Find all solutions to $x^3+(x+1)^3+ \dots + (x+15)^3=y^3$

Find all pairs of integers $(x, y)$ such that $$x^3+(x+1)^3+ \dots + (x+15)^3=y^3$$

What I have tried so far:

The coefficient of $x^3$ is $16$ in the left hand side. It is not useful then to trying bound LHS between, for example, $(ax+b)^3$ and $(ax+c)^3$ and then say that $ax+b<y<ax+c$.

I also tried to use modulo a prime. But it seems unlikely to bound variables this way.

EDIT : Though, it can be factored as $(2x+15)(x^2+15x+120)=(y/2)^3$. LSH factors are almost co-prime and we can say that $x^2+15x+120=3z^3$ or $x^2+15x+120=5z^3$. These are still too difficult to solve!

Any ideas?

Solution 1:

This isn't a complete solution, but I hope it gives you an approach. (It's too long for a comment)

Since we have $$\sum_{r=1}^{n} r^3=\left(\frac { n(n+1)}{2}\right)^2$$

You can write \begin{align} x^3+(x+1)^3+ \dots + (x+15)^3 &=\sum_{r=1}^{x+15} r^3-\sum_{r=1}^{x-1} r^3 \\ &=\left(\frac { (x+15)(x+16)}{2}\right)^2-\left(\frac { x(x-1)}{2}\right)^2\\ &=\left[\left(\frac { (x+15)(x+16)}{2}\right)-\left(\frac { x(x-1)}{2}\right)\right]\left[\left(\frac { (x+15)(x+16)}{2}\right)+\left(\frac { x(x-1)}{2}\right)\right]\\ \end{align}

Simplifying this, we get $$(x^2+15x+120)(2x+15)=\left(\frac y2 \right)^3$$

Solution 2:

You are looking for $(x, y) \in \mathbb{Z}^2$ for which $$ \sum_{i= 0}^{15} (x+i)^3 = y^3, \tag{0}$$ that is, $$ \sum_{i= 0}^{15} \left( x^3 + 3 i x^2 + 3i^2 x + i^3 \right) = y^3, $$ that is, $$ 16 x^3 + 3 \frac{15 (15+1)}{2} x^2 + 3 \frac{15 (15+1)(2 \times 15 + 1)}{6} + \left( \frac{15 (15+1)}{2} \right)^2 = y^3, $$ that is, $$ 16 x^3 + 360 x^2 + 3720 x + 14400 = y^3, $$ which can be written as $$ 8 (2x^3 + 45 x^2 + 465 x + 1800) = y^3 $$ Can you get anywhere from here?