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Question-43971




Question Number 43971 by LYCON TRIX last updated on 19/Sep/18
Answered by MJS last updated on 19/Sep/18
2x^2 +x=3y^2 +y  x=y+a  y^2 −4ay−a(2a+1)=0  y=2a±(√(a(6a+1)))  (√(a(6a+1))) integer ⇒ a(6a+1)=p^2  ⇒ a=q^2  ⇒  ⇒ (√(6q^2 +1)) integer  q_1 =0 ⇒ x_1 =y_1 =0  q_2 =2 ⇒ x_2 =22∧y_2 =18  q_3 =20 ⇒ x_3 =2180∧y_3 =1780  q_4 =198 ⇒ x_4 =213642∧y_4 =174438  Σ_(i=1) ^4 (x_i +y_i )=392080
$$\mathrm{2}{x}^{\mathrm{2}} +{x}=\mathrm{3}{y}^{\mathrm{2}} +{y} \\ $$$${x}={y}+{a} \\ $$$${y}^{\mathrm{2}} −\mathrm{4}{ay}−{a}\left(\mathrm{2}{a}+\mathrm{1}\right)=\mathrm{0} \\ $$$${y}=\mathrm{2}{a}\pm\sqrt{{a}\left(\mathrm{6}{a}+\mathrm{1}\right)} \\ $$$$\sqrt{{a}\left(\mathrm{6}{a}+\mathrm{1}\right)}\:\mathrm{integer}\:\Rightarrow\:{a}\left(\mathrm{6}{a}+\mathrm{1}\right)={p}^{\mathrm{2}} \:\Rightarrow\:{a}={q}^{\mathrm{2}} \:\Rightarrow \\ $$$$\Rightarrow\:\sqrt{\mathrm{6}{q}^{\mathrm{2}} +\mathrm{1}}\:\mathrm{integer} \\ $$$${q}_{\mathrm{1}} =\mathrm{0}\:\Rightarrow\:{x}_{\mathrm{1}} ={y}_{\mathrm{1}} =\mathrm{0} \\ $$$${q}_{\mathrm{2}} =\mathrm{2}\:\Rightarrow\:{x}_{\mathrm{2}} =\mathrm{22}\wedge{y}_{\mathrm{2}} =\mathrm{18} \\ $$$${q}_{\mathrm{3}} =\mathrm{20}\:\Rightarrow\:{x}_{\mathrm{3}} =\mathrm{2180}\wedge{y}_{\mathrm{3}} =\mathrm{1780} \\ $$$${q}_{\mathrm{4}} =\mathrm{198}\:\Rightarrow\:{x}_{\mathrm{4}} =\mathrm{213642}\wedge{y}_{\mathrm{4}} =\mathrm{174438} \\ $$$$\underset{{i}=\mathrm{1}} {\overset{\mathrm{4}} {\sum}}\left({x}_{{i}} +{y}_{{i}} \right)=\mathrm{392080} \\ $$
Commented by MJS last updated on 19/Sep/18
q_5 =1960
$${q}_{\mathrm{5}} =\mathrm{1960} \\ $$
Commented by MJS last updated on 19/Sep/18
q_1 =0  q_2 =2  n≥2: q_(n+1) =10q_n −q_(n−1)
$${q}_{\mathrm{1}} =\mathrm{0} \\ $$$${q}_{\mathrm{2}} =\mathrm{2} \\ $$$${n}\geqslant\mathrm{2}:\:{q}_{{n}+\mathrm{1}} =\mathrm{10}{q}_{{n}} −{q}_{{n}−\mathrm{1}} \\ $$
Commented by MJS last updated on 19/Sep/18
I think not all questions there hold prizes.  some are too easy
$$\mathrm{I}\:\mathrm{think}\:\mathrm{not}\:\mathrm{all}\:\mathrm{questions}\:\mathrm{there}\:\mathrm{hold}\:\mathrm{prizes}. \\ $$$$\mathrm{some}\:\mathrm{are}\:\mathrm{too}\:\mathrm{easy} \\ $$
Commented by LYCON TRIX last updated on 20/Sep/18
Yes , we claim cash prize on selected   questions only ,it′s mentioned in community
$$\mathrm{Yes}\:,\:\mathrm{we}\:\mathrm{claim}\:\mathrm{cash}\:\mathrm{prize}\:\mathrm{on}\:\mathrm{selected}\: \\ $$$$\mathrm{questions}\:\mathrm{only}\:,\mathrm{it}'\mathrm{s}\:\mathrm{mentioned}\:\mathrm{in}\:\mathrm{community} \\ $$

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