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Question Number 63769 by aliesam last updated on 08/Jul/19

Commented by mathmax by abdo last updated on 09/Jul/19
$let prove by recurence n=0 A_0 =0 is divisible by 6 let suppose A_n =n^3 +5n is divisible by 6 ⇒ A_(n+1) =(n+1)^3 +5(n+1) =n^3 +3n^2 +3n +1 +5n +5 =n^3 +5n +3n^2 +3n +6 n^3 +5n =6k( hypothesis of recurence) 3n^2 +3n +6 =3{n^2 +n +2} =3{2k^′ +2}=6{k^′ +1} ⇒ A_(n+1) =6k +6{k^′ +1} =6{k+k^′ +1} ⇒A_(n+1) is divisible by 6 the result is ptoved.$
$l e t p r o v e b y r e c u r e n c e n = 0 A_{0} = 0 i s d i v i s i b l e b y 6$ $l e t s u p p o s e A_{n} = n^{3} + 5 n i s d i v i s i b l e b y 6 \Rightarrow$ $A_{n + 1} = {(n + 1)}^{3} + 5 (n + 1)$ $= n^{3} + 3 n^{2} + 3 n + 1 + 5 n + 5 = n^{3} + 5 n + 3 n^{2} + 3 n + 6$ $n^{3} + 5 n = 6 k (h y p o t h e s i s o f r e c u r e n c e)$ $3 n^{2} + 3 n + 6 = 3 {n^{2} + n + 2} = 3 {2 k^{^{'}} + 2} = 6 {k^{^{'}} + 1} \Rightarrow$ $A_{n + 1} = 6 k + 6 {k^{^{'}} + 1} = 6 {k + k^{^{'}} + 1} \Rightarrow A_{n + 1} i s d i v i s i b l e b y 6$ $t h e r e s u l t i s p t o v e d .$
	Answered by MJS last updated on 08/Jul/19

	$6 ∣ n^{3} + 5 n \Leftrightarrow 2 ∣ n^{3} + 5 n \land 3 ∣ n^{3} + 5 n$ $2 ∣ n^{3} + 5 n$ $1 . n = 2 k$ ${(2 k)}^{3} + 5 \times 2 k = 2 \times (4 k^{3} + 5 k)$ $2 . n = 2 k + 1$ ${(2 k + 1)}^{3} + 5 \times (2 k + 1) = 2 \times (4 k^{3} + 6 k^{2} + 8 k + 1)$ $\Rightarrow 2 ∣ n^{3} + n n$ $3 ∣ n^{3} + 5 n$ $n^{3} + 5 n = n (n^{2} + 5)$ $3 ∣ n \lor 3 ∣ n^{2} + 5$ $1 . n = 3 k \Rightarrow 3 ∣ n$ $2 . n = 3 k + 1$ ${(3 k + 1)}^{2} + 5 = 3 \times (3 k^{2} + 3 k + 2)$ $3 . n = 3 k + 2$ ${(3 k + 2)}^{2} + 5 = 3 \times (3 k^{2} + 4 k + 3)$ $\Rightarrow 3 ∣ n^{3} + 5 n$ $\Rightarrow 6 ∣ n^{3} + 5 n$
	Commented by aliesam last updated on 09/Jul/19

	$g o d b l e s s y o u$
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