2

Here is my question:

Find all solutions to $y^2=x^3+4$.

My attempt:

Rewrite the equation as $(y-2)(y+2)=x^3$. Notice that if $y$ is odd, then $(y-2,y+2)=1$. Hence they are both cubes, but no cubes differ by $4$. Therefore $y$ is even and $x$ is even too.

Write $y=2y'$, then $4y'^2=x^3+4$. Write $x=2x'$, and we finally deduce $$y'^2=2x'^3+1.$$

However, I cannot proceed after this step. Can anyone help? Thanks. The solution should be mainly done with elementary steps.

blastzit
  • 800
  • 1
    We need all rational solutions? Or rather the "simpler situation" of finding all integer solutions? – dan_fulea Aug 04 '18 at 02:33
  • @dan_fulea Only integer solutions. – blastzit Aug 04 '18 at 02:47
  • You can try and continue along the same path: $(y'-1)(y'+1)=2x'^3$. Now, $\gcd(y'-1, y'+1)=2$, so (apart from the trivial $x'=0$ solution) either $y'-1=2a^3$, $y'+1=b^3$, or $y'-1=b^3$, $y'+1=2a^3$ (with $a$ odd and $b$ even in either case, and $\gcd(a,b)=1$); this gives you equations of the form $4b'^3-a^3=\pm1$ (where $b=2b'$), though there's no immediate argument that comes to mind for tackling these. – Steven Stadnicki Aug 04 '18 at 04:33

1 Answers1

1

If you are looking for integer points, then the LMFDB curve 108.a2 (Cremona label 108a1) $\,y^2=x^3+4,\,$ has the information that $\,x=0, y=\pm2 \,$ are the only integer solutions. There is more information in the entry, such as the Mordell-Weil group structure is $\,\mathbb{Z}/3\mathbb{Z}.$

Continuing from your attempt, with $\,x=2x_1, y=2y_1\,$ and your last equation, gives $\,(y_1-1)(y_1+1) = 2x_1^3. \,$ If $\,y_1\,$ is even gives a contradiction, thus $\,y_1\,$ is odd. But now, if $\,x_1\,$ is odd gives a contradiction, thus $\,x_1\,$ is even. Thus, let $\,x_1 = 2x_2, y_1 = 2y_2 + 1.\,$ The equation is now $\,2y_2(2y_2+2) = 8y_2(y_2+1)/2 = 16x_2^3.\,$ This reduces to $\,y_2(y_2+1)/2 = 2x_2^3\,$ where the left side is a triangular number which is twice a cubic number (this is close to but not the same as a cubic triangular number). The obvious solutions are $\,x_2=0,y_2=0\,$ and $\,x_2=0,y_2=-1.$

Somos
  • 35,251
  • 3
  • 30
  • 76