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

Commented by Algoritm last updated on 14/Sep/20

$$\mathrm{x}=1553\mathrm{prove}\mathrm{that}$$

Commented by Algoritm last updated on 14/Sep/20

$$\mathrm{step}\mathrm{by}\mathrm{step}\mathrm{solution}\mathrm{sir}\mathrm{please}$$

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

$$\mathrm{Solution}:(2006!+\frac{4012!}{2006!})=\mathrm{x}\left(\mathrm{mod}4013\right)$$$$\iff {(2006!)}^2-2006!\mathrm{x}+4012!=0\left(\mathrm{mod}4013\right)(\ast )$$$$\iff 4.{(2006!)}^2-4.2006!\mathrm{x}+4.4012!\equiv 0$$$$\iff {(2.2006!-\mathrm{x})}^2-{\mathrm{x}}^2+4.4012!\equiv 0$$$$\iff {(2.2006!-\mathrm{x})}^2={\mathrm{x}}^2-4.4012!\left(\mathrm{mod}4013\right)\left(1\right)$$$$\mathrm{For}\mathrm{x}=1553\mathrm{we}\mathrm{have}$$$${\mathrm{x}}^2={1553}^2=601.4013-4\Rightarrow {1553}^2=-4\left(\mathrm{mod}4013\right).$$$$\mathrm{On}\mathrm{the}\mathrm{other}\mathrm{hands},$$$$\mathrm{Since}4013\mathrm{is}\mathrm{prime}\mathrm{number},\mathrm{by}$$$${\mathrm{Wilson}}^{\prime }\mathrm{s}\mathrm{theorem}\mathrm{we}\mathrm{have}(4012!+1)⋮4013$$$$\mathrm{It}\mathrm{follows}\mathrm{that}4012!=-1\mathrm{mod}\left(4013\right)\left(2\right)$$$$\Rightarrow 4.4012!=-4\left(\mathrm{mod}4013\right)$$$$\mathrm{From}\left(1\right)\left(2\right)\left(3\right)\mathrm{we}\mathrm{get}$$$${(2.2006!-1553)}^2={1553}^2-4.4012!\left(\mathrm{mod}4013\right)$$$$=-4-(-4)=0\left(\mathrm{mod}4013\right)$$$$\Rightarrow {(2.2016!-1553)}^2=0\left(\mathrm{mod}4013\right)$$$$\iff 2.2016!-1553=0\left(\mathrm{mod}4013\right)$$$$\mathrm{The}\mathrm{last}\mathrm{equality}\mathrm{is}\mathrm{always}\mathrm{true}$$$$\mathrm{because}\mathrm{of}2.2016!=2.(1.2.3....2016)$$$$⋮1553$$$$\mathrm{Thus},\mathrm{we}\mathrm{proved}\mathrm{that}\left(1\right)\mathrm{is}\mathrm{true}\mathrm{for}$$$$\mathrm{x}=1553\mathrm{or}(\ast )\mathrm{is}\mathrm{true}\mathrm{for}\mathrm{x}=1553.$$$$\mathrm{Therefore}(2006!+\frac{4012!}{2006!})=1553\left(\mathrm{mod}4013\right)$$$$(\mathbf{Q}.\mathbf{E}.\mathbf{D})$$

Commented by Algoritm last updated on 14/Sep/20

$$\mathrm{Mr}\mathrm{you}\mathrm{have}\mathrm{to}\mathrm{work}\mathrm{without}\mathrm{using}\mathrm{the}\mathrm{proven}\mathrm{answer}\mathrm{depending}\mathrm{on}\mathrm{the}\mathrm{answer}.$$

Commented by Algoritm last updated on 14/Sep/20

$$\mathrm{thanks}$$

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

$$\mathrm{Thank},\mathrm{you}\mathrm{are}\mathrm{welcome}.\mathrm{From}\mathrm{above}\mathrm{result}$$$$\mathrm{we}\mathrm{can}\mathrm{prove}\mathrm{that}{(2016!)}^2+1⋮4013$$$$\mathrm{This}\mathrm{is}\mathrm{really}\mathrm{an}\mathrm{interesting}\mathrm{Question}$$

Commented by Algoritm last updated on 14/Sep/20

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

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

$$2.2006!-1553\equiv 0$$$$\iff {(2.2006!)}^2\equiv {1553}^2\iff 4.2006{!}^2\equiv -4$$$$\iff 2006{!}^2\equiv -1\iff {(2006!)}^2+1=0\left(\mathrm{mod}4013\right)$$$$\iff [2006{!}^2+1]⋮4013$$

Commented by Algoritm last updated on 16/Sep/20

$$\mathrm{sir}\mathrm{please}\mathrm{help}$$

Commented by Algoritm last updated on 16/Sep/20

$$2.(2006!)=\mathrm{x}\left(\mathrm{mod}4013\right)$$

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

$$\mathrm{It}\mathrm{had}\mathrm{in}\mathrm{the}\mathrm{proof}\mathrm{of}\mathrm{above}\mathrm{problem}.$$$$\mathrm{It}\mathrm{is}\mathrm{followed}\mathrm{from}\left(1\right)\left(2\right)\left(3\right)\mathrm{x}=1553$$

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