I'd like to prove that the only integer solutions of $$2x^2+3y^2=z^2$$ is $(0,0,0)$.

By working in $\mathbb{Z}_2$ and $\mathbb{Z_3}$, I have gone as far as proving that in $\mathbb{Z}$, any integer solutions must have $x,y,z$ each being multiples of 3.

But I am not quite sure how to then deduce that in $\mathbb{Z}$, they must therefore be 'zero'-multiples of 3. I do not need a full solution, but a hint or a pointer would be helpful.

  • $\begingroup$ Are you conjecturing that it is true? If so, why do you think so? $\endgroup$ Apr 19, 2016 at 11:58
  • $\begingroup$ The question asks to prove that $(0,0,0)$ is the only integer solution, so I am presuming that the result is true. $\endgroup$
    – Trogdor
    Apr 19, 2016 at 11:59
  • 2
    $\begingroup$ @Trogdor. Yes. You are almost there. If there is a solution with $x,y,z$ not all 0, take a solution with minimum positive $|x|+|y|+|z|$. Such a solution cannot have $x,y,z$ all multiples of 3, because otherwise $x/3,y/3,z/3$ would be a solution with smaller $|x|+|y|+|z|$. $\endgroup$
    – almagest
    Apr 19, 2016 at 12:01
  • $\begingroup$ It might be cleanest to start out saying "Assume we have a non-trivial solution. By dividing by any common factors, we can find a solution so that $x$, $y$, and $z$ have no common factor." $\endgroup$
    – robjohn
    Apr 19, 2016 at 12:42

3 Answers 3



Let $(x_0;y_0;z_0)$ - the smallest solution. Then $$\left(\frac{x_0}3;\frac{y_0}3;\frac{z_0}3 \right) -$$ solution. Contradiction


First prove that x, y, z must be equal to 0 modulo 3 indeed.

Then what happens ? $x=3x'$, $y=3y'$, $z=3z'$ with x', y', z' being integers. What can you show and how to conclude ?

SPOILER : this is the answer

If x, y, z are not (0, 0, 0) you can divide by 3 an infinite number of times and have non 0 integers, which is impossible


HINT.-Using the identity $(a^2-b^2)^2+(2ab)^2=(a^2+b^2)^2$ we can put$$\sqrt2 x=2ts\\\sqrt 3y=t^2-s^2$$ where $t$ and $s$ are quadratic irrational.

Hence $$ts\in \mathbb Q(\sqrt 2)\\t^2-s^2\in \mathbb Q(\sqrt 3)$$ It is not difficult to verify the impossibility of these inclusions.


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