Quadratic Residues form a sub-semigroup, that cannot be acted on by non-residues

Question

I was reading: https://homepages.warwick.ac.uk/~maslan/docs/T-seminar.pdf

Where they go over a proof that $x^2 - 5y^2 = 3z^2$ doesn't have integer solutions $(x,y,z)$ such that $z \ne 0$. Their proof has the following skeleton (outlined from the powerpoint but with more detail...)

$$ x^2 - 5y^2 \equiv 0 \mod 3$$ $$ x^2,y^2 \in \lbrace 0, 1 \rbrace \mod 3, \ 5 \equiv 2 \mod 3 \rightarrow x^2\equiv y^2 \equiv 0 \mod 3 $$ $$ \rightarrow x \equiv y \equiv 0 \mod 3 \rightarrow x^2 \equiv y^2 \equiv 0 \mod 9$$

$$ \rightarrow 3z^2 \equiv 0 \mod 9 \rightarrow z^2 \equiv 0 \mod 3 \rightarrow z^2 \equiv 0 \mod 9$$

$$ \rightarrow x^2 - 5y^2 \equiv 0 \mod 27$$

Now at this point it is implied that it is easy to "deduce" that this chain continues arbitrarily longer as infinite descent. But I reject that being obvious, in order for this "descent" to continue we have to prove that for each odd-power of 3 there doesn't exist a pair of quadratic residues $a,b$ other other than 0 such that $a=5b$. otherwise my step in line 2 won't be able to generalize.

So that leads to a natural conjecture: is it true that for all positive odd $n$ there doesn't exist non-zero quadratic residues $a,b \mod 3^n$ such that $a \equiv 5b$? How do I prove that? It seems like it requires induction but its not clear.

Of course that has a natural generalization too (since we could relabel our "classes" arbitrarily ):

$$ x^2 \equiv a \mod c \\ y^2 \equiv b \mod c \\ a = kb \\ \rightarrow \exists L | L^2 \equiv k \mod c$$

Answer 1

Lemma: if there is any integer solution to $x^2 - 5 y^2 - 3 z^2 = 0$ with not all of $x,y,z$ equal to zero, then there is such a solution with $\gcd(x,y,z) = 1.$

Proof: given a solution $(a,b,c)$, not all zero, divide all by $\gcd(a,b,c)$

Lemma: if we have integers with $$x^2 - 5 y^2 - 3 z^2 \equiv 0 \pmod 9 \; ,$$ then all three of $x,y,z$ are divisible by $3$

Lemma: if we have integers with $$x^2 - 5 y^2 - 3 z^2 \equiv 0 \pmod {25} \; ,$$ then all three of $x,y,z$ are divisible by $5$

Note that the product relation for the Hilbert Norm Residue symbol requires that an indefinite ternary quadratic form be anisotropic over $\mathbb Q_p$ for an even number of finite primes $p$

On the other hand, the prime $2$ presents no obstacle:

   n == x^2 - 3 y^2 - 5 z^2 mod 16
     note x,y,z not all even.

       n         x    y    z

       0         0    1    3
       1         0    3    2
       2         1    3    2
       3         0    0    3
       4         1    0    3
       5         0    3    0
       6         1    3    0
       7         0    2    3
       8         0    1    1
       9         0    1    2
      10         1    1    2
      11         0    0    1
      12         1    0    1
      13         0    1    0
      14         1    1    0
      15         0    2    1

       n         x    y    z

=========================================================

0 and 3 are missing mod 9

   1         0    0    4
   2         1    0    4
   4         0    0    1
   5         1    0    1
   6         0    1    0
   7         0    0    2
   8         1    0    2


0 and 10 and 15 are missing mod 25

   1         1    0    0
   2         0    1    2
   3         0    3    2
   4         1    3    2
   5         0    0    2
   6         1    0    2
   7         0    4    2
   8         0    2    1
   9         1    2    1
  11         2    4    2
  12         0    6    1
  13         0    2    0
  14         1    2    0
  16         2    6    1
  17         0    1    1
  18         0    2    2
  19         1    2    2
  20         0    0    1
  21         1    0    1
  22         0    1    0
  23         0    3    0
  24         1    3    0

=====================================