n-N-prove-9-n-3-n-1-3-n-2-3-

Question Number 21067 by youssoufab last updated on 11/Sep/17

\forall n \in N, p r o v e 9 ∣ [n^{3} + {(n + 1)}^{3} + {(n + 2)}^{3}]

Answered by dioph last updated on 12/Sep/17

n^{3} + {(n + 1)}^{3} + {(n + 2)}^{3} =

= n^{3} + n^{3} + 3 n^{2} + 3 n + 1 + n^{3} + 6 n^{2} + 12 n + 8 =

= 3 n^{3} + 9 n^{2} + 15 n + 9

= 3 n (n^{2} + 5) + 9 (n^{2} + 1) = x

Case 1 : 3 ∣ n

\Rightarrow n = 3 k

\Rightarrow x = 9 k (9 k^{2} + 5) + 9 (9 k^{2} + 1)

\Rightarrow 9 ∣ x

Case 2 : 3 ∤ n

\Rightarrow n^{2} + 5 \equiv n^{2} - 1 \equiv (n + 1) (n - 1)

\equiv 0 (\mod 3)

\Rightarrow n^{2} + 5 = 3 h

\Rightarrow x = 9 n h + 9 (n^{2} + 1)

\Rightarrow 9 ∣ x

Commented by youssoufab last updated on 14/Sep/17

t h a n k s f o r h e l p

Answered by mrW1 last updated on 12/Sep/17

let m = n + 1

\Rightarrow {(m - 1)}^{3} + m^{3} + {(m + 1)}^{3}

= m^{3} - {3 m}^{2} + 3 m - 1 + m^{3} + m^{3} + {3 m}^{2} + 3 m + 1

= {3 m}^{3} + 6 m

= 3 m (m^{2} + 2)

if m = 3 k :

3 m (m^{2} + 2) = 9 k ({9 k}^{2} + 2) \equiv 0 (\mod 9)

if m = 3 k \pm 1 :

3 m (m^{2} + 2) = 3 (3 k \pm 1) ({9 k}^{2} \pm 6 k + 1 + 2)

= 9 (3 k \pm 1) ({3 k}^{2} \pm 2 k + 1) \equiv 0 (\mod 9)

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