Find all integer solutions of equality

Question

Find all integer solutions of equation $$x^3+(x+1)^3+...+(x+7)^3=y^3$$ I've solved it by opening brackets and consideration of signs but I think there is simpler way of solving it .

$$y^3=\sum_{r=1}^{x+7}r^3-\sum_{r=1}^{x-1}r^3$$
$$=\frac{{(x+7)(x+8)}^2-{(x-1)x}^2}4$$ — lab bhattacharjee, Mar 10 '15 at 15:20
@Kugelblitz, $$\sum_{r=n}^mf(r)=\sum_{r=1}^mf(r)-\sum_{r=1}^{n-1}f(r)$$ provided all the sums converge — lab bhattacharjee, Mar 10 '15 at 15:23
@labbhattacharjee , I don't sure that this formula correct for negative x . — Antony Tito, Mar 10 '15 at 15:24
@AntonyTito, The formula in my first comment holds for $x-1\ge0$ — lab bhattacharjee, Mar 10 '15 at 15:30
Note that this seems to be hard, just the case :$x^3+(x+1)^3=y^3$ is not easy (particular case of LFT) which the best we can do is transorme it to $4u^3-v^3=3$ which needs to work in $\mathbb{Z}[\sqrt[3]4]$, your equation goes in the same direction because it can be simplified to $(x+4)(x^2+8x+28)=49y'^3$ which suggest the same thing because after some tests of divisibility we can reduce it to the same problem — Elaqqad, Mar 10 '15 at 15:42

score 1 · Accepted Answer · answered Mar 10 '15 at 15:51

Let $$P(x)=x^3+(x+1)^3+(x+2)^3+(x+3)^3+(x+4)^3+(x+5)^3+(x+6)^3+(x+7)^3$$ so $$P(-x-7)=-P(x)$$

on the other hand we can $$P(x)=8x^3+84x^2+420x+784$$ if $x\ge 0$,we have $$(2x+7)^3=8x^3+84x^2+294x+343<P(x)<8x^3+120x^2+600x+1000=(2x+10)^3$$ so we have $$(2x+7)<y<2x+10$$ so $$P(x)=(2x+8)^3\Longrightarrow -12x^2+36x+272=0$$ this equation no integer roots,

simaler we have $$P(x)-(2x+9)^3=-24x^2-66x+55=0$$ also have no integer roots.

Now we have prove $x\ge 0 $ or $x\le -7$ this equation have no integer roots

But if $-6\le x\le 1$, we can easy to find when $x=-2\Longrightarrow P(-2)=216=6^3$

$x=-3\Longrightarrow P(-3)=64=4^3$

$x=-4,P(-4)=-64=(-4)^3$

$x=-5,P(-5)=216^3=(-6)^3$

Find all integer solutions of equality

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