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Question Number 198103 by mr W last updated on 10/Oct/23
solve for x, y ∈N (√x)+(√y)=(√(2023))
$${solve}\:{for}\:{x},\:{y}\:\in{N} \\ $$$$\sqrt{{x}}+\sqrt{{y}}=\sqrt{\mathrm{2023}} \\ $$
Answered by Safojon last updated on 10/Oct/23
(√7)+16(√7)=(√(2023)) 2(√7)+15(√7)=(√(2023)) 3(√7)+14(√7)=(√(2023)) 4(√7)+13(√7)=(√(2023)) 5(√7)+12(√7)=(√(2023)) 6(√7)+11(√7)=(√(2023)) 7(√7)+10(√7)=(√(2023)) 8(√7)+9(√7)=(√(2023))
$$\sqrt{\mathrm{7}}+\mathrm{16}\sqrt{\mathrm{7}}=\sqrt{\mathrm{2023}} \\ $$$$\mathrm{2}\sqrt{\mathrm{7}}+\mathrm{15}\sqrt{\mathrm{7}}=\sqrt{\mathrm{2023}} \\ $$$$\mathrm{3}\sqrt{\mathrm{7}}+\mathrm{14}\sqrt{\mathrm{7}}=\sqrt{\mathrm{2023}} \\ $$$$\mathrm{4}\sqrt{\mathrm{7}}+\mathrm{13}\sqrt{\mathrm{7}}=\sqrt{\mathrm{2023}} \\ $$$$\mathrm{5}\sqrt{\mathrm{7}}+\mathrm{12}\sqrt{\mathrm{7}}=\sqrt{\mathrm{2023}} \\ $$$$\mathrm{6}\sqrt{\mathrm{7}}+\mathrm{11}\sqrt{\mathrm{7}}=\sqrt{\mathrm{2023}} \\ $$$$\mathrm{7}\sqrt{\mathrm{7}}+\mathrm{10}\sqrt{\mathrm{7}}=\sqrt{\mathrm{2023}} \\ $$$$\mathrm{8}\sqrt{\mathrm{7}}+\mathrm{9}\sqrt{\mathrm{7}}=\sqrt{\mathrm{2023}} \\ $$$$ \\ $$
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)
$$\mathrm{x}+\mathrm{y}=\mathrm{2023}−\mathrm{2}\sqrt{\mathrm{xy}} \\ $$$$\sqrt{\mathrm{xy}}\in\mathbb{N} \\ $$$$\sqrt{\mathrm{ab}}=\mathrm{n}\Rightarrow\mathrm{ab}=\mathrm{n}^{\mathrm{2}} \\ $$$$\mathrm{d}=\mathrm{gcd}\left(\mathrm{a},\mathrm{b}\right)\Rightarrow\mathrm{a}'\mathrm{b}'=\mathrm{c}^{\mathrm{2}} ,\mathrm{a}'\wedge\mathrm{b}'=\mathrm{1} \\ $$$$\mathrm{c}=\underset{\mathrm{i}=\mathrm{1}} {\overset{\mathrm{r}} {\prod}}\mathrm{p}_{\mathrm{i}} ^{\mathrm{a}_{\mathrm{i}} } ,\mathrm{c}^{\mathrm{2}} =\underset{\mathrm{i}=\mathrm{1}} {\overset{\mathrm{n}} {\prod}}\mathrm{p}_{\mathrm{i}} ^{\mathrm{2a}_{\mathrm{i}} } \\ $$$$\Rightarrow\mathrm{b}'=\mathrm{t}^{\mathrm{2}} ,\mathrm{a}'=\mathrm{r}^{\mathrm{2}} \\ $$$$\mathrm{a}=\mathrm{dr}^{\mathrm{2}} ,\mathrm{b}=\mathrm{dt}^{\mathrm{2}} \\ $$$$\mathrm{x}=\mathrm{dr}^{\mathrm{2}} …\mathrm{x}>\mathrm{y} \\ $$$$\mathrm{y}=\mathrm{dt}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{d}\left(\mathrm{r}^{\mathrm{2}} +\mathrm{t}^{\mathrm{2}} +\mathrm{2rt}\right)=\mathrm{2023} \\ $$$$\mathrm{d}\left(\mathrm{r}+\mathrm{t}\right)^{\mathrm{2}} =\mathrm{2023}=\mathrm{7}.