The XYZ Equations

First, we solve if one of $(x,y,z)$ is 0. WLOG, say $x=0$ . From the first equation, $y^2z^2=0$ Hence, another of $(y,z)$ is 0. If two variables are 0, all three equations are satisfied. Therefore, we have $1+3(20)=61$ solutions here.

Now, assume $x,y,z\neq 0$ . Adding the first two equations, and subtracting the third, we get $y^2z^2=xyz$ Similarly, we get $x^2z^2=16xyz$ $x^2y^2=4xyz$ Therefore, $xz=16y$ $xy=4z$ $yz=x$ Multiplying all three, we get that $xyz=64$ Dividing that equation by the above three equations, $x=\pm 8$ $y=\pm 2$ $z=\pm 4$ Also, in the original three equations, the LHS is positive, so an even number of variables can be negative. If 2 are negative, there are 3 solutions. If 0 are negative, there is 1 solution.

The answer is $61+3+1=\boxed{65}$ .

Clearly there are at least 61 solutions, namely (0,0,0) and any ordered triple with two 0's and one other of the 20 numbers -10, -9, ... , -1 and 1, 2, ..., 10. To find others, let $a = x^2y^2$ , $b= y^2z^2$ , $c=z62x^2$ , and $d=xyz$ . We have $a+b=5d, b+c=17d, c+a=20d$ Adding all three gives $a+b+c=21d$ . Now, substituting in values to obtain an equation in $c$ and $d$ gives $c-12d+c-15d+c=21d$ [It is not immediately obvious to me how he arrived at this equation, though we have $21 d = a + b + c = 5d + c$ , which leads to the next conclusion - Calvin] Therefore $c=16d$ , so $x^2z^2=16xyz$ , $y^2z^2=xyz$ , and $x62y^2=4xyz$ . Simplifying these (since $x, y, z$ are non-zero) gives $xz=16y$ , $yz=x$ , and $xy=4z$ . Solving these gives $z=\pm 4, y=\pm 2$ , and $x=\pm 8$ . Since the right side of the original equations are positive, the product of three variables must be positive, they are either all positive, or exactly two are negative. This gives four additional solutions (8,2,4) , (8,-2,-4) , (-8,-2,4) and (-8,2,-4), Therefore there are 65 solutions in total.

Firstly we'll count the zero included solutions. Notice that if $x=0$ , then we must have $yz=0$ from the second equation. So there are three cases- only $y=0$ , only $z=0$ or both of them equal to zero. If only $y=0$ , then $z$ can take any value from -10 to 10 except zero. So there are $20$ such solutions. By symmetry, there are also 20 solutions each of the form $(x,z)=(0,0)$ ; $y\neq 0$ and $(y,z)=(0,0)$ ; $x\neq 0$ Also remember there is a solution $(x,y,z)=(0,0,0)$ So there are total $3\times 20+1=61$ zero included solutions. Now we may assume $xyz\neq 0$ Sum up all the equations and it gives $2(x^2y^2+y^2z^2+z^2x^2)=42xyz$ $\Rightarrow x^2y^2+y^2z^2+z^2x^2=21xyz$ Now from the third equation we derive $y^2z^2=xyz\Rightarrow yz=x$ Similarly from the other two equations we get $zx=16y$ and $xy=4z$ Now multiplying two of these equations at a time and cancelling common terms from both sides we find $x^2=64;y^2=4;z^2=16$ To satisfy the main equations we must have $xyz$ positive because left hand side of each equation is positive. (Being sum of non-zero squares) So the possible solutions are $(x,y,z)=(8,2,4)(8,-2,-4)(-8,2,-4)(-8,-2,4)$ Hence we have exactly $61+4=65$ possible solutions. By (logical) checking we find each of them is correct.

Consider 3 cases.

1. Two of x y z are zero. The other one can be -10 to -1, 1 to 10. Because x and y, y and z, z and x being zero are different orders, so in total there are 20x3=60

2. All of them are zero.

3. None of them are zero.(so that same x, y or z can be cancelled on both sides) Then the equations need to be solved. From 1st equation, y2(x2+z2)=5xyz. -----the 4th equa.

Use the 3rd equation minus 2nd equa. To get y2(x2-z2)=3xyz ------the 5th equa.

Next, 4th equa over 5th equa. (x2+z2)/(x2-z2)=5/3

Solve, x=2z or -2z

Similarly, do this with the 2nd equation to get relationship btw x and y. Result is x=4y or -4y.

