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Question Number 178686 by greougoury555 last updated on 20/Oct/22

Given 2x^2 y^2 +12y^2 =7x^2 +647 for x,y ε Z . Find the remaider if 3x^2 y^4 divide by 11 .

$$\:{Given}\:\mathrm{2}{x}^{\mathrm{2}} {y}^{\mathrm{2}} +\mathrm{12}{y}^{\mathrm{2}} =\mathrm{7}{x}^{\mathrm{2}} +\mathrm{647}\: \\ $$$$\:{for}\:{x},{y}\:\varepsilon\:\mathbb{Z}\:. \\ $$$$\:{Find}\:{the}\:{remaider}\:{if}\:\mathrm{3}{x}^{\mathrm{2}} {y}^{\mathrm{4}} \:{divide}\:{by} \\ $$$$\:\:\mathrm{11}\:. \\ $$

Answered by Rasheed.Sindhi last updated on 20/Oct/22

2x^2 y^2 +12y^2 =7x^2 +647 2y^2 (x^2 +6)=7(x^2 +6)+605 2y^2 (x^2 +6)−7(x^2 +6)=605 (x^2 +6)(2y^2 −7)=5×11^2 ......A Possible values for x^2 +6: 1,5,11,121,55,605,−1,−5,−11,−121,−55,−605 Possible values for x^2 −5^(×) ,−1^(×) ,5^(×) ,115^(×) , determinant (((49)))^✓ ,599^(×) ,−7^(×) ,−11^(×) ,−17^(×) ,−127^(×) ,−61^(×) ,−611^(×) ∵ x∈Z ∴ x^2 =49⇒x=±7 If x^2 +6=(±7)^2 +6=55 then from A 2y^2 −7=605/55=11 2y^2 =18⇒y=±3 (x,y)=(±7,±3) or (±7,∓3) 3x^2 y^4 =3(±7)^2 (±3)^4 =3(±7)^2 (∓3)^4 =11907 11907 (mod 11)= determinant ((5))

$$\mathrm{2}{x}^{\mathrm{2}} {y}^{\mathrm{2}} +\mathrm{12}{y}^{\mathrm{2}} =\mathrm{7}{x}^{\mathrm{2}} +\mathrm{647} \\ $$$$\mathrm{2}{y}^{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{6}\right)=\mathrm{7}\left({x}^{\mathrm{2}} +\mathrm{6}\right)+\mathrm{605} \\ $$$$\mathrm{2}{y}^{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{6}\right)−\mathrm{7}\left({x}^{\mathrm{2}} +\mathrm{6}\right)=\mathrm{605} \\ $$$$\left({x}^{\mathrm{2}} +\mathrm{6}\right)\left(\mathrm{2}{y}^{\mathrm{2}} −\mathrm{7}\right)=\mathrm{5}×\mathrm{11}^{\mathrm{2}} ......{A} \\ $$$${Possible}\:{values}\:{for}\:{x}^{\mathrm{2}} +\mathrm{6}: \\ $$$$\mathrm{1},\mathrm{5},\mathrm{11},\mathrm{121},\mathrm{55},\mathrm{605},−\mathrm{1},−\mathrm{5},−\mathrm{11},−\mathrm{121},−\mathrm{55},−\mathrm{605} \\ $$$${Possible}\:{values}\:{for}\:{x}^{\mathrm{2}} \\ $$$$\overset{×} {−\mathrm{5}},\overset{×} {−\mathrm{1}},\overset{×} {\mathrm{5}},\overset{×} {\mathrm{115}},\:\:\begin{array}{|c|}{\mathrm{49}}\\\hline\end{array}^{\checkmark} ,\overset{×} {\mathrm{599}},\overset{×} {−\mathrm{7}},\overset{×} {−\mathrm{11}},\overset{×} {−\mathrm{17}},\overset{×} {−\mathrm{127}},\overset{×} {−\mathrm{61}},\overset{×} {−\mathrm{611}} \\ $$$$\because\:{x}\in\mathbb{Z}\:\:\:\:\therefore\:\:\:{x}^{\mathrm{2}} =\mathrm{49}\Rightarrow{x}=\pm\mathrm{7} \\ $$$${If}\:{x}^{\mathrm{2}} +\mathrm{6}=\left(\pm\mathrm{7}\right)^{\mathrm{2}} +\mathrm{6}=\mathrm{55}\:{then}\:{from}\:{A} \\ $$$$\mathrm{2}{y}^{\mathrm{2}} −\mathrm{7}=\mathrm{605}/\mathrm{55}=\mathrm{11} \\ $$$$\mathrm{2}{y}^{\mathrm{2}} =\mathrm{18}\Rightarrow{y}=\pm\mathrm{3} \\ $$$$\: \\ $$$$\left({x},{y}\right)=\left(\pm\mathrm{7},\pm\mathrm{3}\right)\:{or}\:\left(\pm\mathrm{7},\mp\mathrm{3}\right) \\ $$$$\:\:\:\:\mathrm{3}{x}^{\mathrm{2}} {y}^{\mathrm{4}} =\mathrm{3}\left(\pm\mathrm{7}\right)^{\mathrm{2}} \left(\pm\mathrm{3}\right)^{\mathrm{4}} =\mathrm{3}\left(\pm\mathrm{7}\right)^{\mathrm{2}} \left(\mp\mathrm{3}\right)^{\mathrm{4}} =\mathrm{11907} \\ $$$$\:\:\:\:\mathrm{11907}\:\left({mod}\:\mathrm{11}\right)=\begin{array}{|c|}{\mathrm{5}}\\\hline\end{array} \\ $$

