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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! ${}{}{}{}{}$ –  mixedmath Apr 1 '12 at 18:53
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1 Answer

up vote 8 down vote accepted

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? –  Jozef Apr 1 '12 at 20:01
    
@Jozef if $y\equiv0\pmod3$, $y^2\equiv0\pmod3$. Otherwise, $y^2\equiv1\pmod3$ by Fermat's Little Theorem. –  Mike Apr 1 '12 at 21:31
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if $y=0 [3]$, then $y^2=0 [3]$, and if $y=1$ or $2[3]$, then $y^2=1[3]$.... –  Louis La Brocante Apr 1 '12 at 22:40
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