\mathrm{17}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{d}=\mathrm{7} \\ $$$$\mathrm{r}−\mathrm{t}=\mathrm{17},\mathrm{r}>\mathrm{t} \\ $$$$\mathrm{r}−\mathrm{t}=\mathrm{17}\Rightarrow\left(\mathrm{r},\mathrm{t}\right)\in\mathrm{S}=\left\{\left(\mathrm{17},\mathrm{0}\right);\right)\left(\mathrm{16},\mathrm{1}\right);\left(\mathrm{15},\mathrm{2}\right);\left(\mathrm{14},\mathrm{3}\right);\left(\mathrm{13},\mathrm{4}\right) \\ $$$$\left.\left(\mathrm{12},\mathrm{5}\right);\left(\mathrm{11},\mathrm{6}\right);\left(\mathrm{10},\mathrm{7}\right);\left(\mathrm{9},\mathrm{8}\right)\right\} \\ $$$$\left(\mathrm{x},\mathrm{y}\right)=\left(\mathrm{7r}^{\mathrm{2}} ,\mathrm{7t}^{\mathrm{2}} \right)\left(\mathrm{r},\mathrm{t}\right)\in\mathrm{S}^{\mathrm{2}} \:\mathrm{withe}\:\mathrm{permutation}\:\left(\mathrm{y},\mathrm{x}\right) \\ $$$$ \\ $$
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)....}
$$ \\ $$$$\:\:\:\:\sqrt{{x}}+\sqrt{{y}}\:=\sqrt{\mathrm{2023}} \\ $$$$\:\:\:\left(\sqrt{{y}}\right)^{\mathrm{2}} \:=\:\left(\sqrt{\mathrm{2023}}−\sqrt{{x}}\right)^{\mathrm{2}} \\ $$$$\:\:\:{y}\:=\:\mathrm{2023}\:−\mathrm{2}\sqrt{\mathrm{2023}{x}}\:+{x} \\ $$$$\:\:\:\:\sqrt{\mathrm{2023}{x}}\:\:\:\:=\:\:\:{integer} \\ $$$$\:\:\:\:\:\sqrt{\mathrm{289}×\mathrm{7}{x}}\:\:=\:\:\:\mathrm{17}\underset{{integer}} {\underbrace{\sqrt{\mathrm{7}{x}}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}\:\:=\:\:\mathrm{7}{m}^{\mathrm{2}} \:\:\:\:;\:\:\:{m}\in{z} \\ $$$$\:{similarly}\:\:\:\:{y}\:\:\:=\:\:\mathrm{7}{n}^{\mathrm{2}} \:\:\:\:;\:\:\:\:{n}\:\in{z} \\ $$$$\:\:\:\:\:\:\:{mn}\in{z} \\ $$$$\:\:\:\:\:\sqrt{\mathrm{7}{m}^{\mathrm{2}} }\:+\:\sqrt{\mathrm{7}{n}^{\mathrm{2}} }\:=\:\mathrm{17}\sqrt{\mathrm{7}} \\ $$$$\:\:\:\:\:{m}+{n}\:=\:\mathrm{17} \\ $$$$\:{m},{n}\:=\:\left\{\left(\mathrm{0},\mathrm{17}\right),\left(\mathrm{1},\mathrm{16}\right),\left(\mathrm{2},\mathrm{15}\right)\:……\right\} \\ $$$${so} \\ $$$$\left({x},{y}\right)\:=\:\left\{\left(\mathrm{0},\mathrm{2023}\right),\left(\mathrm{7},\mathrm{1792}\right)….\right\}\: \\ $$
Commented by mr W last updated on 10/Oct/23
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