Find the integer $x$ such $x^6+x^5+x^4+x^3+x^2+x+1=y^3$

Find the equation integer solution $$\color{red}{y^3=x^6+x^5+x^4+x^3+x^2+x+1}$$

It is obvious $x=0,y=1$ or $x=-1,y=1$ are solutions.

How to find all solutions?

• Just so you know, we can also say $y^3(x-1)=x^7-1$. Not sure if this helps, though. – Noble Mushtak Jun 17 '16 at 3:49
• Somewhat related: here – JMoravitz Jun 17 '16 at 3:52
• FWIW there are no other solutions with $|x|\le10^6$, so given the heuristic evidence in JMoravitz' link I would say you probably have all the solutions there. – Mario Carneiro Jun 17 '16 at 3:56
• Too close to a cube to be a cube, it is no cube. Multiply both sides by $3^6$ and estimate the difference with $(3x+1)^6$. – Jack D'Aurizio Jun 17 '16 at 4:08

Here is an elementary approach, albeit tedious if one computes everything by hand. Our strategy is to find perfect cubes that are close enough to RHS (as Jack D'Aurizio mentioned in the comment).

Denoting $~f(x)=27(x^6+x^5+x^4+x^3+x^2+x+1)$, we have

• Claim 1: If $x\ge3$, then

$$(3x^2+x)^3<f(x)<(3x^2+x+1)^3.$$

• Claim 2: If $x\ge2$, then

$$(3x^2-x)^3<f(-x)<(3x^2-x+1)^3.$$

Back to the original equation, we can easily check the case $x=-1,~0,~1,~2$; when $x\ge3$, by claim 1, we see $(3y)^3=f(x)$ lies between two consecutive cubes, therefore no integer solutions; similarly claim 2 covers the case $x\le-2$. We are done.

• Proof of claim 1:

• $$f(x)-(3x^2+x)^3=27+27x+27x^2+26x^3+18x^4>0,$$
• $$(3x^2+x+1)^3-f(x)=94+6(x-2)(7x+10)+x^2(x-3)(9x+19)>0,$$
• Proof of claim 2:

• $$f(-x)-(3x^2-x)^3=19+21(x-1)+(x-1)^2(18x^2+10x+29)>0;$$
• $$(3x^2-x+1)^3-f(-x)=(x-1)(9x^3+17x^2+2x+26)>0.$$

(1) In hindsight, there is no reason to use elementary methods only. Note the two claims above are essentially same:

$$(3x^2+x)^3<f(x)<(3x^2+x+1)^3$$

for $x\le-2$ or $x\ge3$.

We could simply take the derivative and prove the differences are monotonic on either range...

(2) A bit heavier but more general approach is by Runge's method, e.g., see this paper by Sankaranarayanan and Saradha.