Show that if $a,b,c\in \mathbb{N}$ and ${a^2+b^2+c^2}\over{abc+1}$ is an integer it is the sum of two nonzero squares I was reading about the fascinating problem in the IMO ($1988 $ #$6$) that asks:
Let $a$ and $b$ be positive integers such that $(ab+1) | (a^2+b^2)$. Show that ${a^2+b^2}\over{ab+1}$ is a perfect square.
I wondered what would happen with the natural extension of this problem to three constants. Upon exploration with Mathematica, I found that the integral values of ${a^2+b^2+c^2}\over{abc+1}$ all seemed to be expressible as the sum of two nonzero perfect squares. The converse also seems to be true (given that a number is the sum of two nonzero squares, it seems to be expressible in the form of this fraction). 
So I pose two questions:
If $a,b,c\in \mathbb{N}$ and $k={{a^2+b^2+c^2}\over{abc+1}}$ is an integer, is $k$ the sum of two nonzero squares?
and
Given that $k$ is expressible as the sum of two nonzero perfect squares, do there exist $a,b,c\in\mathbb{N}$ for which ${{a^2+b^2+c^2}\over{abc+1}}=k$?

I PROVED THE CONJECTURE
$$   x^2 + y^2 + z^2 = k + kxyz $$
Notice that we cannot have a solution with $x<0$ 
Got the other part. Let $x \geq y \geq z \geq 1$ be a solution giving $k$ that minimizes $x+y+z$ among positive solutions giving $k.$ Vieta jumping is just the new solution
$$ (x,y,z) \mapsto (kyz-x,y,z)  $$ 
The equation is
$$ x^2 - (kyz)x + y^2 + z^2 - k = 0. $$
The replacement root $x'$ gives $x + x' = kyz. $ The product is $xx' = y^2 + z^2 - k.$
We have minimized $x+y+z$ among positive triples giving $k.$ We get a new triple with $x' = kyz - x.$ As it is impossible to have $x' <0,$ we have an Alternative:
(A) $kyz - x = 0$
(B) $kyz - x \geq x$
We will show that alternative (B) does not occur for, say, $k > 2.$
ASSUME (B), namely $kyz \geq 2x, $ with $x \geq y \geq z \geq 1.$
$$ x^2 + y^2 + z^2 = k + (kyz)x \geq k + (2x)x= k + 2x^2 $$
$$ \color{red}{ y^2 + z^2 \geq k + x^2}.  $$ 
But
$$ y^2 - (kx)yz + z^2 = k - x^2. $$
Add,
$$ 2y^2 - (kx)yz + 2z^2 \geq 2k .  $$
$$ y^2 - \left(\frac{kx}{2} \right)yz + z^2 \geq k .  $$
So, $y \geq z \geq 1$ and
$$ \color{red}{y^2 - \left(\frac{kx}{2} \right)yz + z^2 > 0 }.  $$
From the quadratic formula,
$$  y > \frac{1}{4} \left( kx + \sqrt {k^2 x^2 - 16}  \right) z .  $$
When $k \geq 4,$ we see $y > x,$ a contradiction. When $k=3$ and $z \geq 2,$ we still get $y > x.$ Finally, with $k=3$ and $z=1,$ the condition $y^2 + z^2 \geq k + x^2$ reads $y^2 + 1 \geq 3 + x^2$ or $y^2 \geq 2 + x^2,$ so again $y > x.$
These contradictions tell us that alternative (A) actually holds for $k \geq 3.$ That is, $kyz = x,$ so $kyz - x = 0,$ and we have a new solution $(0,y,z)$ with $y \geq z \geq 1$  and this $x=0$ to
$$ x^2 + y^2 + z^2 = k + kxyz.  $$ As this $x=0,$ we have reached
$$ \color{red}{ y^2 + z^2 = k }  $$
