2^n + n|8^n + n

kunny

2^n + n|8^n + n

**Posted:** Sat Feb 21, 2009 5:52 pm

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Today's calculation of Integral Digest
Hang in there, students.

bambaman

$x^3 + y^3 = (x + y)(x^2 - xy + y^2) \implies 2^n + n|8^n + n^3 \implies 2^n + n | 8^n + n^3 - (8^n + n) = n^3 - n$
And it implies that $n^3 - n = 0$ or $|n^3 - n| \ge 2^n$ .
The first case gives $n = 1$ , which is a solution.
The second case: for $n \ge 10$ , $2^n > n^3 > n^3 - n$ , by induction: $1024 > 1000$ , and $2^n > n^3$ implies that $2^{n + 1} > 2n^3 \ge (n + 1)^3$ because $3n^2 + 3n + 1 \le 3n^2 + 3n^2 + 3n^2 = 9n^2 < n^3$ for $n \ge 10$ .
So we only need to consider $2 \le n \le 9$ . The valid solutions are $n = 2,4,6$ . When we include the 1st case, the set of solutions is $n = 1,2,4,6$ .

**Posted:** Sat Feb 21, 2009 6:27 pm

Mathias_DK

#3

Re: 2^n+n|8^n+n

**Posted:** Wed Feb 25, 2009 8:45 pm