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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!

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

1 Answers 1

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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.

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    Thank you, How can I prove that the solution of $y= \sqrt{3k+2}$ for every $k\in\mathbb{Z}$ is not an integer?2012-04-01
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    @Jozef if $y\equiv0\pmod3$, $y^2\equiv0\pmod3$. Otherwise, $y^2\equiv1\pmod3$ by Fermat's Little Theorem.2012-04-01
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    if $y=0 [3]$, then $y^2=0 [3]$, and if $y=1$ or $2[3]$, then $y^2=1[3]$....2012-04-01