# Diophantine Equation: $x^2 + 3y^2 = 11z^2$

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?

• "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. – David Oct 9 '14 at 5:53
• How would I go about applying modulo 3 or 11? – Kevin_H Oct 10 '14 at 0:32
• 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. – 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 $z$.
• 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? – Kevin_H Oct 10 '14 at 1:03
• 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. – André Nicolas Oct 10 '14 at 4:54