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.

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@Arturo, Nice way! but no, mine is more easier. Rewrite it as $(x+y+z)^2+4^2=18(xy+yz+zx)$ and use the fact that if $3 \mid a^2+b^2$ then $3 \mid a$ and $3 \mid b$. –  Amir Hossein Apr 12 '11 at 15:11
Actually, I have a sign error; doing it correctly I would have gotten the same thing: $(x+y+z)^2 = 18(xy+yz+zx)-16$, so considering modulo $3$ (instead of modulo $7$), you know that $-1$ is not a square modulo $3$ (equivalent to what you have), so no integer solution. –  Arturo Magidin Apr 12 '11 at 15:15
Yes I would like to see a descent proof for this too, I haven't tried yet but if there's no common factors it's going to be very difficult. –  quanta Apr 12 '11 at 16:48

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 $16q^2$.

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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.

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