Question Number 13081 by 433 last updated on 13/May/17
![{ ((x+y+z=[1]_5 )),((xy=[2]_5 )),((yz=[1]_5 )) :} Solve system on Z_5](Q13081.png)
$$\begin{cases}{{x}+{y}+{z}=\left[\mathrm{1}\right]_{\mathrm{5}} }\\{{xy}=\left[\mathrm{2}\right]_{\mathrm{5}} }\\{{yz}=\left[\mathrm{1}\right]_{\mathrm{5}} }\end{cases} \\ $$$${Solve}\:{system}\:{on}\:\mathbb{Z}_{\mathrm{5}} \\ $$
Answered by RasheedSindhi last updated on 14/May/17
![^(Rasheed Soomro) x+y+z+xy+yz=[1]_5 +[2]_5 + [1]_5 y+x+z+y(x+z)=[4]_5 y+(x+z)(1+y)=[4]_5 1+y+(x+z)(1+y)=[4]_5 +1 (1+y)(1+x+z)=[4]_5 +[1]_5 =[0]_5 (1+y)=[0]_5 or(1+x+z)=[0]_5 y=[−1]_5 or x+z=[−1]_5 y=[4]_5 or x+z=[4]_5 yz=[1]_5 [4]z=[16]_5 z= [((16)/4)]_5 =[4]_5 xy=[2]_5 ⇒x×[4]_5 =[2]_5 x×[4]_5 =[12] x=[((12)/4)]_5 =[3]_5 x=[3]_5 ,y=z=[4]_5 For example x=3,y=4,z=4](Q13092.png)
$$\:^{{Rasheed}\:{Soomro}} \\ $$$${x}+{y}+{z}+\mathrm{xy}+\mathrm{yz}=\left[\mathrm{1}\right]_{\mathrm{5}} +\left[\mathrm{2}\right]_{\mathrm{5}} +\:\left[\mathrm{1}\right]_{\mathrm{5}} \\ $$$$\mathrm{y}+\mathrm{x}+\mathrm{z}+\mathrm{y}\left(\mathrm{x}+\mathrm{z}\right)=\left[\mathrm{4}\right]_{\mathrm{5}} \\ $$$$\mathrm{y}+\left(\mathrm{x}+\mathrm{z}\right)\left(\mathrm{1}+\mathrm{y}\right)=\left[\mathrm{4}\right]_{\mathrm{5}} \\ $$$$\mathrm{1}+\mathrm{y}+\left(\mathrm{x}+\mathrm{z}\right)\left(\mathrm{1}+\mathrm{y}\right)=\left[\mathrm{4}\right]_{\mathrm{5}} +\mathrm{1} \\ $$$$\left(\mathrm{1}+\mathrm{y}\right)\left(\mathrm{1}+\mathrm{x}+\mathrm{z}\right)=\left[\mathrm{4}\right]_{\mathrm{5}} +\left[\mathrm{1}\right]_{\mathrm{5}} =\left[\mathrm{0}\right]_{\mathrm{5}} \\ $$$$\left(\mathrm{1}+\mathrm{y}\right)=\left[\mathrm{0}\right]_{\mathrm{5}} \mathrm{or}\left(\mathrm{1}+\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{z}}\right)=\left[\mathrm{0}\right]_{\mathrm{5}} \\ $$$$\boldsymbol{\mathrm{y}}=\left[−\mathrm{1}\right]_{\mathrm{5}} \:\boldsymbol{\mathrm{or}}\:\mathrm{x}+\mathrm{z}=\left[−\mathrm{1}\right]_{\mathrm{5}} \\ $$$$\mathrm{y}=\left[\mathrm{4}\right]_{\mathrm{5}} \:\mathrm{or}\:\mathrm{x}+\mathrm{z}=\left[\mathrm{4}\right]_{\mathrm{5}} \\ $$$$\mathrm{yz}=\left[\mathrm{1}\right]_{\mathrm{5}} \\ $$$$\left[\mathrm{4}\right]\mathrm{z}=\left[\mathrm{16}\right]_{\mathrm{5}} \\ $$$$\mathrm{z}=\:\left[\frac{\mathrm{16}}{\mathrm{4}}\right]_{\mathrm{5}} =\left[\mathrm{4}\right]_{\mathrm{5}} \\ $$$$\mathrm{xy}=\left[\mathrm{2}\right]_{\mathrm{5}} \Rightarrow\mathrm{x}×\left[\mathrm{4}\right]_{\mathrm{5}} =\left[\mathrm{2}\right]_{\mathrm{5}} \\ $$$$\mathrm{x}×\left[\mathrm{4}\right]_{\mathrm{5}} =\left[\mathrm{12}\right] \\ $$$$\mathrm{x}=\left[\frac{\mathrm{12}}{\mathrm{4}}\right]_{\mathrm{5}} =\left[\mathrm{3}\right]_{\mathrm{5}} \\ $$$$\mathrm{x}=\left[\mathrm{3}\right]_{\mathrm{5}} ,\mathrm{y}=\mathrm{z}=\left[\mathrm{4}\right]_{\mathrm{5}} \\ $$$$\mathrm{For}\:\mathrm{example} \\ $$$$\mathrm{x}=\mathrm{3},\mathrm{y}=\mathrm{4},\mathrm{z}=\mathrm{4} \\ $$$$ \\ $$$$ \\ $$
Commented by 433 last updated on 14/May/17
![You forgot x+z=[4]_5 but thank you](Q13094.png)
$${You}\:{forgot}\:{x}+{z}=\left[\mathrm{4}\right]_{\mathrm{5}} \:{but}\:{thank}\:{you} \\ $$
Commented by RasheedSindhi last updated on 14/May/17
![From above answer: { ((x+z=[4]_5 .........(i))),((xy=[2]_5 .............(ii))),((yz=[1]_5 ..............(iii))) :} (i)⇒x=[4]_5 −z (ii)⇒([4]_5 −z)y=[2]_5 [4]_5 y−yz=[2]_5 [4]_5 y−[1]_5 =[2]_5 (∵ yz=[1]_5 (iii) ) [4]_5 y=[2]_5 +[1]_5 =[3]_5 =[8]_5 y=[(8/4)]_5 =[2]_5 ⇒y=[2]_5 (ii)⇒x([2]_5 )=[2]_5 ⇒x=[1]_5 (i)⇒z=[4]_5 −x=[4]_5 −[1]_5 =[3]_5 z=[3]_5 x=[1]_5 , y=[2]_5 , z=[3]_5 For example: x=1 , y=2 , z=3](Q13096.png)
$${From}\:{above}\:{answer}: \\ $$$$\:\:\begin{cases}{\mathrm{x}+\mathrm{z}=\left[\mathrm{4}\right]_{\mathrm{5}} .........\left(\mathrm{i}\right)}\\{\mathrm{xy}=\left[\mathrm{2}\right]_{\mathrm{5}} .............\left(\mathrm{ii}\right)}\\{\mathrm{yz}=\left[\mathrm{1}\right]_{\mathrm{5}} ..............\left(\mathrm{iii}\right)}\end{cases} \\ $$$$\left(\mathrm{i}\right)\Rightarrow\mathrm{x}=\left[\mathrm{4}\right]_{\mathrm{5}} −\mathrm{z} \\ $$$$\left(\mathrm{ii}\right)\Rightarrow\left(\left[\mathrm{4}\right]_{\mathrm{5}} −\mathrm{z}\right)\mathrm{y}=\left[\mathrm{2}\right]_{\mathrm{5}} \\ $$$$\left[\mathrm{4}\right]_{\mathrm{5}} \mathrm{y}−\mathrm{yz}=\left[\mathrm{2}\right]_{\mathrm{5}} \\ $$$$\left[\mathrm{4}\right]_{\mathrm{5}} \mathrm{y}−\left[\mathrm{1}\right]_{\mathrm{5}} =\left[\mathrm{2}\right]_{\mathrm{5}} \:\:\:\:\:\left(\because\:\mathrm{yz}=\left[\mathrm{1}\right]_{\mathrm{5}} \:\:\left(\mathrm{iii}\right)\:\right) \\ $$$$\left[\mathrm{4}\right]_{\mathrm{5}} \mathrm{y}=\left[\mathrm{2}\right]_{\mathrm{5}} +\left[\mathrm{1}\right]_{\mathrm{5}} =\left[\mathrm{3}\right]_{\mathrm{5}} =\left[\mathrm{8}\right]_{\mathrm{5}} \\ $$$$\mathrm{y}=\left[\frac{\mathrm{8}}{\mathrm{4}}\right]_{\mathrm{5}} =\left[\mathrm{2}\right]_{\mathrm{5}} \Rightarrow\mathrm{y}=\left[\mathrm{2}\right]_{\mathrm{5}} \\ $$$$\left(\mathrm{ii}\right)\Rightarrow\mathrm{x}\left(\left[\mathrm{2}\right]_{\mathrm{5}} \right)=\left[\mathrm{2}\right]_{\mathrm{5}} \\ $$$$\:\:\:\:\:\:\:\:\Rightarrow\mathrm{x}=\left[\mathrm{1}\right]_{\mathrm{5}} \\ $$$$\left(\mathrm{i}\right)\Rightarrow\mathrm{z}=\left[\mathrm{4}\right]_{\mathrm{5}} −\mathrm{x}=\left[\mathrm{4}\right]_{\mathrm{5}} −\left[\mathrm{1}\right]_{\mathrm{5}} =\left[\mathrm{3}\right]_{\mathrm{5}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{z}=\left[\mathrm{3}\right]_{\mathrm{5}} \\ $$$$\mathrm{x}=\left[\mathrm{1}\right]_{\mathrm{5}} \:,\:\mathrm{y}=\left[\mathrm{2}\right]_{\mathrm{5}} \:,\:\mathrm{z}=\left[\mathrm{3}\right]_{\mathrm{5}} \\ $$$$\mathrm{For}\:\mathrm{example}: \\ $$$$\mathrm{x}=\mathrm{1}\:,\:\mathrm{y}=\mathrm{2}\:,\:\mathrm{z}=\mathrm{3} \\ $$
Commented by mrW1 last updated on 14/May/17
$${Great}\:{job}! \\ $$
Commented by RasheedSindhi last updated on 16/May/17
$$\mathcal{TH}\alpha{n}\mathcal{X}\:\:\mathcal{S}{ir}! \\ $$
Answered by RasheedSindhi last updated on 14/May/17
![AnOther Way_(−) ^(Rasheed Soomro) x+y+z≡1(mod 5).......I xy≡2(mod 5)..............II yz≡1(mod 5)..............III I+II+III⇒ x+y+z+xy+yz≡4(mod 5) y+(x+z)+y(x+z)≡4(mod 5) y+(x+z)(1+y)≡4(mod 5) 1+y+(x+z)(1+y)≡4+1(mod 5) (1+y)(1+x+z)≡0(mod 5) 1+y≡0 ∨ 1+x+z≡0 (mod 5) y≡−1 ∨ x+z≡−1 (mod 5) y≡4 ∨ x+z≡4 (mod 5).....A { ((y≡4(mod 5)..............(i))),((xy≡2≡12(mod 5)......(ii))) :} (ii)/(i)⇒x≡3(mod 5) { ((y≡4(mod 5)..............(iii))),((yz≡1≡16(mod 5).......(iv))) :} (iv)/(iii)⇒z≡4(mod 5) x,y,z≡3,4,4(mod 5) For example: x=3,y=4,z=4 2nd part x+z≡4(mod 5) [From A] x≡4−z(mod 5).............(v) xy≡2(mod 5) [From II]..(vi) From (v) &(vi): (4−z)y≡2(mod 5) 4y−yz≡2(mof 5) But yz≡1(mod 5) [From III] ∴ 4y−1≡2(mod 5) 4y≡3≡8(mod 5) y≡2(mod 5) II⇒x(2)≡2(mod 5) x≡1(mod 5) From I : z≡4−x(mod 5) z≡4−1(mod 5) z≡3(mod 5) x,y,z≡1,2,3(mod 5)](Q13095.png)
