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I am interested in how to tackle this Diophantine equation:

$$4x^3-y^2=3$$

The solutions I have found so far are $(1,1)$ and $(7,37)$. Are there any more?

I have looked up various material on cubic Diophantines but most of what I’ve found is on equations where the coefficients of $x^3$ and $y^2$ are the same. In this particular problem, if both coefficients were equal to $1$, it would just be a nice Mordell’s equation. But the coefficient of the cubic variable is not $1$ – which is why it’s so frustrating. Still, would I be right in saying that if the solutions were to lie on an elliptic curve, there would only be finitely many of them? What if they don’t lie on an elliptic curve? Will the number of solutions still be finite?

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  • $\begingroup$ I've used a program to test from 1 to 100000 and no other solutions are found. $\endgroup$ – Kenny Lau May 7 '16 at 8:43
  • $\begingroup$ You can use this program in Jelly to find up to an arbitrary range (just change the argument). $\endgroup$ – Kenny Lau May 7 '16 at 8:43
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    $\begingroup$ Some mindless explorations: $$y\equiv1\mbox{ (mod 2)}$$ $$y\equiv1\mbox{ or }8\mbox{ (mod 9)}$$ $$y\equiv1\mbox{ or }17\mbox{ (mod 18)}$$ $$m^2+m+1=x^3\mbox{ where }y=2m+1$$ $\endgroup$ – Kenny Lau May 7 '16 at 9:26
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    $\begingroup$ $4x^3-y^2=3 \Leftrightarrow 4x^3-4=y^2-1 \Leftrightarrow$ $4(x^3-1)=(y-1)(y+1) \Leftrightarrow 4(x-1)(x^2+x+1)=(y-1)(y+1)$. First observation is y must be odd. $\endgroup$ – rtybase May 7 '16 at 9:31
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    $\begingroup$ would'nt it become a Mordell's equation when multiplying both side by 16? $\endgroup$ – René Gy May 7 '16 at 10:11
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Not a solution.

Working successively modulo $4$, $36$, and $108$, the solution set to your problem corresponds to the solution sets of, respectively, \begin{align} 3 &= 4 x_1^3 -(2y_1+1)^2 \\ 3 &= 4 (3 x_3 + 1)^3 - (2 (3 (3 y_3)) + 1)^2 \\ 3 &= 4 (3 (108 x_4 + z_4) + 1)^3 - (2 (3 (3 (3 y_4 + z_4))) + 1)^2 \end{align} for integers $x_1$, $y_1$, $x_3$, $y_3$, $x_4$, $y_4$, and $z_4$, with $0 \leq z_4 < 108$. Searching over intervals for satisfying $(x_4,y_4,z_4)$ triples should cover ground faster than doing so for $(x,y)$ pairs. My own small searches suggest there are now many ways to represent $(1,1)$ and $(7,37)$ in these new variables.

I have the additional relations \begin{align} x_3 &\cong 3 y_3 + 4 y_3^2 + y_3^3 + 2 y_3^4 \pmod{5} \\ x_3 &\cong 10 y_3 + 5 y_3^2 + 7 y_3^3 + 6 y_3^4 + 8 y_3^5 + 4 y_3^6 + 7 y_3^7 + 5 y_3^8 + 7 y_3^9 + 6 y_3^{10} \pmod{11} \end{align} which could, perhaps, accelerate a search on $(x_3,y_3)$ pairs more than the $3$-times sparser search in $(x_4,y_4,z_4)$ above.

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