I am having difficulty solving the following problem:

Prove rigorously that there is no integer solution for the Diophantine Equation $x^2 + 3y^2 = 11z^2$ except when $x = y=z = 0$.

Proving directly would seem far to difficult. I would assume proof by contradiction would be the most efficient way to solve this problem. To keep this simple, I shall assume that $(x,y,z)$ is a primitive integer solution. For any prime number $p$, there is a non-zero solution for the equation $x^2 + 3y^2 = 11z^2\pmod p$. Then simply find a good modulus $p$ to deduce a contradiction.

This is where I have a difficult time, trying to deduce the contradiction. Any suggestions?

Thank you for your time, and thanks in advance for your feedback.

  • 1
    $\begingroup$ "Keep it simple" is a good idea. So, to make the equation as simple as possible, try modulo $3$ or $11$, as either of these will make one of the terms disappear. If they don't work then you might like to have a stab at modulo $4$ or $8$ (btw the modulus does not have to be prime) because both of these have a relatively small number of squares. $\endgroup$ – David Oct 9 '14 at 5:53
  • $\begingroup$ How would I go about applying modulo 3 or 11? $\endgroup$ – Kevin_H Oct 10 '14 at 0:32
  • $\begingroup$ Modulo $3$, see Andre Nicolas' answer. Modulo $11$, similar ideas - find out what are the squares modulo $11$. It might not work - I'm not saying it must work, it's just something that could be worth trying. $\endgroup$ – David Oct 10 '14 at 1:06

Outline: Work modulo $3$. If $z$ is not divisible by $3$, then $11z^2\equiv 2\pmod{3}$. That is impossible, since $x^2$ cannot be congruent to $2$ modulo $3$.

Thus $z$ is divisible by $3$, and therefore $x$ is, and therefore $y$ is.

To use the idea in a formal proof, suppose that there is a non-trivial solution. Then there is a solution with $z\gt 0$ and as small as possible. Use the above reasoning to produce a solution with a smaller positive $z$.

  • $\begingroup$ This is what I have deduced thus far. Since $(x,y,z)$ is a primitive solution, we can assume $\gcd(x,y,z)=1$. Now, $x^2 = 11z^2 - 3y^2$. If I say $11z^2 - 3y^2 \equiv x^2$(mod $p)$ This would imply, by definition, that: $p| (11z^2-3y^2 -x^2) \Rightarrow p|11z^2, p|3y^2,$ and $p|x^2$. However, $\gcd(x,y,z)=1$. Does this imply a contradiction? $\endgroup$ – Kevin_H Oct 10 '14 at 1:03
  • $\begingroup$ You can ask about primitive solution, I prefer to say positive solution with minimal $z$. Please forget about general $p$, my argument uses $p=3$ only. Read the solution carefully, don't do manipulations like in the comment. The point is that $3$ must divide $z$. Say $z=3z_1$. Then we can see that $3$ divides $x$, say $x=3x_1$. So $9x_1^2+3y^2=99z_1^2$. So $3$ divides $y$, say $y=3y_1$. We get $x_1^2+3y_1^2=11z_1^2$, contradicting the minimality of $z$. If you still have trouble tomorrow, I will add more. $\endgroup$ – André Nicolas Oct 10 '14 at 4:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.