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.

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.
