2
$\begingroup$

I'd really love your help with showing that the Diophantine $3x^2+2=y^2+6z^3$ equation has no solutions.

I know that Diophantine equation of the form $ax+by+cz=d$ iff $\gcd(a,b,c) | d$, but how do I deal with the squares?

Any hints? suggestions?

Thanks!

  • 4
    mod 3! ${}{}{}{}{}$2012-04-01

1 Answers 1

8

If $(x,y,z)$ is a solution, then looking modulo $3$ you should have $y^2\equiv2\pmod3$

It is then easy to see that there is no integer satisfying this equation.

  • 1
    if $y=0 [3]$, then $y^2=0 [3]$, and if $y=1$ or $2[3]$, then $y^2=1[3]$....2012-04-01