Question Number 64086 by Rio Michael last updated on 12/Jul/19
$${if}\:\:\:\mathrm{3}{x}\:+\:\mathrm{5}{y}\:=\:\mathrm{1} \\ $$$${use}\:{Bezout}'{s}\:{identity}\:{to}\:{find}\:{the}\:{value}\:{of}\:{x}\:{and}\:{y} \\ $$
Commented by turbo msup by abdo last updated on 13/Jul/19
![let consider congruence [3] (Z/3Z) (e) ⇒3^− x^− +5^− y^− =1^− ⇒ 0+2^− y^− =1^− =−2^− ⇒y^− =−1^− ⇒ y=−1+3k kintegr ⇒ x=((1−5y)/3) =((1+5−15k)/3) =2−5k ⇒ the solution are (2−5k,−1+3k) kfromZ .](Q64090.png)
$${let}\:{consider}\:{congruence}\:\left[\mathrm{3}\right]\:\left({Z}/\mathrm{3}{Z}\right) \\ $$$$\left({e}\right)\:\Rightarrow\overset{−} {\mathrm{3}}\overset{−} {{x}}\:+\overset{−} {\mathrm{5}}\overset{−} {{y}}=\overset{−} {\mathrm{1}}\:\Rightarrow \\ $$$$\mathrm{0}+\overset{−} {\mathrm{2}}\overset{−} {{y}}=\overset{−} {\mathrm{1}}\:=−\overset{−} {\mathrm{2}}\:\Rightarrow\overset{−} {{y}}=−\overset{−} {\mathrm{1}}\:\Rightarrow \\ $$$${y}=−\mathrm{1}+\mathrm{3}{k}\:\:\:{kintegr}\:\Rightarrow \\ $$$${x}=\frac{\mathrm{1}−\mathrm{5}{y}}{\mathrm{3}}\:=\frac{\mathrm{1}+\mathrm{5}−\mathrm{15}{k}}{\mathrm{3}}\:=\mathrm{2}−\mathrm{5}{k}\:\Rightarrow \\ $$$${the}\:{solution}\:{are}\:\left(\mathrm{2}−\mathrm{5}{k},−\mathrm{1}+\mathrm{3}{k}\right) \\ $$$${kfromZ}\:. \\ $$$$ \\ $$
Commented by turbo msup by abdo last updated on 13/Jul/19
$${another}\:{way}\:{by}\:{particular}\:{solution} \\ $$$$\left(\mathrm{2},−\mathrm{1}\right)\:{is}\:{a}\:{osrticular}\:{solution} \\ $$$${we}\:{have}\:\mathrm{3}{x}+\mathrm{5}{y}=\mathrm{1}\:{and} \\ $$$$\mathrm{3}×\mathrm{2}+\mathrm{5}×\left(−\mathrm{1}\right)=\mathrm{1}\:\Rightarrow \\ $$$$\mathrm{3}\left({x}−\mathrm{2}\right)+\mathrm{5}\left({y}+\mathrm{1}\right)=\mathrm{0}\:\Rightarrow \\ $$$$\mathrm{3}\left({x}−\mathrm{2}\right)=−\mathrm{5}\left({y}+\mathrm{1}\right)\:\Rightarrow \\ $$$$\mathrm{5}\:{divide}\:\mathrm{3}\left({x}−\mathrm{2}\right)\:{but}\:\Delta\left(\mathrm{3},\mathrm{5}\right)=\mathrm{1}\Rightarrow \\ $$$$\mathrm{5}\:{divide}\:{x}−\mathrm{2}\:\:\Rightarrow{x}=\mathrm{5}{k}+\mathrm{2} \\ $$$${we}\:{have}\:{y}=\frac{\mathrm{1}−\mathrm{3}{x}}{\mathrm{5}}\:=\frac{\mathrm{1}−\mathrm{15}{k}−\mathrm{6}}{\mathrm{5}} \\ $$$$=−\mathrm{1}−\mathrm{3}{k}\:\:{so}\:\left(\mathrm{5}{k}+\mathrm{2},−\mathrm{1}−\mathrm{3}{k}\right)\:{are} \\ $$$${solution}\:{for}\:{this}\:{equation}. \\ $$
Commented by Rio Michael last updated on 13/Jul/19
$${i}\:{thought}\:{we}\:{are}\:{to}\:{get}\:{constants}\:{as}\:{answers}. \\ $$$${we}\:{use}\:{gcd}\left(\mathrm{3},\mathrm{5}\right)=\mathrm{1}\:{right}?\:{since}\:{they}\:{are}\:{relativey}\:{prime} \\ $$
Commented by mr W last updated on 13/Jul/19
$${see}\:{also}\:{Q}\mathrm{46592} \\ $$
Commented by Rio Michael last updated on 13/Jul/19
$${yes}\:{i}\:{saw}\:{that},{but}\:{must}\:{we}\:{always}\:{put}\:{it}\:{in}\:{a}\:{general}\:{solution} \\ $$$${since}\:{gcd}\left(\mathrm{3},\mathrm{5}\right)=\mathrm{1} \\ $$$${then}\:{we}\:{reverse}\:{right}\: \\ $$$${x}=−\mathrm{2}\:{and}\:{y}=\mathrm{1} \\ $$