Prove that the determinant of the matrix 
$$\begin{pmatrix}
a_{1}^{2}+k & a_1 a_2 & a_1 a_3 &\ldots & a_1 a_n\\
a_2 a_1 & a_{2}^{2}+k & a_2 a_3 &\ldots & a_2 a_n\\
\ldots & \ldots & \ldots & \ldots & \ldots \\
a_n a_1& a_n a_2 & a_n a_3 & \ldots & a_{n}^{2}+k
\end{pmatrix}$$
is divisible by $k^{n-1}$ and find its other factor.
1. **Define the matrix \( A \):**
   Let 
   \[
   A = \begin{pmatrix}
   a_{1}^{2} & a_1 a_2 & a_1 a_3 & \ldots & a_1 a_n \\
   a_2 a_1 & a_{2}^{2} & a_2 a_3 & \ldots & a_2 a_n \\
   \vdots & \vdots & \vdots & \ddots & \vdots \\
   a_n a_1 & a_n a_2 & a_n a_3 & \ldots & a_{n}^{2}
   \end{pmatrix}
   \]
   The given matrix can be written as \( A + kI \), where \( I \) is the identity matrix.

2. **Characteristic Polynomial:**
   The characteristic polynomial of \( A \) is given by \( \phi_A(k) = \det(A - kI) \). However, since we are dealing with \( A + kI \), we need to consider \( \det(A + kI) \).

3. **Rank and Eigenvalues:**
   The rank of \( A \) is either 0 or 1. This is because \( A \) is a matrix of rank 1 unless all \( a_i = 0 \). If \( (a_1, a_2, \ldots, a_n) = (0, 0, \ldots, 0) \), then \( A \) is the zero matrix, and its rank is 0. Otherwise, the rank is 1.

4. **Eigenvalues:**
   - If \( A \) is the zero matrix, all eigenvalues are 0.
   - If \( A \) is not the zero matrix, it has one non-zero eigenvalue and \( n-1 \) zero eigenvalues. This is because \( A \) is a rank-1 matrix, and the non-zero eigenvalue is the trace of \( A \).

5. **Geometric and Algebraic Multiplicities:**
   - The geometric multiplicity of the eigenvalue 0 is \( n-1 \) because the rank of \( A \) is 1.
   - The algebraic multiplicity of the eigenvalue 0 is at least \( n-1 \).

6. **Form of the Characteristic Polynomial:**
   Given that the algebraic multiplicity of 0 is at least \( n-1 \), the characteristic polynomial \( \phi_A(k) \) must be of the form:
   \[
   \phi_A(k) = k^{n-1}(k - \alpha)
   \]
   where \( \alpha \) is the non-zero eigenvalue of \( A \).

7. **Trace of the Matrix:**
   The trace of \( A \) is the sum of its diagonal elements:
   \[
   \text{trace}(A) = a_1^2 + a_2^2 + \cdots + a_n^2
   \]
   Since the trace of a matrix is also the sum of its eigenvalues, we have:
   \[
   \alpha = -(a_1^2 + a_2^2 + \cdots + a_n^2)
   \]

8. **Conclusion:**
   Therefore, the determinant of the given matrix \( A + kI \) is:
   \[
   \det(A + kI) = k^{n-1}(k - (-(a_1^2 + a_2^2 + \cdots + a_n^2))) = k^{n-1}(k + a_1^2 + a_2^2 + \cdots + a_n^2)
   \]

The final answer is \( \boxed{ k^{n-1}(k + a_1^2 + a_2^2 + \cdots + a_n^2) } \).