Menu Close

find-all-solutions-if-exist-of-x-2-5y-2-2016-with-x-y-N-

Tinku Tara 0 Comments
Pin




Question Number 75660 by mr W last updated on 15/Dec/19
find all solutions (if exist) of x^2 +5y^2 =2016 with x,y ∈ N.
$${find}\:{all}\:{solutions}\:\left({if}\:{exist}\right)\:{of} \\ $$$${x}^{\mathrm{2}} +\mathrm{5}{y}^{\mathrm{2}} =\mathrm{2016} \\ $$$${with}\:{x},{y}\:\in\:\mathbb{N}. \\ $$
Commented by mathmax by abdo last updated on 15/Dec/19
5is prime let consider congruence modulo 5 (e)⇒ (x^− )^2 +0 =2016^− =16^− =1^− ⇒x^−^2 −1^− =0^− ⇒ (x^− −1^− )(x+1^− )=0^− Z/5Z is a corps ⇒x^− −1^− =0 or x^− +1=0 ⇒ x=5k+1 or x =5k−1 (k∈Z) now rest to find the values of k wich verify the equation with x^2 ≤2016 and y^2 ≤[((2016)/5)]...
$$\:\mathrm{5}{is}\:{prime}\:{let}\:{consider}\:{congruence}\:{modulo}\:\mathrm{5} \\ $$$$\left({e}\right)\Rightarrow\:\left(\overset{−} {{x}}\right)^{\mathrm{2}} +\mathrm{0}\:=\mathrm{201}\overset{−} {\mathrm{6}}=\mathrm{1}\overset{−} {\mathrm{6}}=\overset{−} {\mathrm{1}}\:\Rightarrow\overset{−^{\mathrm{2}} } {{x}}−\overset{−} {\mathrm{1}}=\overset{−} {\mathrm{0}}\:\Rightarrow \\ $$$$\left(\overset{−} {{x}}−\overset{−} {\mathrm{1}}\right)\left({x}+\overset{−} {\mathrm{1}}\right)=\overset{−} {\mathrm{0}}\:\:\:{Z}/\mathrm{5}{Z}\:{is}\:{a}\:{corps}\:\Rightarrow\overset{−} {{x}}−\overset{−} {\mathrm{1}}=\mathrm{0}\:{or}\:\overset{−} {{x}}+\mathrm{1}=\mathrm{0}\:\Rightarrow \\ $$$${x}=\mathrm{5}{k}+\mathrm{1}\:{or}\:{x}\:=\mathrm{5}{k}−\mathrm{1}\:\:\left({k}\in{Z}\right)\:\:{now}\:{rest}\:{to}\:{find}\:{the}\:{values}\:{of}\:{k} \\ $$$${wich}\:{verify}\:{the}\:{equation}\:\:{with}\:{x}^{\mathrm{2}} \leqslant\mathrm{2016}\:{and}\:{y}^{\mathrm{2}} \leqslant\left[\frac{\mathrm{2016}}{\mathrm{5}}\right]… \\ $$
Answered by vishalbhardwaj last updated on 15/Dec/19
x^2 = 2016−5y^2 ≥0 ⇒ 5y^2 ≤2016 ⇒ y^2 ≤403.20 ⇒ y≤(√(403.20 )) ⇒ 1≤y≤20, y∈ N similarily 5y^2 = 2016−x^2 ⇒ y^2 = ((2016−x^2 )/5) ≥ 0 ⇒ 2016−x^2 ≥ 0 ⇒ x^2 ≤ 2016 ⇒ 1≤x≤44 , x∈ N when x = 4 then y = 20 and when x= 36 then y = 12 ⇒ we have two solutions : (x=4, y=20) and (x=36, y=12)
