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Question Number 197752 by cortano12 last updated on 27/Sep/23

find minimum value of m such that m^(19) = 1800 (mod 2029)

$$\:\mathrm{find}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of}\:\mathrm{m} \\ $$$$\:\mathrm{such}\:\mathrm{that}\:\mathrm{m}^{\mathrm{19}} =\:\mathrm{1800}\:\left(\mathrm{mod}\:\mathrm{2029}\right) \\ $$

Answered by deleteduser1 last updated on 27/Sep/23

m^(2028) =(m^(19) )^(107) m^(−5) ≡1(mod 2029) ⇒m^5 ≡(−229)^(107) ⇒m^5 ≡1183(mod 2029) m^(19) =(((m^5 )^4 )/m)≡1800(mod 2029) ⇒1800m≡(1183)^4 ≡1584(mod 2029) ⇒25m≡22(mod 2029) ⇒25m≡22+2029×7=2425=14225 ⇒m≡569(mod 2029) ⇒min{m}=569

$${m}^{\mathrm{2028}} =\left({m}^{\mathrm{19}} \right)^{\mathrm{107}} {m}^{−\mathrm{5}} \equiv\mathrm{1}\left({mod}\:\mathrm{2029}\right) \\ $$$$\Rightarrow{m}^{\mathrm{5}} \equiv\left(−\mathrm{229}\right)^{\mathrm{107}} \Rightarrow{m}^{\mathrm{5}} \equiv\mathrm{1183}\left({mod}\:\mathrm{2029}\right) \\ $$$${m}^{\mathrm{19}} =\frac{\left({m}^{\mathrm{5}} \right)^{\mathrm{4}} }{{m}}\equiv\mathrm{1800}\left({mod}\:\mathrm{2029}\right) \\ $$$$\Rightarrow\mathrm{1800}{m}\equiv\left(\mathrm{1183}\right)^{\mathrm{4}} \equiv\mathrm{1584}\left({mod}\:\mathrm{2029}\right) \\ $$$$\Rightarrow\mathrm{25}{m}\equiv\mathrm{22}\left({mod}\:\mathrm{2029}\right) \\ $$$$\Rightarrow\mathrm{25}{m}\equiv\mathrm{22}+\mathrm{2029}×\mathrm{7}=\mathrm{2425}=\mathrm{14225} \\ $$$$\Rightarrow{m}\equiv\mathrm{569}\left({mod}\:\mathrm{2029}\right) \\ $$$$\Rightarrow{min}\left\{{m}\right\}=\mathrm{569} \\ $$

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