find n in ordre to have 7 divide n^3 +n−2


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Question Number 31043 by abdo imad last updated on 02/Mar/18

$f i n d n i n o r d r e t o h a v e 7 d i v i d e n^{3} + n - 2$
Commented by MJS last updated on 03/Mar/18
$the sequence of remains of n^3 /7 is ⟨1,1,6,1,6,6,0,...⟩ n/7 is ⟨1,2,3,4,5,6,0,...⟩ their sum−2 is ⟨0,1,0,3,2,3,5,...⟩ so the solution is ⟨1,3,8,10,15,17,...⟩ ={1+7k∣k∈N}∪{3+7k∣k∈N}$
$the sequence of remains of$ $n^{3} / 7 is ⟨ 1, 1, 6, 1, 6, 6, 0, . . . ⟩$ $n / 7 is ⟨ 1, 2, 3, 4, 5, 6, 0, . . . ⟩$ $their sum - 2 is$ $⟨ 0, 1, 0, 3, 2, 3, 5, . . . ⟩$ $so the solution is$ $⟨ 1, 3, 8, 10, 15, 17, . . . ⟩$ $= {1 + 7 k ∣ k \in N} \cup {3 + 7 k ∣ k \in N}$
Commented by rahul 19 last updated on 02/Mar/18

$1 .$
Commented by MJS last updated on 03/Mar/18

$k \in Z is possible too$
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