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Given the equation

$$ t^3=3a^2-3a+13 $$

I want to find integer solutions. WolframAlpha says $t=7,a=-10$ and $t=7,a=11$ are solutions, but how did it compute that?

I thought I wanted to find the intersection of $3a^2-3a+13$ and $t^3$, which gives $\approx 3.289$, but I'm not sure how to turn that into an integer.

I thought maybe solve for $a$, but I'm not sure how that helps me either

$$ a=\frac{3 \pm \sqrt{9-12(13-t^3)}}{6} $$

Burnsba
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    When you got $3.289...$, you solved $a^3=3a^2-3a+13$, which is a different problem. Are you familiar with elliptic curves? – J. W. Tanner Nov 16 '20 at 00:58
  • @J.W.Tanner well, not too much. But I was looking into using Sage (complete the square transform, then use Sage EllipticCurve([0,0,0,0,z]) ->E.integral_points) but this only worked for some of the most simple values (a problem in eclib I think?). It looks like the only other CAS that resolves integral points is Magma, but I don't have access to that. – Burnsba Nov 16 '20 at 01:05
  • It looks like the method is explained here: http://matwbn.icm.edu.pl/ksiazki/aa/aa68/aa6827.pdf ("Computing integral points on elliptic curves"). Does WolframAlpha really go through all that? – Burnsba Nov 16 '20 at 01:20
  • For contrast: given any prime $p = 3 u^2 - 3 uv + 13 v^2$ which are $1 \pmod 3$ but $3,5,6 \pmod 7,$ there is also a representation $p^3 = 3 x^2 - 3 xy + 13 y^2$ with $\gcd(x,y) = 1$ – Will Jagy Nov 16 '20 at 01:43
  • On that note, I am wondering if your question can be settled using unique factorization in the Eisenstein integers. – Will Jagy Nov 16 '20 at 02:18
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    $t^3=3a^2-3a+13$ can be transfomed to $Y^2 = X^3-21168$ with $X=12t$. We can get the integer solutions of $Y^2 = X^3-21168$ using Online Magma Calculator as follows. $E$:=EllipticCurve([$0,0,0,0,-21168$]); IntegralPoints($E$); – Tomita Nov 16 '20 at 02:18
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    Since $21168 = 3\cdot 84^2$, the transformed equation in the previous comment is the Mordell equation $Y^2 = X^3 - 84^2\cdot 3$. This could be rewritten as $X^3 = Y^2 + 84^2 \cdot 3 = (Y+84\sqrt{-3})(Y-84\sqrt{-3})$. Since $\mathbf Z[(1 + \sqrt{-3})/2]$ (the Eisenstein integers, as Will Jagy mentions above) has unique factorization, the fact that the only integral solutions are $(X,Y) = (84,\pm 756)$ is probably provable from that unique factorization after first accounting for how divisible $X$ and $Y$ are by $4$ (showing each is a multiple of $4$). – KCd Nov 16 '20 at 02:52
  • @KCd Thanks. I have a book around here somewhere that does some contest type problems using Gaussian and Eisenstein integers, can't seem to find it. The quadratic form $3x^2 - 3xy+ 13 y^2$ has discriminant $-3 \cdot 7^2,$ factors in Eisenstein, and so on. – Will Jagy Nov 16 '20 at 17:13
  • @KCd found one https://math.stackexchange.com/questions/3044095/what-are-the-positive-integer-solutions-to-x2-x1-y3 – Will Jagy Nov 16 '20 at 17:34
  • @WillJagy I was looking at that, but not sure where the uv polynomial comes from – Burnsba Nov 16 '20 at 17:36
  • @Burnsba from unique factorization, both factors $x-\omega$ and the other must be cubes. So we calculate $(u + v \omega)^3$ Note he is using cube roots of $-1$ rather than $1$ – Will Jagy Nov 16 '20 at 17:39

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