x-y-z-1-5-xy-2-5-yz-1-5-Solve-system-on-Z-5-

Question Number 13081 by 433 last updated on 13/May/17

{\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

F r o m a b o v e a n s w e r :

{\begin{cases} x + z = {[4]}_{5} \dots \dots \dots (i) \\ xy = {[2]}_{5} \dots \dots \dots \dots . (ii) \\ yz = {[1]}_{5} \dots \dots \dots \dots . . (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

\underline{AnOther Way}^{R a s h e e d S o o m r o}

x + y + z \equiv 1 (\mod 5) \dots \dots . I

xy \equiv 2 (\mod 5) \dots \dots \dots \dots . . II

yz \equiv 1 (\mod 5) \dots \dots \dots \dots . . 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) \dots . . A

{\begin{cases} y \equiv 4 (\mod 5) \dots \dots \dots \dots . . (i) \\ xy \equiv 2 \equiv 12 (\mod 5) \dots \dots (ii) \end{cases}

(ii) / (i) \Rightarrow x \equiv 3 (\mod 5)

{\begin{cases} y \equiv 4 (\mod 5) \dots \dots \dots \dots . . (iii) \\ yz \equiv 1 \equiv 16 (\mod 5) \dots \dots . (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) \dots \dots \dots \dots . (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)