{ ((x+y+z=[1]_5 )),((xy=[2]_5 )),((yz=[1]_5 )) :} Solve system on Z


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Question Number 13081 by 433 last updated on 13/May/17
${ ((x+y+z=[1]_5 )),((xy=[2]_5 )),((yz=[1]_5 )) :} Solve system on Z_5$
${\begin{cases} x + y + z = {[1]}_{5} \\ x y = {[2]}_{5} \\ y z = {[1]}_{5} \end{cases}$ $S o l v e s y s t e m o n Z_{5}$
	Answered by RasheedSindhi last updated on 14/May/17

	$^{R a s h e e d S o o m r o}$ $x + y + z + xy + yz = {[1]}_{5} + {[2]}_{5} + {[1]}_{5}$ $y + x + z + y (x + z) = {[4]}_{5}$ $y + (x + z) (1 + y) = {[4]}_{5}$ $1 + y + (x + z) (1 + y) = {[4]}_{5} + 1$ $(1 + y) (1 + x + z) = {[4]}_{5} + {[1]}_{5} = {[0]}_{5}$ $(1 + y) = {[0]}_{5} or (1 + x + z) = {[0]}_{5}$ $y = {[- 1]}_{5} or x + z = {[- 1]}_{5}$ $y = {[4]}_{5} or x + z = {[4]}_{5}$ $yz = {[1]}_{5}$ $[4] z = {[16]}_{5}$ $z = {[\frac{16}{4}]}_{5} = {[4]}_{5}$ $xy = {[2]}_{5} \Rightarrow x \times {[4]}_{5} = {[2]}_{5}$ $x \times {[4]}_{5} = [12]$ $x = {[\frac{12}{4}]}_{5} = {[3]}_{5}$ $x = {[3]}_{5}, y = z = {[4]}_{5}$ $For example$ $x = 3, y = 4, z = 4$
	Commented by 433 last updated on 14/May/17

	$Y o u f o r g o t x + z = {[4]}_{5} b u t t h a n k y o u$
	Commented by RasheedSindhi last updated on 14/May/17
	$From above answer: { ((x+z=[4]_5 .........(i))),((xy=[2]_5 .............(ii))),((yz=[1]_5 ..............(iii))) :} (i)⇒x=[4]_5 −z (ii)⇒([4]_5 −z)y=[2]_5 [4]_5 y−yz=[2]_5 [4]_5 y−[1]_5 =[2]_5 (∵ yz=[1]_5 (iii) ) [4]_5 y=[2]_5 +[1]_5 =[3]_5 =[8]_5 y=[(8/4)]_5 =[2]_5 ⇒y=[2]_5 (ii)⇒x([2]_5 )=[2]_5 ⇒x=[1]_5 (i)⇒z=[4]_5 −x=[4]_5 −[1]_5 =[3]_5 z=[3]_5 x=[1]_5 , y=[2]_5 , z=[3]_5 For example: x=1 , y=2 , z=3$
	$F r o m a b o v e a n s w e r :$ ${\begin{cases} x + z = {[4]}_{5} . . . . . . . . . (i) \\ xy = {[2]}_{5} . . . . . . . . . . . . . (ii) \\ yz = {[1]}_{5} . . . . . . . . . . . . . . (iii) \end{cases}$ $(i) \Rightarrow x = {[4]}_{5} - z$ $(ii) \Rightarrow ({[4]}_{5} - z) y = {[2]}_{5}$ ${[4]}_{5} y - yz = {[2]}_{5}$ ${[4]}_{5} y - {[1]}_{5} = {[2]}_{5} (∵ yz = {[1]}_{5} (iii))$ ${[4]}_{5} y = {[2]}_{5} + {[1]}_{5} = {[3]}_{5} = {[8]}_{5}$ $y = {[\frac{8}{4}]}_{5} = {[2]}_{5} \Rightarrow y = {[2]}_{5}$ $(ii) \Rightarrow x ({[2]}_{5}) = {[2]}_{5}$ $\Rightarrow x = {[1]}_{5}$ $(i) \Rightarrow z = {[4]}_{5} - x = {[4]}_{5} - {[1]}_{5} = {[3]}_{5}$ $z = {[3]}_{5}$ $x = {[1]}_{5}, y = {[2]}_{5}, z = {[3]}_{5}$ $For example :$ $x = 1, y = 2, z = 3$
	Commented by mrW1 last updated on 14/May/17

