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


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Question Number 151256 by pticantor last updated on 19/Aug/21

$A_{n} = 2^{n} + 3^{n} + 4^{n} + 5^{n}$ $B_{n} = 100^{n} + 101^{n} + 102^{n} + 103^{n}$ $1) f i n d v a l u e s o f n w h i l e 7 ∣ A_{n}$ $2) s h o w t h a t B_{n} \equiv A_{n} [7]$
	Answered by Rasheed.Sindhi last updated on 20/Aug/21
	$1) 2^3 ≡1(mod 7)⇒2^6 ≡1(mod 7)⇒2^(6k) ≡1(mod 7) 3^6 ≡1(mod 7) ⇒3^(6k) ≡1(mod 7) 4^3 ≡1(mod 7)⇒4^6 ≡1(mod 7)⇒4^(6k) ≡1(mod 7) 5^6 ≡1(mod 7) ⇒5^(6k) ≡1(mod 7) { ((( 2^(6k) ≡1(mod 7) )×2^u )),((( 3^(6k) ≡1(mod 7) )×3^u )),((( 4^(6k) ≡1(mod 7) )×4^u )),((( 5^(6k) ≡1(mod 7) )×5^u )) :} ;u∈{0,1,2,...,5} ⇒ { ((2^(6k+u) ≡2^u (mod 7))),((3^(6k+u) ≡3^u (mod 7))),((4^(6k+u) ≡4^u (mod 7))),((5^(6k+u) ≡5^u (mod 7))) :} ;u∈{0,1,2,...,5} n=6k+u⇒ A_n =2^n +3^n +4^n +5^n = 2^(6k+u) +3^(6k+u) +4^(6k+u) +5^(6k+u) ≡2^u +3^u +4^u +5^u (mod 7) u=1: 2^1 +3^1 +4^1 +5^1 =14=7×2 ✓ u=2: 2^2 +3^2 +4^2 +5^2 =54⇒7∤54 × ... u=1,3,5 satisfy the above congruence Hence ∴ For n=6k+1,6k+3,6k+5 , 7 ∣ A_n ∀k∈{0,1,2,...}$
	$1)$ $2^{3} \equiv 1 (m o d 7) \Rightarrow 2^{6} \equiv 1 (m o d 7) \Rightarrow 2^{6 k} \equiv 1 (m o d 7)$ $3^{6} \equiv 1 (m o d 7) \Rightarrow 3^{6 k} \equiv 1 (m o d 7)$ $4^{3} \equiv 1 (m o d 7) \Rightarrow 4^{6} \equiv 1 (m o d 7) \Rightarrow 4^{6 k} \equiv 1 (m o d 7)$ $5^{6} \equiv 1 (m o d 7) \Rightarrow 5^{6 k} \equiv 1 (m o d 7)$ ${\begin{cases} (2^{6 k} \equiv 1 (m o d 7)) \times 2^{u} \\ (3^{6 k} \equiv 1 (m o d 7)) \times 3^{u} \\ (4^{6 k} \equiv 1 (m o d 7)) \times 4^{u} \\ (5^{6 k} \equiv 1 (m o d 7)) \times 5^{u} \end{cases}; u \in {0, 1, 2, . . ., 5}$ $\Rightarrow {\begin{cases} 2^{6 k + u} \equiv 2^{u} (m o d 7) \\ 3^{6 k + u} \equiv 3^{u} (m o d 7) \\ 4^{6 k + u} \equiv 4^{u} (m o d 7) \\ 5^{6 k + u} \equiv 5^{u} (m o d 7) \end{cases}; u \in {0, 1, 2, . . ., 5}$ $n = 6 k + u \Rightarrow$ $A_{n} = 2^{n} + 3^{n} + 4^{n} + 5^{n} =$ $2^{6 k + u} + 3^{6 k + u} + 4^{6 k + u} + 5^{6 k + u}$ $\equiv 2^{u} + 3^{u} + 4^{u} + 5^{u} (m o d 7)$ $u = 1 : 2^{1} + 3^{1} + 4^{1} + 5^{1} = 14 = 7 \times 2 ✓$ $u = 2 : 2^{2} + 3^{2} + 4^{2} + 5^{2} = 54 \Rightarrow 7 ∤ 54 \times$ $. . .$ $u = 1, 3, 5 s a t i s f y t h e a b o v e c o n g r u e n c e$ $H e n c e$ $∴ F o r n = 6 k + 1, 6 k + 3, 6 k + 5, 7 ∣ A_{n}$ $\forall k \in {0, 1, 2, . . .}$
	Answered by Rasheed.Sindhi last updated on 20/Aug/21

	$2)$ $100 \equiv 2 (m o d 7) \Rightarrow 100^{n} \equiv 2^{n} (m o d 7) . . . (i)$ $101 \equiv 3 (m o d 7) \Rightarrow 101^{n} \equiv 3^{n} (m o d 7) . . . (i i)$ $102 \equiv 4 (m o d 7) \Rightarrow 102^{n} \equiv 4^{n} (m o d 7) . . . (i i i)$ $103 \equiv 5 (m o d 7) \Rightarrow 103^{n} \equiv 5^{n} (m o d 7) . . . (i v)$ $(i) + (i i) + (i i i) + (i v) :$ $100^{n} + 101^{n} + 102^{n} + 103^{n}$ $\equiv 2^{n} + 3^{n} + 4^{n} + 5^{n} (m o d 7)$ $B_{n} \equiv A_{n} (m o d 7)$
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