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