For how many integers $k$ does the following system of equations has a solution other than $a=b=c=0$ in the set of real numbers? \begin{align*} \begin{cases} a^2+b^2=kc(a+b),\\ b^2+c^2 = ka(b+c),\\ c^2+a^2=kb(c+a).\end{cases}\end{align*}
To determine the number of integer values of \( k \) for which the given system of equations has a nontrivial solution, we start by analyzing the system:

\[
\begin{cases}
a^2 + b^2 = kc(a + b) \\
b^2 + c^2 = ka(b + c) \\
c^2 + a^2 = kb(c + a)
\end{cases}
\]

1. **Subtracting pairs of equations:**

   Subtract the first equation from the second:
   \[
   b^2 + c^2 - (a^2 + b^2) = ka(b + c) - kc(a + b)
   \]
   Simplifying, we get:
   \[
   c^2 - a^2 = ka(b + c) - kc(a + b)
   \]
   \[
   (c - a)(c + a) = k(bc + c^2 - ab - ac)
   \]
   \[
   (c - a)(c + a) = k(c(b - a) + a(c - b))
   \]
   \[
   (c - a)(c + a) = k(c - a)(b - a)
   \]
   Dividing both sides by \( c - a \) (assuming \( c \neq a \)):
   \[
   c + a = k(b - a)
   \]
   Similarly, subtracting the second equation from the third:
   \[
   c^2 + a^2 - (b^2 + c^2) = kb(c + a) - ka(b + c)
   \]
   Simplifying, we get:
   \[
   a^2 - b^2 = kb(c + a) - ka(b + c)
   \]
   \[
   (a - b)(a + b) = k(bc + ac - ab - bc)
   \]
   \[
   (a - b)(a + b) = k(a(b - c) + b(c - a))
   \]
   \[
   (a - b)(a + b) = k(a - b)(c - a)
   \]
   Dividing both sides by \( a - b \) (assuming \( a \neq b \)):
   \[
   a + b = k(c - a)
   \]

2. **Case 1: \( a = b = c \)**

   If \( a = b = c \), substituting into any of the original equations:
   \[
   2a^2 = ka(2a)
   \]
   \[
   2a^2 = 2ka^2
   \]
   Dividing both sides by \( 2a^2 \) (assuming \( a \neq 0 \)):
   \[
   1 = k
   \]
   So, \( k = 1 \).

3. **Case 2: \( a = c \neq b \)**

   If \( a = c \), substituting into the first and third equations:
   \[
   a^2 + b^2 = ka(a + b)
   \]
   \[
   a^2 + a^2 = ka(a + a)
   \]
   \[
   2a^2 = 2ka^2
   \]
   Dividing both sides by \( 2a^2 \) (assuming \( a \neq 0 \)):
   \[
   1 = k
   \]
   So, \( k = 1 \).

4. **Case 3: \( a \neq b \neq c \)**

   For the system to have nontrivial solutions, the determinant of the coefficient matrix must be zero:
   \[
   \begin{vmatrix}
   1 & k & 1 \\
   k & 1 & 1 \\
   1 & 1 & k
   \end{vmatrix} = 0
   \]
   Expanding the determinant:
   \[
   1 \cdot (1 \cdot k - 1 \cdot 1) - k \cdot (k \cdot k - 1 \cdot 1) + 1 \cdot (k \cdot 1 - 1 \cdot 1)
   \]
   \[
   = k - 1 - k^3 + k - k
   \]
   \[
   = k - 1 - k^3
   \]
   Setting the determinant to zero:
   \[
   k^3 - k + 1 = 0
   \]
   Solving for \( k \):
   \[
   (k + 2)(k - 1)^2 = 0
   \]
   The solutions are \( k = -2 \) and \( k = 1 \).

Thus, the integer values of \( k \) for which the system has a nontrivial solution are \( k = 1 \) and \( k = -2 \).

The final answer is \(\boxed{2}\)