si-g-ChineseRemainderTheorm-etermine-polynomial-p-x-such-that-p-x-8-mod-x-1-p-x-24-mod-x-3-p-x-6-mod-x-p-x-0-mod-x-2-

Question Number 67697 by Rasheed.Sindhi last updated on 30/Aug/19

⋓ si ⋒ g ChineseRemainderTheorm

\partial etermine polynomial p (x) such that

p (x) \equiv 8 (\mod x + 1)

p (x) \equiv - 24 (\mod x + 3)

p (x) \equiv 6 (\mod x)

p (x) \equiv 0 (\mod x + 2)

Answered by Rasheed.Sindhi last updated on 01/Sep/19

{\begin{cases} p (x) \equiv 8 (\mod x + 1) \\ p (x) \equiv - 24 (\mod x + 3) \\ p (x) \equiv 6 (\mod x) \\ p (x) \equiv 0 (\mod x + 2) \end{cases}

Let m_{1} = x + 1, m_{2} = x + 3, m_{3} = x, m_{4} = x + 2

Note that (m_{i}, m_{j}) = 1 for \forall i \neq j

(pairwise coprime)

Let M = m_{1} m_{2} m_{3} m_{4} = (x + 1) (x + 3) (x) (x + 2)

M_{1} = \frac{M}{m_{1}} = m_{2} m_{3} m_{4} = (x + 3) (x) (x + 2)

M_{2} = \frac{M}{m_{2}} = m_{1} m_{3} m_{4} = (x + 1) (x) (x + 2)

M_{3} = \frac{M}{m_{3}} = m_{1} m_{2} m_{4} = (x + 1) (x + 3) (x + 2)

M_{4} = \frac{M}{m_{4}} = m_{1} m_{2} m_{3} = (x + 1) (x + 3) (x)

Note that (M_{i}, m_{i}) = 1

∴ M_{i} y \equiv 1 (\mod m_{i})

(x + 3) (x) (x + 2) H_{1} (x) \equiv 1 (\mod x + 1)

(x^{3} + {5 x}^{2} + 6 x) H_{1} (x) \equiv 1 (\mod x + 1)

Let H_{1} (x) = h_{1} (a constant)

(\begin{matrix} - 1) & h_{1} & {5 h}_{1} & {6 h}_{1} & - 1 \\ - h_{1} & - {4 h}_{1} & - {2 h}_{1} \\ h_{1} & {4 h}_{1} & {2 h}_{1} & - {2 h}_{1} - 1 \end{matrix})

- {2 h}_{1} - 1 = 0 \Rightarrow h_{1} = - 1 / 2

(x + 1) (x) (x + 2) H_{2} (x) \equiv 1 (\mod x + 3)

(x^{3} + {3 x}^{2} + 2 x) H_{2} (x) \equiv 1 (\mod x + 3)

If H_{2} (x) = h_{2} (constant)

h_{2} = - 1 / 6 (By similar process)

(x + 1) (x + 3) (x + 2) H_{3} (x) \equiv 1 (\mod x)

(x^{3} + {6 x}^{2} + 11 x + 6) H_{3} (x) \equiv 1 (\mod x)

If H_{3} (x) = h_{3} (constant)

(\begin{matrix} 0) & h_{3} & {6 h}_{3} & {11 h}_{3} & {6 h}_{3} - 1 \\ 0 & 0 & 0 \\ h_{3} & {6 h}_{3} & {11 h}_{3} & {6 h}_{3} - 1 \end{matrix})

{6 h}_{3} - 1 = 0 \Rightarrow h_{3} = 1 / 6

(x + 1) (x + 3) (x) H_{4} (x) \equiv 1 (\mod x + 2)

(x^{3} + {4 x}^{2} + 3 x) H_{4} (x) \equiv 1 (\mod x + 2)

If H_{4} (x) = h_{4}

(\begin{matrix} - 2) & h_{4} & {4 h}_{4} & {3 h}_{4} & - 1 \\ - 2 h 4 & - {4 h}_{4} & {2 h}_{4} \\ h_{4} & {2 h}_{4} & - h_{4} & {2 h}_{4} - 1 \end{matrix})

{2 h}_{4} - 1 = 0 \Rightarrow h_{4} = 1 / 2

x = 8 (x^{3} + {5 x}^{2} + 6 x) (- 1 / 2)

+ (- 24) (x^{3} + {3 x}^{2} + 2 x) (- 1 / 6)

+ 6 (x^{3} + {6 x}^{2} + 11 x + 6) (1 / 6)

+ (0) (x^{3} + {4 x}^{2} + 3 x) (1 / 2)

x = - {4 x}^{3} - {20 x}^{2} - 24 x

+ {4 x}^{3} + {12 x}^{2} + 8 x

+ x^{3} + {6 x}^{2} + 11 x + 6

x = x^{3} - {2 x}^{2} - 5 x + 6

Commented by Prithwish sen last updated on 02/Sep/19

excelent .

Commented by mr W last updated on 03/Sep/19

n i c e c r e a t i o n!

Commented by Rasheed.Sindhi last updated on 04/Sep/19

Thank you both sirs!