Question Number 10755 by Joel576 last updated on 24/Feb/17
$${x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:+\:{xy}\:+\:\mathrm{2}\left({x}\:−\:{y}\right)\:=\:\mathrm{9} \\ $$$$\mathrm{How}\:\mathrm{many}\:\mathrm{solution}\:\mathrm{that}\:\mathrm{fulfilled}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{above}\:? \\ $$$${x},\:{y}\:\in\:\mathbb{N} \\ $$
Answered by mrW1 last updated on 24/Feb/17
$${x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:−\:\mathrm{2}{xy}\:+\:\mathrm{2}\left({x}\:−\:{y}\right)+\mathrm{3}{xy}\:=\:\mathrm{9} \\ $$$$\left({x}−{y}\right)^{\mathrm{2}} +\mathrm{2}\left({x}−{y}\right)+\mathrm{3}{xy}−\mathrm{9}=\mathrm{0} \\ $$$${u}={x}−{y} \\ $$$${v}={xy} \\ $$$$\Rightarrow{u}^{\mathrm{2}} +\mathrm{2}{u}+\mathrm{3}{v}−\mathrm{9}=\mathrm{0} \\ $$$${u}=\frac{−\mathrm{2}\pm\sqrt{\mathrm{4}−\mathrm{4}\left(\mathrm{3}{v}−\mathrm{9}\right)}}{\mathrm{2}}=−\mathrm{1}\pm\sqrt{\mathrm{10}−\mathrm{3}{v}} \\ $$$$\mathrm{10}−\mathrm{3}{v}={k}^{\mathrm{2}} \\ $$$${v}=\frac{\mathrm{10}−{k}^{\mathrm{2}} }{\mathrm{3}}>\mathrm{0} \\ $$$${k}=\mathrm{1}:\:{v}=\frac{\mathrm{10}−\mathrm{1}^{\mathrm{2}} }{\mathrm{3}}=\mathrm{3}\:\:\:{ok} \\ $$$${k}=\mathrm{2}:\:{v}=\frac{\mathrm{10}−\mathrm{2}^{\mathrm{2}} }{\mathrm{3}}=\mathrm{2}\:\:\:{ok} \\ $$$${k}=\mathrm{3}:\:{v}=\frac{\mathrm{10}−\mathrm{3}^{\mathrm{2}} }{\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{3}}\:\:{not}\:{ok} \\ $$$$ \\ $$$${with}\:{v}=\mathrm{3} \\ $$$${u}=−\mathrm{1}\pm\mathrm{1}=\begin{cases}{\mathrm{0}}\\{−\mathrm{2}}\end{cases} \\ $$$$\begin{cases}{{x}−{y}=\mathrm{0}}&{\:\:\:}\\{{xy}=\mathrm{3}}&{\:\:\:\Rightarrow{no}\:{integer}\:{solution}!}\end{cases} \\ $$$$ \\ $$$$\begin{cases}{{x}−{y}=−\mathrm{2}}&{\Rightarrow{x}=\mathrm{1},\:{y}=\mathrm{3}}\\{{xy}=\mathrm{3}}&{\Rightarrow\:{right}\:{solution}!}\end{cases} \\ $$$$ \\ $$$${with}\:{v}=\mathrm{2} \\ $$$${u}=−\mathrm{1}\pm\mathrm{2}=\begin{cases}{\mathrm{1}}\\{−\mathrm{3}}\end{cases} \\ $$$$\begin{cases}{{x}−{y}=\mathrm{1}}&{\Rightarrow{x}=\mathrm{2},\:{y}=\mathrm{1}}\\{{xy}=\mathrm{2}}&{\Rightarrow\:{right}\:{solution}!}\end{cases} \\ $$$$ \\ $$$$\begin{cases}{{x}−{y}=−\mathrm{3}}&{}\\{{xy}=\mathrm{2}}&{\Rightarrow\:{no}\:{integer}\:{solution}!}\end{cases} \\ $$$$ \\ $$$${all}\:{solutions}: \\ $$$$\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}=\begin{pmatrix}{\mathrm{1}}\\{\mathrm{3}}\end{pmatrix}\:{or}\:\begin{pmatrix}{\mathrm{2}}\\{\mathrm{1}}\end{pmatrix} \\ $$
Commented by Joel576 last updated on 24/Feb/17
$${thank}\:{you}\:{very}\:{much} \\ $$