# Integers that fit for $x$ and $y$: $x^3 + x^2 + x + 1 = y^3$

Find all integers $$x$$ and y such that $$x^3 + x^2 + x + 1 = y^3$$ Now I moved the cubes to one side and tried to factor them but some how in thinking I need to try cases for odd and even $$x$$ ? A hint would help i think

Notice that:

$$x^3

because the quadratics $$x^2+x+1$$ and $$x^2+3x+7$$ have negative discriminant. Therefore, the only possibility is

$$(x+1)^3=y^3=x^3+x^2+x+1$$

Can you end it from here?

• Note that your bounds don't work for all $x<0$ and also that you can make the upper bound tighter by $(x+1)^3$ for $x>0$ – Mostafa Ayaz Mar 6 '20 at 20:31
• @MostafaAyaz, which bounds do you mean? $x^2+x+1>0$ and $x^2+3x+7>0$ hold for any real number $x$ (negative discrimnants) – LHF Mar 6 '20 at 20:33
• I mean $x^3+x^2+x+1<(x+2)^3$ can be substituted with $x^3+x^2+x+1<(x+1)^3$ – Mostafa Ayaz Mar 6 '20 at 20:34
• @MostafaAyaz, $x^3+x^2+x+1 < (x+1)^3$ does not hold for $x=0$ and $x=-1$ (which are the solutions of the equation). That's the point of choosing $x^3$ and $(x+2)^3$ which hold for any real $x$. – LHF Mar 6 '20 at 20:35
• Well, you can study those special cases handily and imply that there is no solution for x>0. You also have to change your bounds for $x<0$ a little bit. – Mostafa Ayaz Mar 6 '20 at 20:37

Rather make a case distinction as to whether $$x\gtreqqless 0$$.

If $$x>0$$, then $$(x+1)^3=x^3+3x^2+3x+1>x^3+x^2+x+1>x^3$$

• Atticus it means there cant be any more integers then 0, 1 , -1 , right ? Like squeeze Theorem ? ( good ol calculus) – Randino Andantino mozartino Mar 7 '20 at 7:05

Let $$y = x+d$$ so $$y^3 = x^3 + 3dx^2 + 3d^2x + 1 = x^3 + x^2 + x + 1$$ so

$$3dx^2 + 3d^2x = x^2 + x$$.

If $$x = 0$$ then $$x^3+x^2 + x+1 = 1=y^3$$ would yield $$x=0;y=1$$ as a solution.

But if $$x \ne 0$$ we have:

$$3dx + 3d^2 = x+1$$

$$3d(x+d)= x+1$$

This means $$x+1$$ is a multiple of $$3$$ so let $$x= 3k-1$$.

So we have $$3d(3k+d-1)= 3k$$ and $$d(3k+d-1) =k$$

So $$k(3d-1) = (1-d)d$$. $$3d-1$$ can not be $$0$$ as $$d$$ is an integer so $$k = \frac {(1-d)d}{3d-1}$$.

Now if $$d =0$$ we would have $$k=0$$ and $$x = -1$$ and $$y=-1$$ but that's not a solution $$y^3 = (-1)^3 = -1 \ne 0= (-1)^3 + (-1)^2 + (-1) +1=x^3 + x^2 + x+1$$.

So $$d\ne 0$$.

Gut $$\gcd(3d-1,d) =1$$ so $$3d-1|1-d$$. So $$3d-1 = \pm\gcd(3d-1,1-d)=\gcd(1-d, (3d-1)+3(1-d))=\gcd(1-d,2)=2$$ if $$d$$ is odd; or $$1$$ if $$d$$ is even.

So either $$3d-1=\pm 2$$ so $$d= 1,-\frac 13$$ or $$3d-1=\pm 1$$ and $$d=0,\frac 23$$. But $$d$$ is an integer so $$d=1$$ or $$0$$.

We ruled out $$0$$ so $$d=1$$

and $$k = 0$$. And $$x =-1$$ and $$y= -1 + d = 0$$.

That gives $$x^3+x^2 + x + 1=(-1)^3 + (-1)^2 + (-1) + 1 = 0 =y^3;$$ a solution.

So $$x=0$$ and $$y=1$$, and $$x=-1$$ and $$y=0$$ are the only integer solution.

• Hi fleablood do u happen to go to uw madison ? I think we met when I took a math class there;) – Randino Andantino mozartino Mar 6 '20 at 22:20
• Nope. Never been there. – fleablood Mar 6 '20 at 23:21