Solve in $\mathbb Z^3$. 
Let $x$, $y$ and $z$ be integer numbers. Solve the following equation.
  $$x^2+y^2+z^2=45(xy+xz+yz)$$

My trying.
It's a quadratic equation of $z$ and we need $\Delta=n^2$ for an integer $n$,
but it gives a very ugly expression. 
Thank you!
 A: Did some informal checking. This one appears to be isotropic in $\mathbb Q_2$ and $\mathbb Q_3.$ It is definitely  anisotropic in $\mathbb Q_{11}$ and $\mathbb Q_{47}.$ 
I had done this before. There are integer solutions ($x,y,z$ not all zero) to
$$ A(x^2 + y^2 + z^2) = B (yz + zx + xy)  $$
with $A,B > 0$ and $\gcd(A,B) = 1$  and $B > A$ if and only if both
$$ B - A = r^2 + 3 s^2 $$ and
$$   B + 2 A = u^2 + 3 v^2 $$
You have
$$   45 - 1 = 44 $$
and
$$ 45 + 2 = 47 $$
both of which are $2 \pmod 3$ and cannot be so written.  See
Find a solution: $3(x^2+y^2+z^2)=10(xy+yz+zx)$
Proving necessity: defining
$$ u = -x-y+2z, \; \; \; v = -x+y, \; \; \; w = x+y+z,  $$ we get diagonalization
$$  12 g = (2A+B) u^2 + 3 (2A+B) v^2 -4(B-A) w^2.  $$
Then we use the theorem of Legendre on indefinite ternaries
Proof of Legendre's theorem on the ternary quadratic form 
And, there are no nontrivial solutions to
$$   47 u^2 + 3 \cdot 47 v^2 -4\cdot 44 w^2.  $$
$$   47 u^2 + 141 v^2 -176 w^2.  $$
To be specific, if
$$  47 u^2 + 3 \cdot 47 v^2 -4\cdot 44 w^2 \equiv 0 \pmod {11^2},  $$
then all 
$$ u,v,w \equiv 0 \pmod {11}. $$
As a result, there can be no solutions with $\gcd(x,y,z) = 1,$ hence no nonzero solutions.
If there are any solutions, you get infinitely many by Vieta Jumping, similar to the Markoff numbers $x^2 + y^2 + z^2 = 3xyz.$ Similarly, one may rule out any solutions. I am on the phone, if you cannot work it out I can do something later  https://en.wikipedia.org/wiki/Markov_number
A: For  Will Jagy, I am sorry!
Let $x-y=a$, $y-z=b$ and $z-x=c$.
Hence, $a^2+b^2+c^2\vdots11$ and $a+b+c=0$.
Thus, $a^2+ab+b^2\vdots11$, which says that $a\vdots11$ and $b\vdots11$ and $x\equiv y\equiv z(\mod11)$, 
which gives $x\vdots11$, $y\vdots11$ and $z\vdots11$ (if $x\equiv y\equiv z\equiv r(\mod11)$ then $r^2\vdots11$).
Id est,  an infinite descent ends this problem.
Done!
