To solve the problem of finding the number of triples \((x, y, z)\) such that adding any one of these numbers to the product of the other two results in 2, we can set up the following equations based on the problem statement:
1. \(x + yz = 2\) (Equation 1)
2. \(y + xz = 2\) (Equation 2)
3. \(z + xy = 2\) (Equation 3)
Now, let's solve these equations step by step.
### Step 1: Rearranging the equations
From each equation, we can express one variable in terms of the others:
- From Equation 1:
\[
x = 2 - yz
\]
- From Equation 2:
\[
y = 2 - xz
\]
- From Equation 3:
\[
z = 2 - xy
\]
### Step 2: Substituting values
Now, we can substitute the expression for \(x\) from Equation 1 into Equation 2:
Substituting \(x = 2 - yz\) into Equation 2:
\[
y + (2 - yz)z = 2
\]
Expanding this gives:
\[
y + 2z - y z^2 = 2
\]
Rearranging gives:
\[
-y z^2 + y + 2z - 2 = 0
\]
Factoring out \(y\):
\[
y(1 - z^2) + 2z - 2 = 0
\]
### Step 3: Solving for \(y\)
Rearranging for \(y\):
\[
y(1 - z^2) = 2 - 2z
\]
Thus,
\[
y = \frac{2 - 2z}{1 - z^2}
\]
### Step 4: Substituting \(y\) into Equation 3
Now, we substitute \(y\) back into Equation 3:
\[
z + \left(\frac{2 - 2z}{1 - z^2}\right)x = 2
\]
This will lead us to a more complex equation, but we can also use symmetry in our equations.
### Step 5: Using symmetry
Notice that all three equations are symmetric in \(x\), \(y\), and \(z\). This suggests that \(x = y = z\) might be a solution. Let’s denote \(x = y = z = k\).
Substituting \(k\) into any of the equations:
\[
k + k^2 = 2
\]
This simplifies to:
\[
k^2 + k - 2 = 0
\]
Factoring gives:
\[
(k - 1)(k + 2) = 0
\]
Thus, \(k = 1\) or \(k = -2\).
### Step 6: Finding triples
1. If \(k = 1\), then \((x, y, z) = (1, 1, 1)\).
2. If \(k = -2\), then \((x, y, z) = (-2, -2, -2)\).
### Conclusion
The valid triples \((x, y, z)\) are \((1, 1, 1)\) and \((-2, -2, -2)\). Since the problem asks for the number of triples, we have found two distinct solutions.
Thus, the answer is **2**.
