Integral solutions to the diophantine equation $y^2=x^3+2017$.

Question

Case 1: $y^2=0 \mod{3}$ \begin{align*} & y^2 = 3b \quad (b=3k^2) \implies x^3=3a +2 \\ & 3b = 3a + 2 + 2017 \implies b-a = 673 \\ & k^2 = \frac{a-2}{3} + 225 \\ \end{align*} Then I just took $a =2$ since 225 is already a perfect square. Thus I got the solution $(x,y) = (2,\pm 45)$. This is the only solution I was able to get.

I tried to do the second case, i.e, $y^2 \equiv 1 \mod 3$ as I did Case 1 but did not have any luck. Here are my proceedings.

Let $y^2 = 3b + 1$ and $x^3 = 3a$ where $a = 9k^3$ Then doing the same thing above we get, \begin{equation*} k^3 = \frac{b+3}{9} - 75 \end{equation*}
Now I am stuck, since I cannot find any solutions to the above equation.

$x=14;;y=69$ is another solution. Now the problem is to show that there are no other solutions — Raffaele, Feb 13 '21 at 17:37
@WillJagy Do you think it is useful to observe that the problem of proving that there are only three solutions can be translated modulus $7$ like this? $$y^2\equiv (x+1) (x+2) (x+4)\bmod 7$$ — Raffaele, Feb 13 '21 at 20:33
@Raffaele I don't know. I am looking at the notes by Conrad, https://kconrad.math.uconn.edu/blurbs/gradnumthy/mordelleqn1.pdf which does some examples with at least one integer solution. — Will Jagy, Feb 13 '21 at 20:40

score 1 · Accepted Answer · answered Aug 12 '21 at 11:54

According to a list on Mike Bennet's webpage, the integral solutions to $y^2=x^3+2017$ are the pairs $(x,y)$ $$(-12,\pm17),\qquad(2,\pm45),\qquad(14,\pm69).$$ There is no known general method for solving Mordell equations, i.e. equations of the form $$y^2=x^3+k,$$ where $k$ is a given integer. If there are no solutions, then often one can prove so by simple arguments about modular arithmetic. If there are solutions, sometimes it is possible to find them all (and prove that you have them all) by arguments in the ring of integers of $\Bbb{Q}(\sqrt{k})$. But sometimes more advanced methods are required.

Thanks a lot for that webpage! – Popular Power Aug 12 '21 at 16:52 — Popular Power, Aug 12 '21 at 16:52

score 0 · Answer 2 · answered Feb 13 '21 at 16:18

Comment:

You can try this method:

Take floor of $2017^{\frac 13}=12$

You can check 12 and less to see it gives a solution:

$2017=12^3+17^2=y^2-x^3$

Which gives $y=\pm 17$ and $x=-12$

Among other numbers bellow 12 only 2 gives integer for y:

$2017=45^2-2^3=y^2-x^3$

Which give what you found; $x=2$ and $y=\pm 45$

This method is for small numbers.

Integral solutions to the diophantine equation $y^2=x^3+2017$.

2 Answers2