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

Question

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?

Answer 1

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.

Answer 2

Let $x$ and $y$ be integers such that $4x^3-y^2=3$. Reducing mod $2$ shows that $y$ is odd, so $y=2z+1$ for some integer $z$. Then in the Eisenstein integers $\Bbb{Z}[\omega]$, where $\omega^2+\omega+1=0$, we have \begin{eqnarray*} x^3&=&\frac{y^2+3}{4} =\left(\frac{y+\sqrt{-3}}{2}\right)\left(\frac{y-\sqrt{-3}}{2}\right)\\ &=&\left(\frac{y+1+2\omega}{2}\right)\left(\frac{y-1-2\omega}{2}\right)\\ &=&(z+1+\omega)(z-\omega). \end{eqnarray*} The greatest common divisor of the two factors on the right hand side also divides their sum and difference, which are $y$ and $1+2\omega$, respectively. Note that $1+2\omega$ is prime in $\Bbb{Z}[\omega]$ with $\mathcal{N}(1+2\omega)=3$, and that $y$ is coprime to $3$ as otherwise $$4x^3=y^2+3\equiv3\pmod{9},$$ which is clearly impossible. This shows that the two factors are coprime, and hence that both are perfect cubes in $\Bbb{Z}[\omega]$, up to units. That is to say, there exist $a,b\in\Bbb{Z}$ such that $$z-\omega=\omega^k(a+b\omega)^3,$$ for some $k\in\{0,1,2\}$. Expanding the right hand side then yields \begin{eqnarray*} z-\omega&=&a^3-3ab^2+b^3+(3a^2b-3ab^2)\omega,\\ z-\omega&=&(3ab^2-3a^2b)+(a^3-3a^2b+b^3)\omega,\\ z-\omega&=&-(a^3-3a^2b+b^3)-(a^3-3ab^2+b^3)\omega, \end{eqnarray*} for $k=0,1,2$ respectively. In particular, comparing the coefficients of $\omega$, we get the equations \begin{eqnarray*} -1&=&3a^2b-3ab^2=3ab(a-b),\\ -1&=&a^3-3a^2b+b^2,\\ -1&=&-(a^3-3ab^2+b^3). \end{eqnarray*} The first clearly has no integral solutions, and solutions to the second and third equations are in bijection by $(a,b)\ \leftrightarrow\ (-b,-a)$. So it suffices to find all integers $a$ and $b$ such that $$a^3-3a^2b+b^3=-1.$$ This is a Thue equation, for which there exist effective methods to find all integral solutions. PARI/GP tells me that they are $$(-3,-1),\quad(-1,0),\quad(0,-1),\quad(1,-2),\quad(1,1),\quad(2,3).$$ Then the integral solutions to the third equation are $$(1,3),\quad(0,1),\quad(1,0),\quad(2,-1),\quad(-1,-1),\quad(-3,-2).$$ The corresponding values of $z$ for the second equation are $$0,\quad0,\quad0,\quad18,\quad0,\quad18,$$ corresponding to $y=1,37$. For the third equation they are $$-1,\quad-1,\quad-1,\quad-19,\quad-1,\quad-19,$$ unsurprisingly corresponding to $y=-1,-37$. This shows that you have found all solutions.

Answer 3

If you look at this video below, it shows you how to determine for which primes that p^2 - p + 1 is a cube over the natural numbers.

I.e. p^2 - p + (1 - n^3) = 0, for which values of p this equations holds when n is a natural number. Note: The discriminant of this quadratic is 4x^3 - 3

https://www.youtube.com/watch?v=mYCVZitGsFQ

This problem might seem to be unrelated to the question asked here, but it isn't.

Whilst solving his question, Michael Penn finds when the discriminant 4x^2 - 3 is a perfect square. I.e. When 4n^3 - 3 = y^2 (prefect square).

Now, replacing n by x, and we get the equation 4x^3 - 3 = y^2 (prefect square), which is the required problem.

Michael Penn finds x = 1 and x = 7 are the only natural number values for which 4x^3 - 3 is a prefect square.

We can now find the corresponding values of y for these values of x, which are y = 1 and 37 respectively. I.e. the only solutions are (x,y) = (1,1) or (7,37).

Now, extending the solutions over the integers, we find the only solutions are (x,y) = (1,±1) or (7,±37).