So the ratio of x, y and z is 4 to 1 to 2. Note they can be positive or negative. * Since sum of squares is always positive, so for xyz to be positive, one or three of them must be positive. (can't be two positive and one negative). And given -10 to 10, there are only 8 possible sets, (8,2,4),(8,-2,-4),(-8,2,-4),(-8,-2,4) and (4,1,2),(4,-1,-2),(-4,-1,2),(-4,1,-2).

BUT the last four doesnt work if we plug them in. So the answer is 60+1+4=65.

We first consider the case where none of $x, y, z$ is $0$ .

We first solve the system of equations. We assume that all of $x, y, z$ is positive.

Adding up all of the $3$ equations, we get $2x^2y^2+2x^2z^2+2y^2z^2=42xyz$ , so $x^2y^2+x^2z^2+y^2z^2=21xyz$ . Hence $x^2y^2=4xyz$ , $x^2z^2=16xyz$ , and $y^2z^2=xyz$ .

We now find the ratio of $x, y, z$ . $x^2y^2:x^2z^2:y^2z^2=4:16:1$ , so we could have $x^2:y^2:z^2=16:1:4$ , so $x:y:z=4:1:2$ . We let $x=4k, y=k, z=2k$ . Then from the first equation we have $20k^4=40k^3$ , so $k=2$ . Thus we have $(x,y,z)=(8,2,4)$ .

If not all of them are positive, then the only thing changed would be that $2$ of them can be negative. So there are $3$ ways in this case.

If at least one of them is $0$ , then it would follow that another of them would be zero. Consider the variable that will be zero, wlog let it be $x$ . Then $y^2z^2=0$ , so at least one more would be zero. There are $(20)(3)=60$ ways ( $3$ ways to choose the one that would be not zero, and 20 ways to choose the value of it) for there to be two zeros, and $1$ way for there to be one zero.

Hence there are $3+1+60+1=\boxed{65}$ possible such ordered pairs of integers.,

Add all equation, so we obtain

$\sum_{cyc} x^2y^2 = 21xyz$ . And then subtract this one to the first, second, and third equation. So, we obtain that :

$x^2z^2 = 16xyz, x^2y^2 = 4xyz$ , and $y^2z^2 = xyz$

And, then if one of $(x, y, z) = 0$ , WLOG $x = 0$ . We obtain that $y^2z^2 = xyz = 0$ . That is mean one of $(y, z)$ should be $0$ . And then, devide into $2$ cases, that is :

Case $1$ : If there is $0$ . So, WLOG $x = y = 0$ . Since $-10 \le z \le 10$ . So, the possibilities values of $(x, y, z) = 21$ . Since there is three possibilities of $(x = y = 0$ , or $x=z=0$ , or $y = z = 0$ . So, total is $63$ solution. But, $(0, 0, 0)$ Counted $3$ times. So, there is $63 - 2 = 61$ possibilities for this case

Case $2$ : $x, y, z \ne 0$ . So, from $x^2z^2 = 16xyz, x^2y^2 = 4xyz$ , and $y^2z^2 = xyz$ . We obtain that :

$xz = 16y$ , and $xy = 4z$ , and $yz = x$ . Multiplies all of expression, so, we obtain that $xyz = 64$ . And, divide it into these expression. So, we obtain that :

$y^2 = 4$ , and $z^2 = 16$ , and $x^2 = 64$ . So, we obtain the ordered solution for this case is : $(2, 4, 8), (2, -4, -8), (-2, 4, -8), (-2, -4, 8)$ . So, there is $4$ solution.

Totally, there is $61 + 4 = \boxed{5}$ solution.

We first label the equations as $(1), (2), (3)$ based on their order above.

$(1)+(2)-(3) \Rightarrow 2y^2z^2=2xyz$ , or $yz(yz-x)=0*$ . $(1)+(3)-(2) \Rightarrow 2x^2y^2=8xyz$ , or $xy(xy-4z)=0**$ . $(2)+(3)-(1) \Rightarrow 2z^2x^2=32xyz$ , or $zx(zx-16y)=0***$ .

First, we consider solutions where at least one of $x,y,z$ is 0.

In $*,y=0$ is a solution. Substituting this in $(2)$ gives $zx=0$ , which in turn gives $x=0, z=0,$ , or $x=z=0$ . Also, $z=0$ is a solution. Substituting this in $(1)$ gives $xy=0$ , which in turn gives $x=0, y=0,$ , or $x=y=0$ . Do this for $**$ and $***$ , and notice that aside from $(0,0,0)$ , the solutions that we will obtain are of the form $(a,0,0), (0,a,0), (0,0,a),$ , where $-10\le a\le 10,a\not=0$ . These make 61 solutions.

