2

How can we solve $x^3 - 2y^2 =54$ Diophantine equation in natural numbers? Modular arithmetic method is not useful maybe?

Martin Sleziak
  • 53,687
Analysis
  • 171
  • Try modulo with $n$ such that $\phi(n)=$lcm$(2,3)$ – lab bhattacharjee Sep 05 '14 at 19:13
  • i am not number theorist i have no idea on elliptic curves. I try to solve elementary way. But i shared because i thought that maybe i didnt see important secret in the problem – Analysis Sep 05 '14 at 19:24
  • No secrets in the theory of elliptic curves: compare with http://math.stackexchange.com/questions/703137/cubic-diophantine-equation, http://math.stackexchange.com/questions/669620/a-cubic-diophantine-equation-in-two-variables, etc. – Dietrich Burde Sep 05 '14 at 19:34
  • no i mean "trick" to solve the problem for examples how we get a quick contradiction and to show unique solution (6,9)? – Analysis Sep 05 '14 at 19:40
  • The fact that the ring of the Eisenstein integers is a UFD leads to a solution, too. – Jack D'Aurizio Sep 05 '14 at 19:40

1 Answers1

3

Since $x$ must be even, $x=2z$, we have: $$ 4z^3- y^2 = 27 \tag{1}$$ or: $$z^3 = \frac{y+3\sqrt{-3}}{2}\cdot\frac{y-3\sqrt{-3}}{2}=w\cdot\overline{w},\tag{2}$$ where $w$ lies in the ring of integers of $\mathbb{Z}[\sqrt{-3}]$, that is a UFD. This implies that, if $w$ and $\overline{w}$ are coprime, $w=\frac{y-1}{2}+3\omega$ must be a cube in the Eisenstein integers $\mathbb{Z}[\omega]$ (notice that $y$ must be odd, so $\frac{y-1}{2}\in\mathbb{Z}$), but: $$ (a+b\omega)^3 = a^3+3a^2 b\omega + 3ab^2\omega^2 +b^3 = (a^3+b^3-3ab^2)+3ab(a-b)\omega\tag{3}$$ and $3ab(a-b)=3$ has no integer solutions. The only possibility left is that $w=\sqrt{-3} u^3$ and $\overline{w}= -3v^3$ with $\gcd(u,v)=1$. This happens for $(x,y)=(6,9)$.

Jack D'Aurizio
  • 353,855