{2023}$
	Commented by mr W last updated on 10/Oct/23
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	Answered by witcher3 last updated on 10/Oct/23
	$x+y=2023−2(√(xy)) (√(xy))∈N (√(ab))=n⇒ab=n^2 d=gcd(a,b)⇒a′b′=c^2 ,a′∧b′=1 c=Π_(i=1) ^r p_i ^a_i ,c^2 =Π_(i=1) ^n p_i ^(2a_i ) ⇒b′=t^2 ,a′=r^2 a=dr^2 ,b=dt^2 x=dr^2 ...x>y y=dt^2 ⇒d(r^2 +t^2 +2rt)=2023 d(r+t)^2 =2023=7.17^2 ⇒d=7 r−t=17,r>t r−t=17⇒(r,t)∈S={(17,0);)(16,1);(15,2);(14,3);(13,4) (12,5);(11,6);(10,7);(9,8)} (x,y)=(7r^2 ,7t^2 )(r,t)∈S^2 withe permutation (y,x)$
	$x + y = 2023 - 2 \sqrt{xy}$ $\sqrt{xy} \in N$ $\sqrt{ab} = n \Rightarrow ab = n^{2}$ $d = \gcd (a, b) \Rightarrow a^{'} b^{'} = c^{2}, a^{'} \land b^{'} = 1$ $c = \underset{i = 1}{\prod^{r}} p_{i}^{a_{i}}, c^{2} = \underset{i = 1}{\prod^{n}} p_{i}^{{2 a}_{i}}$ $\Rightarrow b^{'} = t^{2}, a^{'} = r^{2}$ $a = {dr}^{2}, b = {dt}^{2}$ $x = {dr}^{2} . . . x > y$ $y = {dt}^{2}$ $\Rightarrow d (r^{2} + t^{2} + 2 rt) = 2023$ $d {(r + t)}^{2} = 2023 = 7 . 17^{2}$ $\Rightarrow d = 7$ $r - t = 17, r > t$ $r - t = 17 \Rightarrow (r, t) \in S = {(17, 0);) (16, 1); (15, 2); (14, 3); (13, 4)$ $(12, 5); (11, 6); (10, 7); (9, 8)}$ $(x, y) = ({7 r}^{2}, {7 t}^{2}) (r, t) \in S^{2} withe permutation (y, x)$
	Commented by mr W last updated on 10/Oct/23
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	Answered by universe last updated on 10/Oct/23
	$(√x)+(√y) =(√(2023)) ((√y))^2 = ((√(2023))−(√x))^2 y = 2023 −2(√(2023x)) +x (√(2023x)) = integer (√(289×7x)) = 17(√(7x))_(integer) x = 7m^2 ; m∈z similarly y = 7n^2 ; n ∈z mn∈z (√(7m^2 )) + (√(7n^2 )) = 17(√7) m+n = 17 m,n = {(0,17),(1,16),(2,15) ......} so (x,y) = {(0,2023),(7,1792)....}$
	$\sqrt{x} + \sqrt{y} = \sqrt{2023}$ ${(\sqrt{y})}^{2} = {(\sqrt{2023} - \sqrt{x})}^{2}$ $y = 2023 - 2 \sqrt{2023 x} + x$ $\sqrt{2023 x} = i n t e g e r$ $\sqrt{289 \times 7 x} = 17 \underset{i n t e g e r}{\underset{⏟}{\sqrt{7 x}}}$ $x = 7 m^{2}; m \in z$ $s i m i l a r l y y = 7 n^{2}; n \in z$ $m n \in z$ $\sqrt{7 m^{2}} + \sqrt{7 n^{2}} = 17 \sqrt{7}$ $m + n = 17$ $m, n = {(0, 17), (1, 16), (2, 15) . . . . . .}$ $s o$ $(x, y) = {(0, 2023), (7, 1792) . . . .}$
	Commented by mr W last updated on 10/Oct/23
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