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Question-196117




Question Number 196117 by sonukgindia last updated on 18/Aug/23
Answered by Rasheed.Sindhi last updated on 18/Aug/23
m^2 (1+3n^2 )−10(3n^2 +1)=517−10  (m^2 −10)(3n^2 +1)=507  (m^2 −10)(3n^2 +1)=3×13^2   ....
$${m}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{3}{n}^{\mathrm{2}} \right)−\mathrm{10}\left(\mathrm{3}{n}^{\mathrm{2}} +\mathrm{1}\right)=\mathrm{517}−\mathrm{10} \\ $$$$\left({m}^{\mathrm{2}} −\mathrm{10}\right)\left(\mathrm{3}{n}^{\mathrm{2}} +\mathrm{1}\right)=\mathrm{507} \\ $$$$\left({m}^{\mathrm{2}} −\mathrm{10}\right)\left(\mathrm{3}{n}^{\mathrm{2}} +\mathrm{1}\right)=\mathrm{3}×\mathrm{13}^{\mathrm{2}} \\ $$$$…. \\ $$
Answered by gatocomcirrose last updated on 18/Aug/23
m^2 +3m^2 n^2 −30n^2 =517  m^2 (1+3n^2 )−10(1+3n^2 )+10=517  (m^2 −10)(3n^2 +1)=507=3.13^2    { ((m^2 −10=1, 3n^2 +1=507 )),((m^2 −10=3, 3n^2 +1=169)),((m^2 −10=13, 3n^2 +1=39)),((m^2 −10=39, 3n^2 +1=13)),((m^2 −10=169, 3n^2 +1=3)),((m^2 −10=507, 3n^2 +1=1)) :}  ⇒(m,n)={(7,2), (7,−2), (−7,2), (−7,−2)}
$$\mathrm{m}^{\mathrm{2}} +\mathrm{3m}^{\mathrm{2}} \mathrm{n}^{\mathrm{2}} −\mathrm{30n}^{\mathrm{2}} =\mathrm{517} \\ $$$$\mathrm{m}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{3n}^{\mathrm{2}} \right)−\mathrm{10}\left(\mathrm{1}+\mathrm{3n}^{\mathrm{2}} \right)+\mathrm{10}=\mathrm{517} \\ $$$$\left(\mathrm{m}^{\mathrm{2}} −\mathrm{10}\right)\left(\mathrm{3n}^{\mathrm{2}} +\mathrm{1}\right)=\mathrm{507}=\mathrm{3}.\mathrm{13}^{\mathrm{2}} \\ $$$$\begin{cases}{\mathrm{m}^{\mathrm{2}} −\mathrm{10}=\mathrm{1},\:\mathrm{3n}^{\mathrm{2}} +\mathrm{1}=\mathrm{507}\:}\\{\mathrm{m}^{\mathrm{2}} −\mathrm{10}=\mathrm{3},\:\mathrm{3n}^{\mathrm{2}} +\mathrm{1}=\mathrm{169}}\\{\mathrm{m}^{\mathrm{2}} −\mathrm{10}=\mathrm{13},\:\mathrm{3n}^{\mathrm{2}} +\mathrm{1}=\mathrm{39}}\\{\mathrm{m}^{\mathrm{2}} −\mathrm{10}=\mathrm{39},\:\mathrm{3n}^{\mathrm{2}} +\mathrm{1}=\mathrm{13}}\\{\mathrm{m}^{\mathrm{2}} −\mathrm{10}=\mathrm{169},\:\mathrm{3n}^{\mathrm{2}} +\mathrm{1}=\mathrm{3}}\\{\mathrm{m}^{\mathrm{2}} −\mathrm{10}=\mathrm{507},\:\mathrm{3n}^{\mathrm{2}} +\mathrm{1}=\mathrm{1}}\end{cases} \\ $$$$\Rightarrow\left(\mathrm{m},\mathrm{n}\right)=\left\{\left(\mathrm{7},\mathrm{2}\right),\:\left(\mathrm{7},−\mathrm{2}\right),\:\left(−\mathrm{7},\mathrm{2}\right),\:\left(−\mathrm{7},−\mathrm{2}\right)\right\} \\ $$

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