For how many positive integers $m$ the number $m^3+5m^2+3$ is a perfect cube?

What if $m$ is a negative integer?

It intuitively seems to me that no positive integer can satisfy the given condition, but I can't prove it mathematically.

  • $\begingroup$ Maybe if you write the closed form of the Perfect Cube, you can move forward in the proof. $\endgroup$ – NoChance Jun 30 at 13:41
  • $\begingroup$ Do you mean $(x + y)^3$ ? $\endgroup$ – Schiele Jun 30 at 13:46
  • $\begingroup$ Yes, but now you have 2 "better" solutions to choose from. $\endgroup$ – NoChance Jun 30 at 14:01
  • $\begingroup$ Just as a curiosity, it has infinitely many rational solutions $m$... $\endgroup$ – xarles Jun 30 at 22:51
  • $\begingroup$ @xarles Give that infinite family of solutions, please. $\endgroup$ – Parcly Taxel Jul 3 at 7:51

We only need to check a finite number of cases: when $|m|$ becomes large enough $f(m)=5m^2+3$ will always lie between $g(m)=3m^2+3m+1=(m+1)^3-m^3$ and $h(m)=6m^2+12m+8=(m+2)^3-m^3$, and so $m^3+5m^2+3$ is between two consecutive cubes.

$h(m)-f(m)$ is positive for $m\le-12$ and $m\ge0$, while $f(m)-g(m)$ is always positive. Therefore, we only need check $-11\le m\le-1$ inclusive, and this reveals that no $m$ is such that $m^3+5m^2+3$ is a perfect cube.

  • $\begingroup$ But why it is important that $h(m) - f(m)$ and $f(m)-g(m)$ are positive? $\endgroup$ – Schiele Jun 30 at 14:54
  • 3
    $\begingroup$ @Schiele We want $g(m)<f(m)<h(m)$, so that when we add $m^3$ we get that the original expression is bounded between two consecutive cubes, and thus cannot be a cube itself. $\endgroup$ – Parcly Taxel Jun 30 at 14:56

So we have for some $n$: $$m^3+5m^2+3=n^3$$ Say $n$ and $m$ differ for $k$ then $n=m+k$ so we have $$5m^2+3 = 3m^2k+3mk^2+k^3$$ and now we have a quadratic equation in $m$ with an integer parameter $k$: $$\boxed{m^2(3k-5)+3mk^2+(k^3-3)=0}$$

Now the discirminant is a perfect square so $$9k^4-4(3k-5)(k^3-3)=d^2$$ for some integer $d$. Now check when this polynomial $$p(k)=-3k^4+20k^3+36k-60$$ is positive (and a perfect square) and you are (almost) done.

Since (I used here Am-Gm inequality) $$20k^3+36k\geq 3k^4+60 >2k^4+\underbrace{k^4+3+9+3}_{\geq 36|k|}$$

we have $$k^3(10-k)>0\implies k\in \{1,2,3...,9\}$$

  • 1
    $\begingroup$ Isn't $p(k)=-3k^4+20k^3+36k-60$ ? $\endgroup$ – Schiele Jun 30 at 16:42
  • $\begingroup$ Yes, I correct it know $\endgroup$ – Aqua Jun 30 at 16:48
  • $\begingroup$ I have another question if that doesn't bother you: proving that the discriminant is positive for $k ∈ { 1, 2, 3, ..., 9 }$ means that the equation has solution, but what are the $m$ that solve the equation ? $\endgroup$ – Schiele Jun 30 at 16:57
  • $\begingroup$ For each $k$ you must solve the quadrtic equation in the box. But the discriminant must be perfect square! $\endgroup$ – Aqua Jun 30 at 17:03
  • $\begingroup$ Is it of any help? $\endgroup$ – Aqua Jun 30 at 17:25

Assume, for a contradiction,that it's a perfect cube and notice : $m^3 < m^3 +5m^2+3 < (m+2)^3$ so we must have $m^3+5m^2+3=(m+1)^3$ but that equation has no integer solutions.


HINT.-If $m^3+5m^2+3=n^3$ then $n=m+h$ where $h$ is a natural integer and $$5m^2+3=3m^2h+3mh^2+h^3$$ $$(5-3h)m^2-3h^2m+(3-h^3)=0$$ and the discriminant of this quadratic equation in $m$ is equal to $$\Delta=-3h^4+20h^3+36h-60$$ we have that $$\Delta\ge0\iff h=2,3,4,5,6$$ On the other hand $$\sqrt \Delta$$ is not integer for these five values of $h$.


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