Commented by cortano1 last updated on 21/Oct/22

typo ⇒2y^2 =18 ; y=± 3 (x,y)= (±7 ,±3) then 3x^2 y^4 =3×49×81=11907 11907 = 1082×11+5=5(mod 11)

$$\:\mathrm{typo}\:\Rightarrow\mathrm{2y}^{\mathrm{2}} =\mathrm{18}\:;\:\mathrm{y}=\pm\:\mathrm{3} \\ $$$$\left(\mathrm{x},\mathrm{y}\right)=\:\left(\pm\mathrm{7}\:,\pm\mathrm{3}\right) \\ $$$$\mathrm{then}\:\mathrm{3x}^{\mathrm{2}} \mathrm{y}^{\mathrm{4}} =\mathrm{3}×\mathrm{49}×\mathrm{81}=\mathrm{11907} \\ $$$$\:\mathrm{11907}\:=\:\mathrm{1082}×\mathrm{11}+\mathrm{5}=\mathrm{5}\left(\mathrm{mod}\:\mathrm{11}\right) \\ $$

Commented by Rasheed.Sindhi last updated on 20/Oct/22

Thank you sir,I′ve corrected now.

$$\mathcal{T}{hank}\:{you}\:{sir},{I}'{ve}\:\:{corrected}\:{now}. \\ $$

Answered by greougoury555 last updated on 21/Oct/22

2x^2 y^2 −7x^2 = 647−12y^2 x^2 (2y^2 −7)=605+42−12y^2 x^2 (2y^2 −7)=605+6(7−2y^(2)) x^2 (2y^2 −7)+6(2y^2 −7)=605 (2y^2 −7)(x^2 +6)=11×55 { ((2y^2 −7=11⇒y^2 =9)),((x^2 +6=55⇒x=49)) :} ⇒3x^2 y^4 = 3×49×81=11097 ⇒((11097)/(11)) = 1082+(5/(11))

$$\:\mathrm{2}{x}^{\mathrm{2}} {y}^{\mathrm{2}} −\mathrm{7}{x}^{\mathrm{2}} =\:\mathrm{647}−\mathrm{12}{y}^{\mathrm{2}} \\ $$$$\:{x}^{\mathrm{2}} \left(\mathrm{2}{y}^{\mathrm{2}} −\mathrm{7}\right)=\mathrm{605}+\mathrm{42}−\mathrm{12}{y}^{\mathrm{2}} \\ $$$$\:{x}^{\mathrm{2}} \left(\mathrm{2}{y}^{\mathrm{2}} −\mathrm{7}\right)=\mathrm{605}+\mathrm{6}\left(\mathrm{7}−\mathrm{2}{y}^{\left.\mathrm{2}\right)} \right. \\ $$$$\:{x}^{\mathrm{2}} \left(\mathrm{2}{y}^{\mathrm{2}} −\mathrm{7}\right)+\mathrm{6}\left(\mathrm{2}{y}^{\mathrm{2}} −\mathrm{7}\right)=\mathrm{605} \\ $$$$\:\left(\mathrm{2}{y}^{\mathrm{2}} −\mathrm{7}\right)\left({x}^{\mathrm{2}} +\mathrm{6}\right)=\mathrm{11}×\mathrm{55} \\ $$$$\:\begin{cases}{\mathrm{2}{y}^{\mathrm{2}} −\mathrm{7}=\mathrm{11}\Rightarrow{y}^{\mathrm{2}} =\mathrm{9}}\\{{x}^{\mathrm{2}} +\mathrm{6}=\mathrm{55}\Rightarrow{x}=\mathrm{49}}\end{cases} \\ $$$$\:\Rightarrow\mathrm{3}{x}^{\mathrm{2}} {y}^{\mathrm{4}} \:=\:\mathrm{3}×\mathrm{49}×\mathrm{81}=\mathrm{11097} \\ $$$$\Rightarrow\frac{\mathrm{11097}}{\mathrm{11}}\:=\:\mathrm{1082}+\frac{\mathrm{5}}{\mathrm{11}} \\ $$

Commented by Rasheed.Sindhi last updated on 21/Oct/22

Typo: 3×49×81=11907

$$\mathcal{T}{ypo}:\:\mathrm{3}×\mathrm{49}×\mathrm{81}=\mathrm{11907} \\ $$

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