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

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.

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

• But why it is important that $h(m) - f(m)$ and $f(m)-g(m)$ are positive? – Schiele Jun 30 at 14:54
• @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. – 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 $$k^3(10-k)>0\implies k\in \{1,2,3...,9\}$$

• Isn't $p(k)=-3k^4+20k^3+36k-60$ ? – Schiele Jun 30 at 16:42
• Yes, I correct it know – Aqua Jun 30 at 16:48
• 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 ? – Schiele Jun 30 at 16:57
• For each $k$ you must solve the quadrtic equation in the box. But the discriminant must be perfect square! – Aqua Jun 30 at 17:03
• Is it of any help? – 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 $$h$$.