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Find infinitely many sets of positive integers $n,a,b,c$ such that $$n^3<a<b<c<(n+1)^3\quad\hbox{and}\quad \hbox{$abc$ is a cube},$$ or prove that this cannot be done.

Motivation: the corresponding problem for squares has no solution. That is, if $$n^2<a<b<(n+1)^2\ ,$$ then $ab$ is not a square. This is quite easy to prove. By (not entirely random) trial and error I have found one solution of the cube problem: $$n=5\ ,\quad a=128\ ,\quad b=144\ ,\quad c=162\ .$$

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A computational exploration

For a fixed value of n, there appear to be many solutions. The numbes of solutions for n = 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 are 3, 4, 6, 9, 14, 24, 18, 21, 26, 34 respectively.

For example, for n = 5 we have following solutions: [[196, 192, 126], [162, 144, 128], [200, 160, 128], [196, 189, 128], [169, 156, 144], [196, 168, 144], [192, 168, 147], [200, 180, 162], [196, 182, 169]]

One interesting this to note is that these numbers have relatively small prime factors - no number appearing in one of these solutions has a prime factor greater than 13.

These solutions come from the following Python code (is_perfect_cube comes from https://stackoverflow.com/a/23622115/1556369). Unfortunately the running time of this code goes like O(n^6) so it's hard to run searches for large n.

def is_perfect_cube(n):
    c = int(n**(1/3.))
    return (c**3 == n) or ((c+1)**3 == n)

def solutions(n):
   return [[a, b, c] for c in range(n**3 + 1, (n+1)**3) for b in range(c+1, (n+1)**3) for a in range(b+1, (n+1)**3) if is_perfect_cube(a*b*c)]

An infinite family

The existence of [196, 182, 169] at the end of the list for $n = 5$, and of $[1296, 1260, 1225]$ at the end of a similar list for $n = 10$, suggests a possible infinite family to me. These are $[14^2, 14 \times 13, 13^2]$ and $[36^2, 36 \times 35, 35^2]$ respectively. So if we have $j, k$ such that $n^3 < j^2 < k^2 < (n+1)^3$, then $j^2, jk, k^2$ is the $a, b, c$ we seek.

Taking square roots, we need $n^{3/2} < j < k < (n+1)^{3/2}$. So to have a solution with a given value of $n$, it suffices to have $(n+1)^{3/2} - n^{3/2} \ge 3$ (therefore there will be at least two integers between the real numbers $n^{3/2}$ and $(n+1)^{3/2}$. This holds for $n \ge 4$. So for each $n \ge 4$, we can find two squares between $n^3$ and $(n+1)^3$ and construct a solution from them. If we adopt the convention that we pick the smallest such squares, we get the solutions

n = 4, a = 81, b = 90, c = 100

n = 5, a = 144, b = 156, c = 169

n = 6, a = 225, b = 240, c = 256

and so on.

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