How can I show $z^2 = 6x^2 + 2y^2$ has no non-trivial integer solutions?

Question

EDIT: I missed a part of the question as there was a typo in my notes (this is part of my working on a proof using the Bruck–Ryser–Chowla theorem which was presented incorrectly in my version of the lecture notes). I have updated the question and shown some new working.

I was thinking of supposing that $x^2$, $y^2$ & $z^2$ have no common factors (because if they did, just divide out by that factor), and then considering possible solutions to $z^2 - 2y^2 = 6x^2$ modulo 6. See, we know $z^2 \equiv 2y^2 \mod 6$ which gives us six options.

$z^2 \equiv 2y^2 \equiv 0 \mod 6$
$z^2 \equiv 2y^2 \equiv 1 \mod 6$
$z^2 \equiv 2y^2 \equiv 2 \mod 6$
$z^2 \equiv 2y^2 \equiv 3 \mod 6$
$z^2 \equiv 2y^2 \equiv 4 \mod 6$
$z^2 \equiv 2y^2 \equiv 5 \mod 6$

In each case, I was thinking I could arrive to a contradiction, but I'm having a little trouble with the last five cases.

Let $z^2 = 6 \alpha$ and $2y^2 = 6 \beta$. Then $x^2$, $y^2$ & $z^2$ all have a factor of 3 which is impossible.
Let $z^2 = 6 \alpha + 1$ and $2y^2 = 6 \beta + 1$ ...
Let $z^2 = 6 \alpha + 2$ and $2y^2 = 6 \beta + 2$ ...
Let $z^2 = 6 \alpha + 3$ and $2y^2 = 6 \beta + 3$ ...
Let $z^2 = 6 \alpha + 4$ and $2y^2 = 6 \beta + 4$ ...
Let $z^2 = 6 \alpha + 5$ and $2y^2 = 6 \beta + 5$ ...

Any comments on the general proof structure or help with the last two cases would be greatly appreciated! Cheers.

The last case is easy to disprove because there exist no $n^2 \equiv 2 \mod 3$. — orlp, Oct 03 '17 at 07:08
Thanks! I didn't know about that. I'll have to check that out. — Tristan Batchler, Oct 03 '17 at 07:10
One strategy could be to try to squeeze $6x^2+y^2$ strictly between two perfect squares, and then check the ones in between. I might try it on this problem when I have time. — Mårten W, Oct 03 '17 at 07:33
@user202729 sorry! There was a typo in my notes. I have updated the question and I would appreciate it if you maybe had another look. Cheers. — Tristan Batchler, Oct 03 '17 at 08:38
I always use this formula. https://math.stackexchange.com/questions/1513733/solving-a-diophantine-equation-of-the-form-x2-ay2-byz-cz2-with-the-co/1514030#1514030 — individ, Oct 03 '17 at 08:41
You can rule out three cases because, for example, $2y^2$ is even and $6k+1$ is odd. — Empy2, Oct 03 '17 at 14:02

pisco · Answer 1 · 2017-10-03T14:48:34.653

2

Assume $(x,y,z)$ is a solution to $z^2 = 6x^2 + 2y^2 $ with smallest positive $|z|$, then $\gcd(x,y) = 1$. Putting $z=2z_1$ gives $$2z_1^ 2 = 3x^2 + y^2 $$ Since $x,y$ are both odd, RHS is divisible by $4$, so $z_1$ is even, that is $8\mid (2z_1^2)$. Hence $3x^2 + y^2 \equiv 0 \pmod{8}$. But the square of an odd number modolo $8$ must be $1$, a contradiction.

Hence $(0,0,0)$ is the only solution.

edited Oct 03 '17 at 14:48

answered Oct 03 '17 at 08:52

pisco

19,748

If x,y are odd, then z^2=6+2=0 (mod 8) – Empy2 Oct 03 '17 at 14:01
@Michael Thanks for pointing this out, I edited my answer. – pisco Oct 03 '17 at 14:47

How can I show $z^2 = 6x^2 + 2y^2$ has no non-trivial integer solutions?

1 Answers1