# Proving there are no rational cubic roots between two $f^3$ and $(f+1)^3$

Consider the difference between two consecutive positive cubes $$f^3$$ and $$(f+1)^3$$ is given by $$3f^2 + 3f + 1$$. If $$f$$ is a positive integer, there are no integer solutions for $$\sqrt[3]{3f^2 + 3f + 1}$$ unless $$f=0$$, as this directly follows from FLT.

I'm interested instead in the case where $$f = \frac{a}{b}$$ is not an integer but a rational number greater than one, and proving there are no rational cubic roots in this case. I don't think FLT applies as it is not integer, nor can I use rational root theorem. I think I have an outline of a proof that there are no rational solutions even in this case, but I am not sure if I'm approaching it correctly, so I'll outline it briefly here. First, we assume there is an irreducible rational solution $$p/q$$, then

$$p^3 = q^3\left( \frac{3a^2}{b^2} + \frac{3a}{b} + 1 \right)$$

As $$p^3$$ is an integer, then multiplication by integer $$q^3$$ must convert the RHS to an integer, so I think one can write that $$q^3 = mb^2$$ where $$m$$ is a positive whole number. It follows that

$$p^3 = m(3a^2 + 3ab + b^2)$$

But the bracketed term on the the RHS is an integer, so this implies that $$p^3 = nm$$. But this would mean that $$p^3/q^3$$ is not irreducible (common factor $$m$$), a contradiction to what we stated, suggesting there's no rational solution.

The problem is that even if this is correct, this wouldn't work in the case of $$m=1$$ in its current form. Is the logic of what I've done so far reasonable, and is there a way to better do this or patch the $$m=1$$ case?

• I can definitely see why we must have that $q=1$ or else that $q^3=mb^2$ for some positive integer $m$ (at least, we can conclude that $m$ is positive if we further assume that $q$ is positive). I'm not seeing offhand how to rule out that $q=1,$ though. I also can't see off-hand how to patch the $m=1$ issue. Commented May 19, 2023 at 1:30
• If q = 1 we're back to the integer case on RHS, and I think this also implies b = 1 so there's integers on LHS and FLT applies here
– DRG
Commented May 19, 2023 at 7:26
• Your title doesn't seem to summarize the question you go on to consider. Shouldn't it be "Proving that there are no rational cube roots of $(f+1)^3-f^3$ if $f$ is rational"? Commented May 19, 2023 at 21:11
• Yes @AdamBailey - should I change it?
– DRG
Commented May 19, 2023 at 21:48
• @DRG Yes, editing your title to show what you are really asking would be helpful - eg to those who might be searching the site in future. Commented May 22, 2023 at 8:52

If $$\left(\frac{a}{b} + 1\right)^3 - \left(\frac{a}{b}\right)^3 = \left(\frac{c}{d}\right)^3$$ for some rational numbers $$a/b$$ and $$c/d$$, you can multiply everything by $$b^3d^3$$ to get integers, $$(ad + bd)^3 - (ad)^3 = (bc)^3$$ thus but Fermat-Wiles theorem, we must have $$bc = 0$$ or $$ad = 0$$ or $$(a + b)d = 0$$. $$b \neq 0$$ and $$d \neq 0$$ so $$a = 0$$ (i.e. $$f = a/b = 0$$) or $$a = -b$$ (i.e. $$f = -1$$) or $$c = 0$$ i.e. $$(f + 1)^3 = f^3$$ which is impossible.
• Note that the impossibility of $x^3+y^3=z^3$ in nonzero integers was proved long before Wiles' work. Commented May 19, 2023 at 9:58