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In the post Finding all solutions to $y^3 = x^2 + x + 1$ with $x,y$ integers larger than $1$, the single positive integer solution $(x,y)=(18,7)$ is found using algebraic integers.

In one of the comments (Finding all solutions to $y^3 = x^2 + x + 1$ with $x,y$ integers larger than $1$, the OP indicated the method suggested (i.e., the “theory of elliptic curves”, torsion groups, etc.) was “far too advanced for [his] level of understanding”. This inspired me to try finding a totally elementary approach. I've made it to a certain stage, and wanted some advice on how to proceed.

Beginning with the original equation $$y^3 = x^2+x+1,\tag{1}$$ I added $x^3$ to both sides and factored, obtaining \begin{align} x^3+y^3 &= x^3+x^2+x+1 \\ (x+y)(x^2-xy+y^2) &= (x^2+1)(x+1).\tag{2} \end{align} Since we're looking for $x,y > 1$, and both factors on the right-hand side of (2) are positive, so are both on the left-hand side. Hence there exist positive integers $a,b,c,d$ such that \begin{align} x^2+1 &= ab, \\ x+1 &= cd, \\ x+y &= ac, \\ x^2-xy+y^2 &= bd. \end{align} Knowing the solution a priori, I'm now faced with trying to show that $(a,b,c,d)=(25,13,1,19)$. Using a few simple congruence and divisibility arguments, it can fairly easily be shown that $c=1$, and hence we have \begin{align} x^2+1 &= ab, \tag{3.1} \\ x+1 &= d, \tag{3.2} \\ x+y &= a, \tag{3.3} \\ x^2-xy+y^2 &= bd. \tag{3.4} \end{align} But now I'm running in circles. Evidently $(x,y)=(d-1,a-d+1)$, and I can prove other results like $(y-1)\mid x$ and $y \mid (b+1)$, but I can't seem to take it across the goal line. Any help would be greatly appreciated.

EDIT: From $a^2-bd=3xy$ and $d^2-ab=2x$, we have $$ (a-d)(a+d+b)=x(3y-2). \tag{4} $$ By the form of the left-hand side of (3.4) and the fact that $b,d$ are odd and relatively prime [because $x^2+1$ and $x+1$ are], we deduce $b\equiv d\equiv 1\!\pmod{6}$. Then (3.1) implies $a \equiv 1\!\pmod{6}$; in fact, by the form of (3.1), we have also $a \equiv b \equiv 1\!\pmod{12}$. In any case, $a+b+d \equiv 3\!\pmod{6}$, and (4) now implies $18 \mid x$.

EDIT: Another thread on the same question is https://mathoverflow.net/questions/56338/is-n-m-18-7-the-only-positive-solution-to-n2-n-1-m3.

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  • $\begingroup$ math.stackexchange.com/questions/624792/… is a great example of an elliptic curve problem which has a stunningly elegant and elementary solution. The other solutions, below the highest upvoted answer, show other less elementary ways of obtaining the same solution. I'm hoping there's something similar to be found here, since the equations have similarities. $\endgroup$ – Kieren MacMillan Aug 20 '14 at 21:39
  • $\begingroup$ Note that it is easy to see that $y$ must be odd. You could also try $y^3-1=(y-1)(y^2+y+1)=x(x+1)$ $\endgroup$ – Mark Bennet Aug 20 '14 at 22:13
  • $\begingroup$ @MarkBennet: Yes, I can show that $y$ is odd, and also that $x$ is even. But I haven't yet found anything helpful from that. As for the other factorization, it only confirms my congruence and divisibility findings (e.g., $(y-1)\mid x$), but I haven't found anything new there. $\endgroup$ – Kieren MacMillan Aug 20 '14 at 23:36
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    $\begingroup$ Multiply by 64 to get $Y^3=X^2+48$, where $Y=4y$ and $X=8x+4$. This is a "Mordell equation", that is, an equation of the form $u^2=v^3+k$. There is an enormous literature on these --- Google should at least get you started. For some values of $k$, there are elementary ways to find all the solutions. $\endgroup$ – Gerry Myerson Aug 21 '14 at 10:46
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    $\begingroup$ I seem to have forgotten a lot of mathematics in the last three years. $\endgroup$ – Gerry Myerson Aug 22 '14 at 6:57
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So I've found an elementary and fairly easy method of proof that the only two positive integer solutions are $(x,y)=(0,1)$ and $(x,y)=(18,7)$, with the only two negative integer solutions being $(x,y)=(-1,1)$ and $(x,y)=(-19,7)$.

Using the logic as shown in the question, we quickly deduce that $(y-1) \mid x$, say $x = (y-1)w$ for integer $w$. Substituting into the original equation and solving for $y$ yields $$ y = \frac{w^2-1 \pm \sqrt{w^4-6w^2+4w-3}}{2}. \tag{$\star$} $$ Evidently, $w^4-6w^2+4w-3$ must be an integer square. Now observe that $$ (w^4-6w^2+4w-3)-(w^2-3)^2 = 4w-12 \tag{1} $$ and $$ (w^2-2)^2-(w^4-6w^2+4w-3) = 2w^2-4w+7. \tag{2} $$ The right-hand side of (2) is positive for all integer $w$, and the right-hand side of (1) is positive for $w>3$. So for $w > 3$ $$ (w^2-3)^2 < w^4-6w^2+4w-3 < (w^2-2)^2, $$ and the expression cannot be a square. For $w < 0$, we determine that $$ (w^2-4)^2 < w^4-6w^2+4w-3 < (w^2-3)^2, $$ and again it cannot be a square. Hence we need only consider $0 \le w \le 3$, and it is quickly found that the only solution is $w=3$. By substitution into $(\star)$, we find $y=1$ or $y=7$, as claimed.

p.s. Thanks to Will Jagy for the elegant bounds-based solution to the last step!

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