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Question Number 151256 by pticantor last updated on 19/Aug/21
A_n =2^n +3^n +4^n +5^n B_n =100^n +101^n +102^n +103^n 1)find values of n while 7∣A_n 2) show that B_n ≡A_n [7 ]
$$\:{A}_{{n}} =\mathrm{2}^{{n}} +\mathrm{3}^{{n}} +\mathrm{4}^{{n}} +\mathrm{5}^{{n}} \\ $$$${B}_{{n}} =\mathrm{100}^{{n}} +\mathrm{101}^{{n}} +\mathrm{102}^{{n}} +\mathrm{103}^{{n}} \\ $$$$\left.\mathrm{1}\right)\boldsymbol{{find}}\:\boldsymbol{{values}}\:\boldsymbol{{of}}\:\boldsymbol{{n}}\:\boldsymbol{{while}}\:\mathrm{7}\mid\boldsymbol{{A}}_{\boldsymbol{{n}}} \\ $$$$\left.\mathrm{2}\right)\:\boldsymbol{{show}}\:\boldsymbol{{that}}\:\boldsymbol{{B}}_{\boldsymbol{{n}}} \equiv\boldsymbol{{A}}_{\boldsymbol{{n}}} \left[\mathrm{7}\:\right] \\ $$
Answered by Rasheed.Sindhi last updated on 20/Aug/21
1) 2^3 ≡1(mod 7)⇒2^6 ≡1(mod 7)⇒2^(6k) ≡1(mod 7) 3^6 ≡1(mod 7) ⇒3^(6k) ≡1(mod 7) 4^3 ≡1(mod 7)⇒4^6 ≡1(mod 7)⇒4^(6k) ≡1(mod 7) 5^6 ≡1(mod 7) ⇒5^(6k) ≡1(mod 7) { ((( 2^(6k) ≡1(mod 7) )×2^u )),((( 3^(6k) ≡1(mod 7) )×3^u )),((( 4^(6k) ≡1(mod 7) )×4^u )),((( 5^(6k) ≡1(mod 7) )×5^u )) :} ;u∈{0,1,2,...,5} ⇒ { ((2^(6k+u) ≡2^u (mod 7))),((3^(6k+u) ≡3^u (mod 7))),((4^(6k+u) ≡4^u (mod 7))),((5^(6k+u) ≡5^u (mod 7))) :} ;u∈{0,1,2,...,5} n=6k+u⇒ A_n =2^n +3^n +4^n +5^n = 2^(6k+u) +3^(6k+u) +4^(6k+u) +5^(6k+u) ≡2^u +3^u +4^u +5^u (mod 7) u=1: 2^1 +3^1 +4^1 +5^1 =14=7×2 ✓ u=2: 2^2 +3^2 +4^2 +5^2 =54⇒7∤54 × ... u=1,3,5 satisfy the above congruence Hence ∴ For n=6k+1,6k+3,6k+5 , 7 ∣ A_n ∀k∈{0,1,2,...}
$$\left.\mathrm{1}\right) \\ $$$$\mathrm{2}^{\mathrm{3}} \equiv\mathrm{1}\left({mod}\:\mathrm{7}\right)\Rightarrow\mathrm{2}^{\mathrm{6}} \equiv\mathrm{1}\left({mod}\:\mathrm{7}\right)\Rightarrow\mathrm{2}^{\mathrm{6}{k}} \equiv\mathrm{1}\left({mod}\:\mathrm{7}\right) \\ $$$$\mathrm{3}^{\mathrm{6}} \equiv\mathrm{1}\left({mod}\:\mathrm{7}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\mathrm{3}^{\mathrm{6}{k}} \equiv\mathrm{1}\left({mod}\:\mathrm{7}\right) \\ $$$$\mathrm{4}^{\mathrm{3}} \equiv\mathrm{1}\left({mod}\:\mathrm{7}\right)\Rightarrow\mathrm{4}^{\mathrm{6}} \equiv\mathrm{1}\left({mod}\:\mathrm{7}\right)\Rightarrow\mathrm{4}^{\mathrm{6}{k}} \equiv\mathrm{1}\left({mod}\:\mathrm{7}\right) \\ $$$$\mathrm{5}^{\mathrm{6}} \equiv\mathrm{1}\left({mod}\:\mathrm{7}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\mathrm{5}^{\mathrm{6}{k}} \equiv\mathrm{1}\left({mod}\:\mathrm{7}\right) \\ $$$$\begin{cases}{\left(\:\mathrm{2}^{\mathrm{6}{k}} \equiv\mathrm{1}\left({mod}\:\mathrm{7}\right)\:\right)×\mathrm{2}^{{u}} }\\{\left(\:\mathrm{3}^{\mathrm{6}{k}} \equiv\mathrm{1}\left({mod}\:\mathrm{7}\right)\:\right)×\mathrm{3}^{{u}} }\\{\left(\:\mathrm{4}^{\mathrm{6}{k}} \equiv\mathrm{1}\left({mod}\:\mathrm{7}\right)\:\right)×\mathrm{4}^{{u}} }\\{\left(\:\mathrm{5}^{\mathrm{6}{k}} \equiv\mathrm{1}\left({mod}\:\mathrm{7}\right)\:\right)×\mathrm{5}^{{u}} }\end{cases}\:\:;{u}\in\left\{\mathrm{0},\mathrm{1},\mathrm{2},…,\mathrm{5}\right\} \\ $$$$ \\ $$$$\Rightarrow\begin{cases}{\mathrm{2}^{\mathrm{6}{k}+{u}} \equiv\mathrm{2}^{{u}} \left({mod}\:\mathrm{7}\right)}\\{\mathrm{3}^{\mathrm{6}{k}+{u}} \equiv\mathrm{3}^{{u}} \left({mod}\:\mathrm{7}\right)}\\{\mathrm{4}^{\mathrm{6}{k}+{u}} \equiv\mathrm{4}^{{u}} \left({mod}\:\mathrm{7}\right)}\\{\mathrm{5}^{\mathrm{6}{k}+{u}} \equiv\mathrm{5}^{{u}} \left({mod}\:\mathrm{7}\right)}\end{cases}\:\:;{u}\in\left\{\mathrm{0},\mathrm{1},\mathrm{2},…,\mathrm{5}\right\} \\ $$$${n}=\mathrm{6}{k}+{u}\Rightarrow \\ $$$${A}_{{n}} =\mathrm{2}^{{n}} +\mathrm{3}^{{n}} +\mathrm{4}^{{n}} +\mathrm{5}^{{n}} = \\ $$$$\mathrm{2}^{\mathrm{6}{k}+{u}} +\mathrm{3}^{\mathrm{6}{k}+{u}} +\mathrm{4}^{\mathrm{6}{k}+{u}} +\mathrm{5}^{\mathrm{6}{k}+{u}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\equiv\mathrm{2}^{{u}} +\mathrm{3}^{{u}} +\mathrm{4}^{{u}} +\mathrm{5}^{{u}} \left({mod}\:\mathrm{7}\right) \\ $$$${u}=\mathrm{1}:\:\mathrm{2}^{\mathrm{1}} +\mathrm{3}^{\mathrm{1}} +\mathrm{4}^{\mathrm{1}} +\mathrm{5}^{\mathrm{1}} =\mathrm{14}=\mathrm{7}×\mathrm{2}\:\checkmark \\ $$$${u}=\mathrm{2}:\:\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} =\mathrm{54}\Rightarrow\mathrm{7}\nmid\mathrm{54}\:× \\ $$$$… \\ $$$${u}=\mathrm{1},\mathrm{3},\mathrm{5}\:{satisfy}\:{the}\:{above}\:{congruence} \\ $$$${Hence} \\ $$$$\therefore\:{For}\:{n}=\mathrm{6}{k}+\mathrm{1},\mathrm{6}{k}+\mathrm{3},\mathrm{6}{k}+\mathrm{5}\:,\:\mathrm{7}\:\mid\:{A}_{{n}} \\ $$$$\:\:\:\:\:\:\:\forall{k}\in\left\{\mathrm{0},\mathrm{1},\mathrm{2},…\right\} \\ $$
Answered by Rasheed.Sindhi last updated on 20/Aug/21
2) 100≡2(mod 7)⇒100^n ≡2^n (mod 7)...(i) 101≡3(mod 7)⇒101^n ≡3^n (mod 7)...(ii) 102≡4(mod 7)⇒102^n ≡4^n (mod 7)...(iii) 103≡5(mod 7)⇒103^n ≡5^n (mod 7)...(iv) (i)+(ii)+(iii)+(iv): 100^n +101^n +102^n +103^n ≡2^n +3^n +4^n +5^n (mod 7) B_n ≡A_n (mod 7)
$$\left.\mathrm{2}\right) \\ $$$$\mathrm{100}\equiv\mathrm{2}\left({mod}\:\mathrm{7}\right)\Rightarrow\mathrm{100}^{{n}} \equiv\mathrm{2}^{{n}} \left({mod}\:\mathrm{7}\right)…\left({i}\right) \\ $$$$\mathrm{101}\equiv\mathrm{3}\left({mod}\:\mathrm{7}\right)\Rightarrow\mathrm{101}^{{n}} \equiv\mathrm{3}^{{n}} \left({mod}\:\mathrm{7}\right)…\left({ii}\right) \\ $$$$\mathrm{102}\equiv\mathrm{4}\left({mod}\:\mathrm{7}\right)\Rightarrow\mathrm{102}^{{n}} \equiv\mathrm{4}^{{n}} \left({mod}\:\mathrm{7}\right)…\left({iii}\right) \\ $$$$\mathrm{103}\equiv\mathrm{5}\left({mod}\:\mathrm{7}\right)\Rightarrow\mathrm{103}^{{n}} \equiv\mathrm{5}^{{n}} \left({mod}\:\mathrm{7}\right)…\left({iv}\right) \\ $$$$\left({i}\right)+\left({ii}\right)+\left({iii}\right)+\left({iv}\right): \\ $$$$\mathrm{100}^{{n}} +\mathrm{101}^{{n}} +\mathrm{102}^{{n}} +\mathrm{103}^{{n}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\equiv\mathrm{2}^{{n}} +\mathrm{3}^{{n}} +\mathrm{4}^{{n}} +\mathrm{5}^{{n}} \left({mod}\:\mathrm{7}\right) \\ $$$$\:\:\:\:\:\:\:\:\:{B}_{{n}} \equiv{A}_{{n}} \left({mod}\:\mathrm{7}\right) \\ $$

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