# Numbers preceded by perfect cube and followed by perfect square

I read a puzzle about finding a natural number that is preceded by a perfect square and followed by a perfect cube. The answer was $$26$$, which is preceded by $$25=5^2$$ and followed by $$27=3^3$$. Are there any more numbers satisfying this property?

I have checked using python the first million natural numbers and there are none within that range.

Edit: I have found an answer in the site. I have edited the title for the following alternative question.

What about the reversed property: a number $$n$$ such that $$n-1$$ is a perfect cube and $$n+1$$ is a perfect square? In this case I haven't found any number at all.

For the first question, $$n - 1 = y^2$$ and $$n+1 = x^3$$, so $$n = y^2 + 1 = x^3 - 1$$, so $$y^2 = x^3 - 2$$. Fermat famously challenged British mathematicians to explain why the only solution in positive integers is $$(x,y) = (3,5)$$. Its solutions in integers are $$(x,y) = (3,\pm 5)$$ and one way to show that uses properties of the ring $$\mathbf Z[\sqrt{-2}]$$ (namely, that it has unique factorization and its only units are $$\pm 1$$). This ring gets involved by rewriting $$y^2 = x^3 - 2$$ as $$x^3 = y^2 + 2 = (y+\sqrt{-2})(y-\sqrt{-2})$$. You put a link to this approach in an edit to your post.

For the second question, $$n - 1 = x^3$$ and $$n+1 = y^2$$, so $$n = x^3 + 1 = y^2 - 1$$, so $$y^2 = x^3 + 2$$. Rewrite this as $$x^3 = y^2 - 2 = (y+\sqrt{2})(y-\sqrt{2})$$ in $$\mathbf Z[\sqrt{2}]$$, which is a PID (it is Euclidean). Show $$x$$ and $$y$$ are odd and, using that condition, show $$y+\sqrt{2}$$ and $$y-\sqrt{2}$$ are relatively prime in $$\mathbf Z[\sqrt{2}]$$. Therefore since their product is a cube, each is a cube up to unit multiple, so $$y+\sqrt{2} = (a+b\sqrt{2})^3u$$ for some unit $$u \in \mathbf Z[\sqrt{2}]$$. On the right side, unit cubes can be absorbed into $$(a+b\sqrt{2})^3$$, so we only need to use as $$u$$ representatives of the units modulo cubes of units. It turns out that the units of $$\mathbf Z[\sqrt{2}]$$ are $$\pm(1+\sqrt{2})^{\mathbf Z}$$, so representatives for the units modulo unit cubes are $$1, 1+\sqrt{2}$$, and $$(1+\sqrt{2})^{-1} = -(1-\sqrt{2})$$. We can ignore the factor $$-1$$ since it's a unit cube, so we're reduced to three equations: $$y+\sqrt{2} = (a+b\sqrt{2})^3,$$ $$y+\sqrt{2} = (a+b\sqrt{2})^3(1+\sqrt{2}),$$ $$y+\sqrt{2} = (a+b\sqrt{2})^3(1-\sqrt{2}).$$

The first equation, after equating coefficients of $$\sqrt{2}$$ on both sides, implies $$1 = 3a^2b + 2b^3 = b(3a^2 + 2b^2).$$ Therefore $$b$$ has to be $$\pm 1$$, but neither choice leads to an integer for $$a$$. So we get no solutions this way to the original equation.

The second equation, after equating coefficients of $$\sqrt{2}$$ on both sides, implies $$1 = a^3 + 3a^2b + 6ab^2 + 2b^3.$$ It turns out that the only integral solution to this is $$(a,b) = (1,0)$$, but that is not easy! (I don't have a reference for you off the top of my head -- google "cubic Thue equations".) Using this fact, $$y+\sqrt{2} = (a+b\sqrt{2})^3(1+\sqrt{2}) = 1+\sqrt{2}$$. so $$y = 1$$ and $$x^3 = y^2- 2 = -1$$. Thus $$(x,y)= (-1,1)$$.

The third equation, after equating coefficients of $$\sqrt{2}$$ on both sides, implies $$1 = (-a)^3 + 3(-a)^2b + 6(-a)b^2 + 2b^3,$$ which is the same as the previous equation with $$-a$$ in place of $$a$$. Therefore $$(a,b) = (-1,0)$$, which leads to $$y = -1$$ and then $$x^3 = y^2 - 2 = -1$$, so $$(x,y) = (-1,-1)$$.

In summary, if we accept that the only solution of $$1 = a^3 + 3a^2b + 6ab^2 + 2b^3$$ in $$\mathbf Z$$ is $$(a,b) = (1,0)$$ then the above argument shows the integral solutions of $$y^2 = x^3 + 2$$ are $$(x,y) = (-1,\pm 1)$$.

One lesson from this is that when $$y^2 = x^3 + k$$ really has integral solutions and you want to prove that you found all of them, the case when $$k > 0$$ is typically more difficult than the case when $$k < 0$$.

• Do you have a proof or link to a proof of the second statement?
– Javi
Commented May 15, 2022 at 14:38
• I added some more details for the second statement, but not a full argument. It is a genuinely more subtle issue than solving $y^2 = x^3 - 2$ in $\mathbf Z$.
– KCd
Commented May 15, 2022 at 22:26