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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

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Notice that:

$$x^3<x^3+x^2+x+1<x^3+2x^2+4x+8=(x+2)^3$$

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?

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  • $\begingroup$ 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 $x>0$ $\endgroup$ – Mostafa Ayaz Mar 6 '20 at 20:31
  • $\begingroup$ @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) $\endgroup$ – LHF Mar 6 '20 at 20:33
  • $\begingroup$ I mean $x^3+x^2+x+1<(x+2)^3$ can be substituted with $x^3+x^2+x+1<(x+1)^3$ $\endgroup$ – Mostafa Ayaz Mar 6 '20 at 20:34
  • $\begingroup$ @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 $x$. $\endgroup$ – LHF Mar 6 '20 at 20:35
  • $\begingroup$ 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. $\endgroup$ – Mostafa Ayaz Mar 6 '20 at 20:37
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Rather make a case distinction as to whether $x\gtreqqless 0$.

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

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

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

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