Prove that there is a perfect cube between n and 3n for any integer n≥10

Question

I was solving one of the Number theory problems from Mathematical Olympiad Challenges, And the problem goes like :

Prove that there is a perfect cube between $n$ and $3n$ for any integer $n\geq 10$.

And experimenting I found that if $a^3$ lies b/w, $n$ and $3n$ ($n≥19$) then it also lies b/w $m$ and $3m$ , where $m<a^3$ ; Therefore if the statement to be proved is true then taking $m=a^3$ implies b/w $m$ and $3m$ there lies $(a+1)^3$ i.e., $(a+1)^3$ lies b/w $a^3$ and $3a^3$ , So therefore to prove this we need to prove an equivalent inequality i.e.,

$n^3 <(n+1)^3 < 3n^3$ ; for any integer $n\geq10$

Answer 1

Let us assume $n < k^3 < 3n$

This is true for $n=10, k = 3$.

Now by construction, we know there is cube in $k^3 - 1 < k^3 < 3k^3 - 3$ So we are covered for all $n$ from $n$ to $k^3 - 1$.

Now we just have to prove that $k^3 < (k+1)^3 < 3k^3$

Obviously $k^3 < (k+1)^3$ and the second inequality is true for $k \ge 3$

And we can continue this forever.

Answer 2

I would like to prove the statement using contradiction by assuming, on the contrary that there exists a natural number $n \geq 10$ such that there is no perfect cube lying between $n$ and $3n$. Then for any natural number $k$,

$$ k^{3}<n<3 n<(k+1)^{3},$$ which implies that for all natural number $k$,

$$(k+1)^{3}-k^3>2n \Leftrightarrow 3k^2+3k+(1-2n)>0 \tag*{(*)} $$

Using that fact that $$\Delta_{(*)}<0 ,$$ we have $$ 9<12(1-2 n)\Leftrightarrow n<\frac{1}{8} , $$

which contradicts to that $n\geq 10$ and hence prove the statement.

More precisely, the statement is true for any $n\in N \backslash\{1,2,8,9\}.$