	$G r e a t j o b!$
	Commented by RasheedSindhi last updated on 16/May/17

	$TH α n X S i r!$
	Answered by RasheedSindhi last updated on 14/May/17
	$AnOther Way_(−) ^(Rasheed Soomro) x+y+z≡1(mod 5).......I xy≡2(mod 5)..............II yz≡1(mod 5)..............III I+II+III⇒ x+y+z+xy+yz≡4(mod 5) y+(x+z)+y(x+z)≡4(mod 5) y+(x+z)(1+y)≡4(mod 5) 1+y+(x+z)(1+y)≡4+1(mod 5) (1+y)(1+x+z)≡0(mod 5) 1+y≡0 ∨ 1+x+z≡0 (mod 5) y≡−1 ∨ x+z≡−1 (mod 5) y≡4 ∨ x+z≡4 (mod 5).....A { ((y≡4(mod 5)..............(i))),((xy≡2≡12(mod 5)......(ii))) :} (ii)/(i)⇒x≡3(mod 5) { ((y≡4(mod 5)..............(iii))),((yz≡1≡16(mod 5).......(iv))) :} (iv)/(iii)⇒z≡4(mod 5) x,y,z≡3,4,4(mod 5) For example: x=3,y=4,z=4 2nd part x+z≡4(mod 5) [From A] x≡4−z(mod 5).............(v) xy≡2(mod 5) [From II]..(vi) From (v) &(vi): (4−z)y≡2(mod 5) 4y−yz≡2(mof 5) But yz≡1(mod 5) [From III] ∴ 4y−1≡2(mod 5) 4y≡3≡8(mod 5) y≡2(mod 5) II⇒x(2)≡2(mod 5) x≡1(mod 5) From I : z≡4−x(mod 5) z≡4−1(mod 5) z≡3(mod 5) x,y,z≡1,2,3(mod 5)$
	$\underline{AnOther Way}^{R a s h e e d S o o m r o}$ $x + y + z \equiv 1 (\mod 5) . . . . . . . I$ $xy \equiv 2 (\mod 5) . . . . . . . . . . . . . . II$ $yz \equiv 1 (\mod 5) . . . . . . . . . . . . . . III$ $I + II + III \Rightarrow$ $x + y + z + xy + yz \equiv 4 (\mod 5)$ $y + (x + z) + y (x + z) \equiv 4 (\mod 5)$ $y + (x + z) (1 + y) \equiv 4 (\mod 5)$ $1 + y + (x + z) (1 + y) \equiv 4 + 1 (\mod 5)$ $(1 + y) (1 + x + z) \equiv 0 (\mod 5)$ $1 + y \equiv 0 \lor 1 + x + z \equiv 0 (\mod 5)$ $y \equiv - 1 \lor x + z \equiv - 1 (\mod 5)$ $y \equiv 4 \lor x + z \equiv 4 (\mod 5) . . . . . A$ ${\begin{cases} y \equiv 4 (\mod 5) . . . . . . . . . . . . . . (i) \\ xy \equiv 2 \equiv 12 (\mod 5) . . . . . . (ii) \end{cases}$ $(ii) / (i) \Rightarrow x \equiv 3 (\mod 5)$ ${\begin{cases} y \equiv 4 (\mod 5) . . . . . . . . . . . . . . (iii) \\ yz \equiv 1 \equiv 16 (\mod 5) . . . . . . . (iv) \end{cases}$ $(iv) / (iii) \Rightarrow z \equiv 4 (\mod 5)$ $x, y, z \equiv 3, 4, 4 (\mod 5)$ $For example :$ $x = 3, y = 4, z = 4$ $2 nd part$ $x + z \equiv 4 (\mod 5) [From A]$ $x \equiv 4 - z (\mod 5) . . . . . . . . . . . . . (v)$ $xy \equiv 2 (\mod 5) [From II] . . (vi)$ $From (v) & (vi) :$ $(4 - z) y \equiv 2 (\mod 5)$ $4 y - yz \equiv 2 (mof 5)$ $But yz \equiv 1 (\mod 5) [From III]$ $∴ 4 y - 1 \equiv 2 (\mod 5)$ $4 y \equiv 3 \equiv 8 (\mod 5)$ $y \equiv 2 (\mod 5)$ $II \Rightarrow x (2) \equiv 2 (\mod 5)$ $x \equiv 1 (\mod 5)$ $From I : z \equiv 4 - x (\mod 5)$ $z \equiv 4 - 1 (\mod 5)$ $z \equiv 3 (\mod 5)$ $x, y, z \equiv 1, 2, 3 (\mod 5)$
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