Now we consider solutions where $x,y,z\not=0$ using the equations $yz=x, xy=4z, zx=16y$ . Substitute $yz=x$ into $zx=16y$ and we get $y=\pm2$ .

If $y=2$ , then $2z=x,zx=32$ , and $x=8,z=4$ or $x=-8,z=-4$ .

If $y=-2$ , then $-2z=x,zx=-32$ , and $x=8,z=-4$ or $x=-8,z=4$ .

All in all, there are 65 solutions.

First, take the cases when x, y or z is zero. Say x=0. Then $y^2 z^2=0$ by equation 2. This means that either $y=0$ and z can be any number between -10 and 10 inclusive, or vice versa. Examining the cases y=0 and z=0, it can be seen in the same way that it is also possible that y and z are equal to zero and x can be any number between -10 and 10 inclusive.

In total this makes 61 possibilities (20 cases each when two variables are zero and the other is not zero, and 1 case when all three variables are zero).

Then, we examine the case when x, y and z are not zero. Then for all 3 equations, we can divide both sides by $xyz$ . Let $A=\frac{xy}{z}$ , $B=\frac{yz}{x}$ , $C=\frac{zx}{y}$ . Then we get:

$A+B=5$
$B+C=17$
$C+A=20$

Solving them yields A=4, B=1, C=16. Substituting back expressions for A, B and C and solving, we get:

x = 8 or -8
y = 2 or -2
z = 2 or -2

However, only an even number of the variables can be negative since $xyz$ must be non-negative (this can be seen in the original set of equations where the left hand side of each equation is a sum of squares and must thus be non-negative). So, either x and y are negative, y and z are negative, z and x are negative, or none of them are negative. This makes 4 solutions.

In total, there are 61+4=65 solutions.

Label the given equations (1), (2), (3) in order.

(1) + (2) - (3): $xyz = y^2z^2 \Rightarrow yz(x-yz)=0$ - (4)

(1) + (3) - (2): $4xyz = x^2y^2 \Rightarrow xy(4z-xy)=0$ - (5)

(2) + (3) - (1): $16xyz = x^2z^2 \Rightarrow xz(16y-xz)=0$ - (6)

Case 1: At least one of $x,y,z$ is 0 and WLOG assume $x=0$

Equations (5), (6) are satisfied and Equation (4) simplifies to $-y^2z^2=0 \Rightarrow y=0 \text{ or } z=0$ and the case is similar for $y=0 \text{ and } z=0$

Case 2: All of $x,y,z \neq 0$

The equations above simplify to $16y=xz$ , $4z=xy$ , $x=yz$ which after multiplying give us $64xyz=x^2y^2z^2 \Rightarrow 64=xyz$ and dividing this equation by the 3 equations above gives us $y^2=4$ , $z^2=16$ , $x^2=64$ and since $xyz > 0$ then either all 3 are positive or exactly 2 of them are negative

Combining the above cases we have that all the solutions are $(x,y,z) = (0,0,k),(0,k,0),(k,0,0),(0,0,0)$ , $(8,2,4),(-8,-2,4),(-8,2,-4),(8,-2,-4)$ where $k \neq 0$ and since $k \in [-10,10]$ there are $10-(-10) = 20$ values since it cannot be $0$ so there are a total of $20+20+20+5=65$ solutions

Adding Equations (1), (2) and (3), and dividing both sides by 2, we have $x^2y^2+y^2z^2+z^2x^2=21xyz$ . Using this fact in Equ (1), (2) and (3) respectively and simplifying, we have-

$16xyz=z^2x^2$

$4xyz=x^2y^2$

$xyz=yz$

If one of the variables is 0, then one of the other variables is also 0. The third variable can take any value in $[-10,10]$ . So there are 21 possibilities. In total, when atleast two of the three variables are 0, there are 21*3-2 possibilities. We subtract by 2 because we want to count the case where all the variables are 0 exactly once.

If $x \ne 0, y \ne 0, z \ne 0$ , then we can simplify the above three equations to-

$xz=\pm 32, xy = \pm 16, yz=\pm 8$ .

All three variables can be positive or if one of the variables is negative, then another variable also needs to be negative. Dividing the 3-d space in 8 quadrants, the solution to the above can lie in any 4 quadrants. In the $x \gt 0, y \gt 0, z \gt 0$ quadrant, there is exactly one solution $x=16, y=1, z=2$ . Each of the variables can take positive or negative value (giving one potential solution for each of the 8 quadrants) with the restriction that if one of them is negative, then exactly one other is also negative resulting in exactly 4 solutions.

The total number of solutions is 21*3-2+4 = 65.