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p-2-3q-2-11907-p-q-Z-find-p-amp-q-




Question Number 84461 by Tony Lin last updated on 13/Mar/20
p^2 +3q^2 =11907, p,q∈Z,find p&q
$${p}^{\mathrm{2}} +\mathrm{3}{q}^{\mathrm{2}} =\mathrm{11907},\:{p},{q}\in\mathbb{Z},{find}\:{p\&q} \\ $$
Commented by mr W last updated on 13/Mar/20
i got only two solutions:  p=0, q=±63  p=±108, q=±9
$${i}\:{got}\:{only}\:{two}\:{solutions}: \\ $$$${p}=\mathrm{0},\:{q}=\pm\mathrm{63} \\ $$$${p}=\pm\mathrm{108},\:{q}=\pm\mathrm{9} \\ $$
Commented by Tony Lin last updated on 13/Mar/20
p^2 +3q^2 =3^5 ×7^2   ⇒p^2 =3(63+q)(63−q), q≤63  ⇒p^2 =3^5 (7+(q/9))(7−(q/9))  ∵ p&q∈Z  ∴ 9∣q  →check that (7+(q/9))(7−(q/9))=3×n^2 ,n∈Z  (7+1)(7−1)=3×4^2   (7+7)(7−7)=3×0^2   ⇒ { ((p=0, q=±63)),((p=±108, q=±9)) :}
$${p}^{\mathrm{2}} +\mathrm{3}{q}^{\mathrm{2}} =\mathrm{3}^{\mathrm{5}} ×\mathrm{7}^{\mathrm{2}} \\ $$$$\Rightarrow{p}^{\mathrm{2}} =\mathrm{3}\left(\mathrm{63}+{q}\right)\left(\mathrm{63}−{q}\right),\:{q}\leqslant\mathrm{63} \\ $$$$\Rightarrow{p}^{\mathrm{2}} =\mathrm{3}^{\mathrm{5}} \left(\mathrm{7}+\frac{{q}}{\mathrm{9}}\right)\left(\mathrm{7}−\frac{{q}}{\mathrm{9}}\right) \\ $$$$\because\:{p\&q}\in\mathbb{Z} \\ $$$$\therefore\:\mathrm{9}\mid{q} \\ $$$$\rightarrow{check}\:{that}\:\left(\mathrm{7}+\frac{{q}}{\mathrm{9}}\right)\left(\mathrm{7}−\frac{{q}}{\mathrm{9}}\right)=\mathrm{3}×{n}^{\mathrm{2}} ,{n}\in\mathbb{Z} \\ $$$$\left(\mathrm{7}+\mathrm{1}\right)\left(\mathrm{7}−\mathrm{1}\right)=\mathrm{3}×\mathrm{4}^{\mathrm{2}} \\ $$$$\left(\mathrm{7}+\mathrm{7}\right)\left(\mathrm{7}−\mathrm{7}\right)=\mathrm{3}×\mathrm{0}^{\mathrm{2}} \\ $$$$\Rightarrow\begin{cases}{{p}=\mathrm{0},\:{q}=\pm\mathrm{63}}\\{{p}=\pm\mathrm{108},\:{q}=\pm\mathrm{9}}\end{cases} \\ $$

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