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?

  • $\begingroup$ First choose $$x+x+15=2y$$ $\endgroup$ – lab bhattacharjee Aug 7 '17 at 5:24
  • $\begingroup$ I think I can't understand this! Is that a variable change? How did you express $x$ and $y$ in terms of each other? $\endgroup$ – user450041 Aug 7 '17 at 5:32
  • $\begingroup$ The sum $1^3+\cdots+x^3$ is a square, i.e. $(x(x+1)/2)^2$. Does that help? $\endgroup$ – Colescu Aug 7 '17 at 5:41
  • $\begingroup$ @YuxiaoXie I'm not sure how to utilize that! $\endgroup$ – user450041 Aug 7 '17 at 5:48
  • 1
    $\begingroup$ One solution is $x=-7$, $y=8$. $\endgroup$ – Robert Israel Aug 7 '17 at 6:26

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 $$(x^2+15x+120)(2x+15)=\left(\frac y2 \right)^3$$

  • $\begingroup$ In fact, this is the easily deduced simplified problem and concerns binary cubic forms which are well known to be very difficult. In particular, making $2x+15=\pm1$, one finds by chance the other factor equal to 64 then the solutions $(x,y)=(-7,8), (-8,-8)$; similarly making $2x+15=z^3$ one can find integer solutions for this last equation but the corresponding other factor does not work. All this is just trials and errors without "true" mathematics. $\endgroup$ – Piquito Aug 7 '17 at 18:09
  • $\begingroup$ @Piquito If this is easy mathematics, please convince me. At least that there are a finite number of cases to check... $\endgroup$ – ama Aug 7 '17 at 19:37
  • $\begingroup$ And also, a product can be a cube without each number being a cube. For exemple, for $x=1$ $(x^2+15x+273)(2x+15)=17^3$. So your assumption $2x+15=z^3$ depends on the number $120$ instead of $273$. I don't see why. $\endgroup$ – ama Aug 7 '17 at 19:52
  • $\begingroup$ @ama:1)This change of the problem is just the use of the difference of two squares (knowing the formula giving $\left(\frac { n(n+1)}{2}\right)^2$). 2) Nobody has said that necessarily $2x+15$ must be a cube. $\endgroup$ – Piquito Aug 7 '17 at 20:17
  • $\begingroup$ The factorisation is not the problem. What follows in your comment is the problem. Our life is to short to check out all the values of $2x+15$. $\endgroup$ – ama Aug 7 '17 at 21:35

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?

  • $\begingroup$ It begins here! What to do next? $\endgroup$ – user450041 Aug 7 '17 at 5:53
  • $\begingroup$ You could have factorized directly... $\endgroup$ – Colescu Aug 7 '17 at 5:53
  • $\begingroup$ After factorizing still there is trouble! $\endgroup$ – user450041 Aug 7 '17 at 5:54

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