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.
 A: 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 $16q^2$.
A: Here is another approach and solution.
Using the identity   $X^2 + Y^2 + Z^2 = (X + Y + Z)^2 - 2(XY + XZ + YZ)$
Substituting and simplifying we get
$$(X + Y + Z)^2 + 16 = 18(XY + XZ + YZ)$$
But we know from Fermat and others that the sum of $2$ squares
is not divisible by primes of the form $4N+3$ unless both squares are
themselves divisible by such prime.  Since $16$ is not divisible by
any primes of the form $4N+3$, we see that the left side of he equation
can not be divisible by $3$ while the right side is divisible by $3$.  The
contradiction concludes the proof.
