Question-113578

Question Number 113578 by Algoritm last updated on 14/Sep/20

Commented by Algoritm last updated on 14/Sep/20

x = 1553 prove that

Commented by Algoritm last updated on 14/Sep/20

step by step solution sir please

Answered by 1549442205PVT last updated on 14/Sep/20

Solution : (2006! + \frac{4012!}{2006!}) = x (\mod 4013)

\Leftrightarrow {(2006!)}^{2} - 2006! x + 4012! = 0 (\mod 4013) (*)

\Leftrightarrow 4 . {(2006!)}^{2} - 4.2006! x + 4.4012! \equiv 0

\Leftrightarrow {(2.2006! - x)}^{2} - x^{2} + 4.4012! \equiv 0

\Leftrightarrow {(2.2006! - x)}^{2} = x^{2} - 4.4012! (\mod 4013) (1)

For x = 1553 we have

x^{2} = 1553^{2} = 601.4013 - 4 \Rightarrow 1553^{2} = - 4 (\mod 4013) .

On the other hands,

Since 4013 is prime number, by

{Wilson}^{'} s theorem we have (4012! + 1) ⋮ 4013

It follows that 4012! = - 1 \mod (4013) (2)

\Rightarrow 4.4012! = - 4 (\mod 4013)

From (1) (2) (3) we get

{(2.2006! - 1553)}^{2} = 1553^{2} - 4.4012! (\mod 4013)

= - 4 - (- 4) = 0 (\mod 4013)

\Rightarrow {(2.2016! - 1553)}^{2} = 0 (\mod 4013)

\Leftrightarrow 2.2016! - 1553 = 0 (\mod 4013)

The last equality is always true

because of 2.2016! = 2 . (1.2.3 \dots . 2016)

⋮ 1553

Thus, we proved that (1) is true for

x = 1553 or (*) is true for x = 1553 .

Therefore (2006! + \frac{4012!}{2006!}) = 1553 (\mod 4013)

(\boldsymbol Q . \boldsymbol E . \boldsymbol D)

Commented by Algoritm last updated on 14/Sep/20

Mr you have to work without using the proven answer depending on the answer .

Commented by Algoritm last updated on 14/Sep/20

thanks

Commented by 1549442205PVT last updated on 14/Sep/20

Thank, you are welcome . From above result

we can prove that {(2016!)}^{2} + 1 ⋮ 4013

This is really an interesting Question

Commented by Algoritm last updated on 14/Sep/20

sir 2016!^{2} + 1 ⋮ 4013 ? ? ? please explain in detale

Commented by 1549442205PVT last updated on 15/Sep/20

2.2006! - 1553 \equiv 0

\Leftrightarrow {(2.2006!)}^{2} \equiv 1553^{2} \Leftrightarrow 4.2006!^{2} \equiv - 4

\Leftrightarrow 2006!^{2} \equiv - 1 \Leftrightarrow {(2006!)}^{2} + 1 = 0 (\mod 4013)

\Leftrightarrow [2006!^{2} + 1] ⋮ 4013

Commented by Algoritm last updated on 16/Sep/20

sir please help

Commented by Algoritm last updated on 16/Sep/20

2 . (2006!) = x (\mod 4013)

Commented by 1549442205PVT last updated on 19/Sep/20

It had in the proof of above problem .

It is followed from (1) (2) (3) x = 1553