15. The integer $x$ makes $x^{2}+x+13$ a multiple of 121, the number of such integers $x$ is ( ).
A. 0
B. 1
C. 2
D. Infinite
15. A. Suppose there exists an $x$ satisfying the condition, then $x^{2}+x+3=121 k(k \in \mathbf{N}), \therefore x^{2}+x+(3-121 k)=0$. Since $x$ is an integer, the discriminant must be a perfect square, i.e.,
$1-4(3-121 k)=n^{2}(n \in \mathbf{N})$, i.e., $11(44 k-1)=n^{2}, \therefore \quad 11 \times(44 k-1), \therefore \quad 11^{2} \times n^{2}$. But since $11 \mid n^{2}$ and 11 is a prime number, $11 \mid n$, thus $11^{2} \mid n^{2}$, which is a contradiction! Therefore, the answer is $\mathrm{A}$.