Question Number 61605 by Mr X pcx last updated on 05/Jun/19
$${solve}\:\:{at}\:{Z}^{\mathrm{2}} \:\:\:\:\mathrm{2}{x}\:+\mathrm{5}{y}\:=\mathrm{4} \\ $$
Commented by mr W last updated on 05/Jun/19
$${x}=\mathrm{2}−\mathrm{5}{n} \\ $$$${y}=\mathrm{2}{n} \\ $$$${n}\in{Z} \\ $$$${is}\:{this}\:{correct}? \\ $$
Commented by MJS last updated on 05/Jun/19
$$\mathrm{let}\:{y}=\mathrm{2}{a}+\mathrm{2}{b}\mathrm{i} \\ $$$$\mathrm{2}{x}+\mathrm{10}{a}+\mathrm{10}{b}\mathrm{i}=\mathrm{4} \\ $$$${x}=\mathrm{2}−\mathrm{5}{a}−\mathrm{5}{b}\mathrm{i} \\ $$$$\mathrm{it}'\mathrm{s}\:\mathrm{correct},\:\mathrm{but}\:\mathrm{we}'\mathrm{re}\:\mathrm{in}\:\mathbb{Z}^{\mathrm{2}} =\mathbb{R}^{\mathrm{4}} \\ $$$$\begin{pmatrix}{{x}_{{real}} }\\{{x}_{{imag}} }\\{{y}_{{real}} }\\{{y}_{{imag}} }\end{pmatrix}\:\:=\begin{pmatrix}{\mathrm{2}}\\{\mathrm{0}}\\{\mathrm{0}}\\{\mathrm{0}}\end{pmatrix}\:+{a}×\begin{pmatrix}{−\mathrm{5}}\\{\mathrm{0}}\\{\mathrm{2}}\\{\mathrm{0}}\end{pmatrix}\:+{b}×\begin{pmatrix}{\mathrm{0}}\\{−\mathrm{5}}\\{\mathrm{0}}\\{\mathrm{2}}\end{pmatrix} \\ $$
Commented by MJS last updated on 05/Jun/19
$$...\mathrm{so}\:\mathrm{this}\:\mathrm{is}\:\mathrm{a}\:\mathrm{plane}\:\mathrm{not}\:\mathrm{a}\:\mathrm{line} \\ $$
Commented by mr W last updated on 05/Jun/19
$${thanks}\:{alot}\:{sir}! \\ $$
Commented by maxmathsup by imad last updated on 05/Jun/19
$${let}\:{consider}\:{the}\:{congruence}\:{modulo}\:\mathrm{2}\:\left(\mathrm{2}\:{is}\:{prime}\right) \\ $$$$\left({e}\right)\:\Rightarrow\:\overset{−} {\mathrm{2}}\overset{−} {{x}}\:+\overset{−} {\mathrm{5}}\overset{−} {{y}}=\overset{−} {\mathrm{4}}\:\Rightarrow\mathrm{0}\:+\overset{−} {{y}}\:=\mathrm{0}\:\Rightarrow{y}\:=\mathrm{2}{k}\:\:\:\:\:\left({k}\:{from}\:{Z}\right) \\ $$$$\mathrm{2}{x}+\mathrm{5}{y}\:=\mathrm{4}\:\Rightarrow\mathrm{2}{x}\:+\mathrm{5}\left(\mathrm{2}{k}\right)\:=\mathrm{4}\:\Rightarrow{x}+\mathrm{5}{k}\:=\mathrm{2}\:\Rightarrow{x}\:=−\mathrm{5}{k}\:+\mathrm{2}\:\:{so}\:{the}\:{solution} \\ $$$${are}\:\left(−\mathrm{5}{k}+\mathrm{2}\:,\mathrm{2}{k}\right)\:\:{with}\:{k}\:\in{Z}\:. \\ $$$${remark}\:{we}\:{can}\:{consider}\:{congruence}\:{modulo}\:\mathrm{5}\:\left(\:{Z}/\mathrm{5}{Z}\right) \\ $$
Commented by maxmathsup by imad last updated on 05/Jun/19
$${correct}\:{sir} \\ $$
Commented by kaivan.ahmadi last updated on 06/Jun/19
$$\mathrm{2}{x}\overset{\mathrm{5}} {\equiv}\mathrm{4}\Rightarrow\mathrm{6}{x}\overset{\mathrm{5}} {\equiv}\mathrm{12}\Rightarrow{x}\overset{\mathrm{5}} {\equiv}\mathrm{2}\Rightarrow{x}=\mathrm{5}{k}+\mathrm{2},\:{k}\in{Z} \\ $$$${and} \\ $$$$\mathrm{2}\left(\mathrm{5}{k}+\mathrm{2}\right)+\mathrm{5}{y}=\mathrm{4}\Rightarrow\mathrm{10}{k}+\mathrm{4}+\mathrm{5}{y}=\mathrm{4}\Rightarrow{y}=−\mathrm{2}{k} \\ $$$$ \\ $$