$$\underset{−} {\mathrm{AnOther}\:\mathrm{Way}}\:\:\:\:\:\:\:\:\:^{{Rasheed}\:{Soomro}} \\ $$$$\mathrm{x}+\mathrm{y}+\mathrm{z}\equiv\mathrm{1}\left(\mathrm{mod}\:\mathrm{5}\right).......\mathrm{I} \\ $$$$\mathrm{xy}\equiv\mathrm{2}\left(\mathrm{mod}\:\mathrm{5}\right)..............\mathrm{II} \\ $$$$\mathrm{yz}\equiv\mathrm{1}\left(\mathrm{mod}\:\mathrm{5}\right)..............\mathrm{III} \\ $$$$\mathrm{I}+\mathrm{II}+\mathrm{III}\Rightarrow \\ $$$$\:\:\:\:\:\mathrm{x}+\mathrm{y}+\mathrm{z}+\mathrm{xy}+\mathrm{yz}\equiv\mathrm{4}\left(\mathrm{mod}\:\mathrm{5}\right) \\ $$$$\:\:\mathrm{y}+\left(\mathrm{x}+\mathrm{z}\right)+\mathrm{y}\left(\mathrm{x}+\mathrm{z}\right)\equiv\mathrm{4}\left(\mathrm{mod}\:\mathrm{5}\right) \\ $$$$\mathrm{y}+\left(\mathrm{x}+\mathrm{z}\right)\left(\mathrm{1}+\mathrm{y}\right)\equiv\mathrm{4}\left(\mathrm{mod}\:\mathrm{5}\right) \\ $$$$\mathrm{1}+\mathrm{y}+\left(\mathrm{x}+\mathrm{z}\right)\left(\mathrm{1}+\mathrm{y}\right)\equiv\mathrm{4}+\mathrm{1}\left(\mathrm{mod}\:\mathrm{5}\right) \\ $$$$\left(\mathrm{1}+\mathrm{y}\right)\left(\mathrm{1}+\mathrm{x}+\mathrm{z}\right)\equiv\mathrm{0}\left(\mathrm{mod}\:\mathrm{5}\right) \\ $$$$\mathrm{1}+\mathrm{y}\equiv\mathrm{0}\:\vee\:\mathrm{1}+\mathrm{x}+\mathrm{z}\equiv\mathrm{0}\:\left(\mathrm{mod}\:\mathrm{5}\right) \\ $$$$\mathrm{y}\equiv−\mathrm{1}\:\vee\:\mathrm{x}+\mathrm{z}\equiv−\mathrm{1}\:\left(\mathrm{mod}\:\mathrm{5}\right) \\ $$$$\mathrm{y}\equiv\mathrm{4}\:\vee\:\mathrm{x}+\mathrm{z}\equiv\mathrm{4}\:\left(\mathrm{mod}\:\mathrm{5}\right).....\boldsymbol{\mathrm{A}} \\ $$$$\begin{cases}{\mathrm{y}\equiv\mathrm{4}\left(\mathrm{mod}\:\mathrm{5}\right)..............\left(\mathrm{i}\right)}\\{\mathrm{xy}\equiv\mathrm{2}\equiv\mathrm{12}\left(\mathrm{mod}\:\mathrm{5}\right)......\left(\mathrm{ii}\right)}\end{cases} \\ $$$$\left(\mathrm{ii}\right)/\left(\mathrm{i}\right)\Rightarrow\mathrm{x}\equiv\mathrm{3}\left(\mathrm{mod}\:\mathrm{5}\right) \\ $$$$\begin{cases}{\mathrm{y}\equiv\mathrm{4}\left(\mathrm{mod}\:\mathrm{5}\right)..............\left(\mathrm{iii}\right)}\\{\mathrm{yz}\equiv\mathrm{1}\equiv\mathrm{16}\left(\mathrm{mod}\:\mathrm{5}\right).......