### 100 is the only square that is the sum of 4 consecutive (positive) cubes.

The OEIS article on the number $$100$$ opens with an interesting factoid:

"$$100$$ is the square of $$10$$, and the smallest square that is the sum of four [positive] consecutive cubes: $$1^3 + 2^3 + 3^3 + 4^3 = 100$$."

In fact, it is the only one. To see this, let's look at the equation

\begin{align*} y^2 &= x^3 + (x + 1)^3 + (x + 2)^3 + (x + 3)^3 \\ &= 4x^3 + 18x^2 + 42x + 36. \end{align*}

Let $$X = 4x + 6, Y = 4y$$; the above equation then reduces to

$Y^2 = X^3 + 60X.$

Note that a positive integer solution in $$(x, y)$$ will give a positive integer solution in $$(X, Y)$$, though the converse is not true.

Now, generally speaking, whenever one sees an equation of the form $$Y^2 = X^3 + aX + b$$, one has an elliptic curve. Well, we have the extra condition $$4a^3 + 27b^2 \ne 0$$ to get rid of problematic cases like $$y^2 = x^3$$, which are referred to as singular curves; we'll see the logic of this later on. I will not go too deeply into specifics here, as that is not quite the objective of this post, but elliptic curves are the object of much study in modern number theory. The website L-Functions and Modular Forms Database (LMFDB) contains information about many elliptic curves over $$\mathbb{Q}$$ and some other number fields. This specific elliptic curve is indeed on LMFDB, with label 14400.cq2; the integral points $$(X, Y)$$ are given by

$(0,0), (6, \pm 24), (10, \pm 40), (240, \pm 3720).$

Not all of these correspond to positive integer points $$(x, y)$$ in the original equation; only $$(10, 40)$$, which corresponds to $$(1, 10)$$, does. This completes the proof of the claim.

(Yes, I am aware I can submit this to the OEIS Wiki. I have indeed put in an edit for it with this reasoning attached several months ago, but it has not been reviewed.)

It's natural to ask, of course, about $$n$$ consecutive cubes summing to a square for other values of $$n$$? We have the identity

$1^3 + 2^3 + \dots + n^3 = (1 + 2 + \dots + n)^2,$

which has a few nice proofs here. Does this happen otherwise? The above shows that for $$n = 4$$, it does not. The case $$n = 1$$ is a bit of an anomaly here, as you just have all the perfect sixth powers; elliptic curve methods do not work and are not needed here, here, because you get $$y^2 = x^3$$,  which fails the nonsingularity condition $$4a^3 + 27b^2 \ne 0$$.

More generally, as Stroeker outlines here, starting with the equation

\begin{align*}y^2 &= x^2 + (x + 1)^2 + \dots + (x + n - 1)^2 \\ &= nx^3 + \frac{3}{2}n(n-1)x^2 + n\left(n - \frac{1}{2}\right)(n - 1)x + \frac{1}{4}n^2(n - 1)^2 \end{align*}

the change of variables $$X = nx + \frac{1}{2}n(n-1), Y = ny$$ turns this into

$Y^2 = X^3 + \frac{1}{4}n^2(n^2 - 1).$

One can check that, just like before, if $$x , y$$ are integers, so are $$X, Y$$.

A similar analysis for $$n = 2$$, which I'll leave to the reader, yields that indeed the only square that is the sum of two consecutive cubes is $$3^2 = 1^3 + 2^3$$. But, $$n = 3$$ is a bit more interesting. We have the curve $$Y^2 = X^3 + 18X$$, which has two integral points that give us positive integer solutions. Aside from the point $$(6, 18)$$, which recovers the typical solution, the point $$(72, 612)$$ gives us

$204^2 = 23^3 + 24^3 + 25^3$

which Cassels showed, using some basic algebraic number theory, is indeed is the last solution of its kind. Indeed, we can confirm Cassels' result by looking this curve up on LMFDB.

In general, does it ever happen that a perfect square is a sum of $$n$$ consecutive cubes infinitely often? The answer, for $$n > 1$$, is no. This is because by a theorem of Siegel, elliptic curves over $$\mathbb{Q}$$ (and number fields in general) have only finitely many integral points.

### A silly little derivation of $$\zeta(2)$$
(This is a cleaned-up and somewhat expanded version of this Twitter thread .) What follows is a silly little proof that $\zeta(2) = \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$ where $$\zeta$$ is the Riemann zeta function. Consider the integral $I := \int_0^1 \frac{\log(1 - x + x^2)}{x(x - 1)} \, dx.$ We have, by using partial fractions and performing some other algebraic manipulations, \begin{align*} I &= -\int_0^1 \! \frac{\log(1 - x + x^2)}{x} \, dx - \int_0^1 \! \frac{\log(1 - x + x^2)}{1 - x} \, dx \\ &= -2\int_0^1 \! \frac{\log(1 - x + x^2)}{x} & (x \mapsto 1 - x ) \\ &= 2\left( \int_0^1 \! \frac{\log(1 + x)}{x} \, dx - \int_0^1 \! \frac{\log(1 + x^3)}{x} \, dx \right) \\ &= \frac{4}{3}\int_0^1 \frac{\log(1 + x)}{x} \, dx & (x \mapsto x^{1/3}). \end{align*} To evaluate this integral, we take the Maclaurin series: $\int_0^1 \! \frac{\log(1 + x)}{x} \, dx = \int_0^1 \! \sum_{n=1}^{\infty} \frac{(-1)^nx^{n-1}}{n} \, dx$ Since for