$$\mathrm{x}^{\mathrm{2}} =\:\mathrm{2016}−\mathrm{5y}^{\mathrm{2}} \geqslant\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{5y}^{\mathrm{2}} \leqslant\mathrm{2016} \\ $$$$\Rightarrow\:\mathrm{y}^{\mathrm{2}} \leqslant\mathrm{403}.\mathrm{20} \\ $$$$\Rightarrow\:\mathrm{y}\leqslant\sqrt{\mathrm{403}.\mathrm{20}\:}\:\:\Rightarrow\:\:\mathrm{1}\leqslant\mathrm{y}\leqslant\mathrm{20},\:\:\mathrm{y}\in\:\mathrm{N} \\ $$$$\mathrm{similarily}\:\:\mathrm{5y}^{\mathrm{2}} \:=\:\mathrm{2016}−\mathrm{x}^{\mathrm{2}} \\ $$$$\:\Rightarrow\:\:\mathrm{y}^{\mathrm{2}} \:=\:\frac{\mathrm{2016}−\mathrm{x}^{\mathrm{2}} }{\mathrm{5}}\:\geqslant\:\mathrm{0} \\ $$$$\Rightarrow\:\:\mathrm{2016}−\mathrm{x}^{\mathrm{2}} \:\geqslant\:\mathrm{0} \\ $$$$\Rightarrow\:\:\mathrm{x}^{\mathrm{2}} \:\leqslant\:\mathrm{2016}\:\:\Rightarrow\:\:\:\mathrm{1}\leqslant\mathrm{x}\leqslant\mathrm{44}\:,\:\mathrm{x}\in\:\mathrm{N} \\ $$$$\mathrm{when}\:\mathrm{x}\:=\:\mathrm{4}\:\mathrm{then}\:\mathrm{y}\:=\:\mathrm{20} \\ $$$$\mathrm{and}\:\mathrm{when}\:\mathrm{x}=\:\mathrm{36}\:\mathrm{then}\:\mathrm{y}\:=\:\mathrm{12} \\ $$$$\Rightarrow\:\mathrm{we}\:\mathrm{have}\:\mathrm{two}\:\mathrm{solutions}\::\:\left(\mathrm{x}=\mathrm{4},\:\mathrm{y}=\mathrm{20}\right) \\ $$$$\mathrm{and}\:\left(\mathrm{x}=\mathrm{36},\:\mathrm{y}=\mathrm{12}\right) \\ $$
Commented by mr W last updated on 15/Dec/19
thanks alot sir.
$${thanks}\:{alot}\:{sir}. \\ $$
Commented by MJS last updated on 15/Dec/19
one is missing: x=44, y=4
$$\mathrm{one}\:\mathrm{is}\:\mathrm{missing}:\:{x}=\mathrm{44},\:{y}=\mathrm{4} \\ $$
Answered by mind is power last updated on 15/Dec/19
x^2 +5y^2 =2016 x^2 +5y^2 =0(4) ⇔x^2 +y^2 =0(4) x^2 =0,1(4)⇒x^2 =y^2 ≡0(4) ⇔x=2k y=2t ⇒4(k^2 +5t^2 )=2016⇒k^2 +5t^2 =504≡0(4) like previous congrunce⇒ ⇒k=2s,t=2z ⇒t^2 +5z^2 =126 if z≡0(2)⇒t≡0(2) ⇒t^2 +5z^2 =126≡0(4) absurde⇒z=2y+1 t^2 =126−5z^2 ⇒z≤(√((126)/5))⇒z≤5⇒z∈{1,3,5} z=1⇒t^2 =121⇒t=11 z=3⇒t^2 =81⇒t=9 z=5⇒t^2 =1⇒t=1 z=1⇒t=2⇒y=4 t=11⇒x=44 z=5⇒y=20 t=1⇒x=4 z=3⇒y=12⇒x=36 S={(4,20);(44,4);(36,12)}
