Let a, x, y, z be real numbers satisfying the equations
`ax+ay=z`
`x+ay=z`
`x+ay=az,` where x, y, z are not all zero, then the number of the possible values of a is

Answer

To solve the problem, we need to analyze the given equations and find the values of \( a \).

### Given Equations:
1. \( ax + ay = z \)
2. \( x + ay = z \)
3. \( x + ay = az \)

### Step 1: Rearranging the Equations
We can rearrange the equations to express them in a standard form:
1. \( ax + ay - z = 0 \)
2. \( x + ay - z = 0 \)
3. \( x + ay - az = 0 \)

### Step 2: Writing the System in Matrix Form
We can write the above equations in matrix form as follows:
\[ \begin{bmatrix} a & a & -1 \\ 1 & a & -1 \\ 1 & a & -a \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \]

### Step 3: Finding the Determinant
To find the values of \( a \) for which the system has non-trivial solutions (i.e., \( x, y, z \) are not all zero), we need to set the determinant of the coefficient matrix to zero:
\[ \Delta = \begin{vmatrix} a & a & -1 \\ 1 & a & -1 \\ 1 & a & -a \end{vmatrix} \]

### Step 4: Calculating the Determinant
Calculating the determinant using cofactor expansion:
\[ \Delta = a \begin{vmatrix} a & -1 \\ a & -a \end{vmatrix} - a \begin{vmatrix} 1 & -1 \\ 1 & -a \end{vmatrix} - 1 \begin{vmatrix} 1 & a \\ 1 & a \end{vmatrix} \]

Calculating the 2x2 determinants:
1. \( \begin{vmatrix} a & -1 \\ a & -a \end{vmatrix} = a(-a) - a(-1) = -a^2 + a = a - a^2 \)

2. \( \begin{vmatrix} 1 & -1 \\ 1 & -a \end{vmatrix} = 1(-a) - 1(-1) = -a + 1 = 1 - a \)

3. \( \begin{vmatrix} 1 & a \\ 1 & a \end{vmatrix} = 1 \cdot a - 1 \cdot a = 0 \)

Substituting these back into the determinant:
\[ \Delta = a(a - a^2) - a(1 - a) - 0 \]
\[ = a^2 - a^3 - a + a^2 \]
\[ = 2a^2 - a^3 - a \]

### Step 5: Setting the Determinant to Zero
Setting the determinant to zero gives:
\[ -a^3 + 2a^2 - a = 0 \]
Factoring out \( -a \):
\[ -a(a^2 - 2a + 1) = 0 \]
This simplifies to:
\[ -a(a - 1)^2 = 0 \]

### Step 6: Finding the Values of \( a \)
From the factored equation, we have:
1. \( -a = 0 \) which gives \( a = 0 \)
2. \( (a - 1)^2 = 0 \) which gives \( a = 1 \)

### Conclusion
The possible values of \( a \) are \( 0 \) and \( 1 \). Therefore, the number of possible values of \( a \) is **2**.