On integer solutions of the equation $x^2+y^2+z^2=16(xy+yz+zx-1)$ Here is the question:
Question. Show that the equation
$$x^2+y^2+z^2=16(xy+yz+zx-1)$$
does no have integer solutions.
I know a nice and easy (actually, an obvious) way to solve this problem. But I'm just wondering can we solve this using infinite descent method? I remember I saw a solution using that method, but it was wrong.

The following is an approach that is not by infinite descent, but imitates at the beginning descent approaches to similar problems.
Any square is congruent to $0$, $1$, or $4$ modulo $8$. It follows easily that in any solution of the given equation, $x$, $y$, and $z$ must be even, say $x=2r$, $y=2s$, $z=2t$. Substitute and simplify. We get
$$r^2+s^2+t^2=16(rs+st+tr) -4$$
Using more or less the same idea, we observe that $r$, $s$, and $t$ must be even. Let $r=2u$, $s=2v$, $t=2w$. Substitute and simplify. We get
$$u^2+v^2+w^2=16(uv+vw+wu)-1$$
Now the descending stops. The right-hand side is congruent to $-1$ modulo $8$, but no sum of $3$ squares can be.
ADDED: I have found a way to make the descent infinite, for proving a stronger result. Look at the equation
$$x^2+y^2+z^2=16(xy+yz+zx)-16q^2$$
We want to show that the only solution is the trivial one $x=y=z=q=0$. The argument is more or less the same as the one above, except that when (after $2$ steps) we reach $-q^2$, we observe that there is a contradiction if $q$ is odd, so now let $q$ be even, and the descent continues. It would probably be more attractive to use $8$ than $16$, and $4q^2$ instead of $16q^2$.
