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Try this beautiful problem from American Invitational Mathematics Examination, HANOI, 2018 based on **Inequality.**

Find the number of integers that satisfy the inequality \(n^{4}-n^{3}-3n^{2}-3n-17 \lt 0\).

- is 4
- is 6
- is 8
- cannot be determined from the given information

Algebra

Theory of Equations

Inequality

But try the problem first...

Answer: is 4.

Source

Suggested Reading

HANOI, 2018

Inequalities (Little Mathematical Library) by Korovkin

First hint

We have \((n+1)^{3}+16 \gt n^{4} \geq 0\) which implies \(n \geq -3\).

Second Hint

For \(n \geq 4\) we have \(n^{4}-(n+1)^{3}\) \(\geq 3n^{3}-3n^{2}-3n-1\) \(\geq 12n^{2}-3n^{2}-3n-1\) \(=n(n-3)+8n^{2}-1 \gt 16\).

Final Step

Then \(-3 \leq n \leq 3\). By directly calculation we obtain n=-1,0,1,2 that is 4 such integers.

- https://www.cheenta.com/cubes-and-rectangles-math-olympiad-hanoi-2018/
- https://www.youtube.com/watch?v=ST58GTF95t4&t=140s

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