solve-at-Z-2-2x-5y-4-

Question Number 61605 by Mr X pcx last updated on 05/Jun/19

s o l v e a t Z^{2} 2 x + 5 y = 4

Commented by mr W last updated on 05/Jun/19

x = 2 - 5 n

y = 2 n

n \in Z

i s t h i s c o r r e c t ?

Commented by MJS last updated on 05/Jun/19

let y = 2 a + 2 b i

2 x + 10 a + 10 b i = 4

x = 2 - 5 a - 5 b i

{it}^{'} s correct, but {we}^{'} re in Z^{2} = R^{4}

(\begin{matrix} x_{r e a l} \\ x_{i m a g} \\ y_{r e a l} \\ y_{i m a g} \end{matrix}) = (\begin{matrix} 2 \\ 0 \\ 0 \\ 0 \end{matrix}) + a \times (\begin{matrix} - 5 \\ 0 \\ 2 \\ 0 \end{matrix}) + b \times (\begin{matrix} 0 \\ - 5 \\ 0 \\ 2 \end{matrix})

Commented by MJS last updated on 05/Jun/19

\dots so this is a plane not a line

Commented by mr W last updated on 05/Jun/19

t h a n k s a l o t s i r!

Commented by maxmathsup by imad last updated on 05/Jun/19

l e t c o n s i d e r t h e c o n g r u e n c e m o d u l o 2 (2 i s p r i m e)

(e) \Rightarrow \bar{2} \bar{x} + \bar{5} \bar{y} = \bar{4} \Rightarrow 0 + \bar{y} = 0 \Rightarrow y = 2 k (k f r o m Z)

2 x + 5 y = 4 \Rightarrow 2 x + 5 (2 k) = 4 \Rightarrow x + 5 k = 2 \Rightarrow x = - 5 k + 2 s o t h e s o l u t i o n

a r e (- 5 k + 2, 2 k) w i t h k \in Z .

r e m a r k w e c a n c o n s i d e r c o n g r u e n c e m o d u l o 5 (Z / 5 Z)

Commented by maxmathsup by imad last updated on 05/Jun/19

c o r r e c t s i r

Commented by kaivan.ahmadi last updated on 06/Jun/19

2 x \overset{5}{\equiv} 4 \Rightarrow 6 x \overset{5}{\equiv} 12 \Rightarrow x \overset{5}{\equiv} 2 \Rightarrow x = 5 k + 2, k \in Z

a n d

2 (5 k + 2) + 5 y = 4 \Rightarrow 10 k + 4 + 5 y = 4 \Rightarrow y = - 2 k