\left(\mathrm{iv}\right)}\end{cases} \\ $$$$\left(\mathrm{iv}\right)/\left(\mathrm{iii}\right)\Rightarrow\mathrm{z}\equiv\mathrm{4}\left(\mathrm{mod}\:\mathrm{5}\right) \\ $$$$ \\ $$$$\mathrm{x},\mathrm{y},\mathrm{z}\equiv\mathrm{3},\mathrm{4},\mathrm{4}\left(\mathrm{mod}\:\mathrm{5}\right) \\ $$$$\mathrm{For}\:\mathrm{example}: \\ $$$$\mathrm{x}=\mathrm{3},\mathrm{y}=\mathrm{4},\mathrm{z}=\mathrm{4} \\ $$$$\mathrm{2}\boldsymbol{\mathrm{nd}}\:\boldsymbol{\mathrm{part}} \\ $$$$\mathrm{x}+\mathrm{z}\equiv\mathrm{4}\left(\mathrm{mod}\:\mathrm{5}\right)\:\:\left[\mathrm{From}\:\boldsymbol{\mathrm{A}}\right] \\ $$$$\mathrm{x}\equiv\mathrm{4}−\mathrm{z}\left(\mathrm{mod}\:\mathrm{5}\right).............\left(\mathrm{v}\right) \\ $$$$\mathrm{xy}\equiv\mathrm{2}\left(\mathrm{mod}\:\mathrm{5}\right)\:\:\left[\mathrm{From}\:\mathrm{II}\right]..\left(\mathrm{vi}\right) \\ $$$$\mathrm{From}\:\left(\mathrm{v}\right)\:\&\left(\mathrm{vi}\right): \\ $$$$\left(\mathrm{4}−\mathrm{z}\right)\mathrm{y}\equiv\mathrm{2}\left(\mathrm{mod}\:\mathrm{5}\right) \\ $$$$\mathrm{4y}−\mathrm{yz}\equiv\mathrm{2}\left(\mathrm{mof}\:\mathrm{5}\right) \\ $$$$\:\mathrm{But}\:\mathrm{yz}\equiv\mathrm{1}\left(\mathrm{mod}\:\mathrm{5}\right)\:\left[\mathrm{From}\:\mathrm{III}\right] \\ $$$$\therefore\:\:\mathrm{4y}−\mathrm{1}\equiv\mathrm{2}\left(\mathrm{mod}\:\mathrm{5}\right) \\ $$$$\:\:\:\:\:\:\mathrm{4y}\equiv\mathrm{3}\equiv\mathrm{8}\left(\mathrm{mod}\:\mathrm{5}\right) \\ $$$$\:\:\:\:\:\:\:\mathrm{y}\equiv\mathrm{2}\left(\mathrm{mod}\:\mathrm{5}\right) \\ $$$$\mathrm{II}\Rightarrow\mathrm{x}\left(\mathrm{2}\right)\equiv\mathrm{2}\left(\mathrm{mod}\:\mathrm{5}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{x}}\equiv\mathrm{1}\left(\boldsymbol{\mathrm{mod}}\:\mathrm{5}\right) \\ $$$$\mathrm{From}\:\mathrm{I}\::\:\mathrm{z}\equiv\mathrm{4}−\mathrm{x}\left(\mathrm{mod}\:\mathrm{5}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{z}\equiv\mathrm{4}−\mathrm{1}\left(\mathrm{mod}\:\mathrm{5}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{z}\equiv\mathrm{3}\left(\mathrm{mod}\:\mathrm{5}\right) \\ $$$$\mathrm{x},\mathrm{y},\mathrm{z}\equiv\mathrm{1},\mathrm{2},\mathrm{3}\left(\mathrm{mod}\:\mathrm{5}\right) \\ $$$$ \\ $$