Diophantine equations and the number 4

I'll try to solve both problems at once. So let's start with the equation $x^2 + y^2 + z^2 + 1 = mxyz.$

First let's do a little $2$ -adic search. Clearly $x,y,z$ can't all be even, and if exactly one of them is even this is no good either (left side is odd, right side is even). If two of $x,y,z$ are even, then the left side is $2$ mod $4$ but the right side is divisible by $4$ , so that's also impossible. So $x,y,z$ are all odd. Looking mod $8$ we see that this implies $m \equiv 4$ mod $8$ .

The key thing to note about this particular equation is that if $(x,y,z)$ is a solution, then so is $(x,y,mxy-z)$ . This gives us a way of generating solutions from other solutions by replacing one of the coordinates. As we can apply this process iteratively, note that given any solution we can construct, via this process, a "nice" solution $(x,y,z)$ such that $x \le y \le z \le mxy/2$ : if this is not satisfied for a given solution, then we replace $z$ by $mxy-z$ and reorder the new coordinates--the point is that this makes the sum of the coordinates strictly smaller, and we can't decrease this sum indefinitely, so the process must terminate eventually.

For this nice solution, the inequality implies $x^2 + y^2 + 1 > (m/2)xyz$ . But the left side is less than $2y^2 + 1$ , so we get $2y^2 + 1 > (m/2)xyz \ge (m/2)xy^2$ .

If $m > 6$ then we get $2y^2 + 1 > 3xy^2 \ge 3y^2$ which is impossible. So I'm done with the extra credit problem since $m \equiv 4$ mod $8$ means that $m=4$ is the only possibility left.

If $m = 4$ this becomes $2y^2 + 1 > 2xy^2$ , so clearly we must have $x = 1$ . Plugging into one of the inequalities above, we get $2y^2 + 1 > 2yz$ , so $y + \frac1{2y} > z$ , but since $z \ge y$ we must have $z = y$ . Plugging back into the original equation we get $2y^2 + 2 = 4y^2$ so $y = z = 1$ ; that is, the unique "nice" solution is $x=y=z=1$ .

So every solution can be obtained from $(1,1,1)$ by applying $z \mapsto 4xy-z$ (or $y \mapsto 4xz-y$ or $x \mapsto 4yz-x$ ) and reordering, iteratively. Applying it once gives $(1,1,3)$ . Applying it again gives $(1,3,11)$ . Now we can either change the $3$ or the $1$ . To get a minimal solution with coordinates all $> 1$ , we should apply it to the $1$ , so we get $(3,11,131)$ . Hence the answer is $\fbox{131}$ .

I reached the point where all the three of x, y and z have to be odd. Then solved for z in terms of x and y by considering the given equation as a quadratic equation. Thereafter it was hit and trial to arrive at 3, 11 and 131. I used an excel sheet to confirm my answer was in fact the minimum value of z.

The given solution to the bonus question seems good though I could not understand all of it. But it is an ingenious and brilliant observation that x, y, z also implies x, y, mxy-z is a solution. Of course, 1,1,1 is a trivial solution and the further 3, 11, 131 can be easily derived from that.