$$\mathrm{x}^{\mathrm{2}} +\mathrm{5y}^{\mathrm{2}} =\mathrm{2016} \\ $$$$\mathrm{x}^{\mathrm{2}} +\mathrm{5y}^{\mathrm{2}} =\mathrm{0}\left(\mathrm{4}\right) \\ $$$$\Leftrightarrow\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} =\mathrm{0}\left(\mathrm{4}\right) \\ $$$$\mathrm{x}^{\mathrm{2}} =\mathrm{0},\mathrm{1}\left(\mathrm{4}\right)\Rightarrow\mathrm{x}^{\mathrm{2}} =\mathrm{y}^{\mathrm{2}} \equiv\mathrm{0}\left(\mathrm{4}\right) \\ $$$$\Leftrightarrow\mathrm{x}=\mathrm{2k}\:\mathrm{y}=\mathrm{2t} \\ $$$$\Rightarrow\mathrm{4}\left(\mathrm{k}^{\mathrm{2}} +\mathrm{5t}^{\mathrm{2}} \right)=\mathrm{2016}\Rightarrow\mathrm{k}^{\mathrm{2}} +\mathrm{5t}^{\mathrm{2}} =\mathrm{504}\equiv\mathrm{0}\left(\mathrm{4}\right)\:\mathrm{like}\:\mathrm{previous}\:\mathrm{congrunce}\Rightarrow \\ $$$$\Rightarrow\mathrm{k}=\mathrm{2s},\mathrm{t}=\mathrm{2z}\:\Rightarrow\mathrm{t}^{\mathrm{2}} +\mathrm{5z}^{\mathrm{2}} =\mathrm{126}\:\:\:\mathrm{if}\:\mathrm{z}\equiv\mathrm{0}\left(\mathrm{2}\right)\Rightarrow\mathrm{t}\equiv\mathrm{0}\left(\mathrm{2}\right)\: \\ $$$$\Rightarrow\mathrm{t}^{\mathrm{2}} +\mathrm{5z}^{\mathrm{2}} =\mathrm{126}\equiv\mathrm{0}\left(\mathrm{4}\right)\:\mathrm{absurde}\Rightarrow\mathrm{z}=\mathrm{2y}+\mathrm{1} \\ $$$$\mathrm{t}^{\mathrm{2}} =\mathrm{126}−\mathrm{5z}^{\mathrm{2}} \Rightarrow\mathrm{z}\leqslant\sqrt{\frac{\mathrm{126}}{\mathrm{5}}}\Rightarrow\mathrm{z}\leqslant\mathrm{5}\Rightarrow\mathrm{z}\in\left\{\mathrm{1},\mathrm{3},\mathrm{5}\right\} \\ $$$$\mathrm{z}=\mathrm{1}\Rightarrow\mathrm{t}^{\mathrm{2}} =\mathrm{121}\Rightarrow\mathrm{t}=\mathrm{11} \\ $$$$\mathrm{z}=\mathrm{3}\Rightarrow\mathrm{t}^{\mathrm{2}} =\mathrm{81}\Rightarrow\mathrm{t}=\mathrm{9} \\ $$$$\mathrm{z}=\mathrm{5}\Rightarrow\mathrm{t}^{\mathrm{2}} =\mathrm{1}\Rightarrow\mathrm{t}=\mathrm{1} \\ $$$$\mathrm{z}=\mathrm{1}\Rightarrow\mathrm{t}=\mathrm{2}\Rightarrow\mathrm{y}=\mathrm{4} \\ $$$$\mathrm{t}=\mathrm{11}\Rightarrow\mathrm{x}=\mathrm{44} \\ $$$$\mathrm{z}=\mathrm{5}\Rightarrow\mathrm{y}=\mathrm{20} \\ $$$$\mathrm{t}=\mathrm{1}\Rightarrow\mathrm{x}=\mathrm{4} \\ $$$$\mathrm{z}=\mathrm{3}\Rightarrow\mathrm{y}=\mathrm{12}\Rightarrow\mathrm{x}=\mathrm{36} \\ $$$$\mathrm{S}=\left\{\left(\mathrm{4},\mathrm{20}\right);\left(\mathrm{44},\mathrm{4}\right);\left(\mathrm{36},\mathrm{12}\right)\right\} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by mr W last updated on 15/Dec/19
thanks alot sir!
$${thanks}\:{alot}\:{sir}! \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *