I3.1 If $P$ is a prime number and the roots of the equation $x^{2}+2(P+1) x+P^{2}-P-14=0$ are integers, find the least value of $P$.
$$
\begin{array}{l}
\Delta=4(P+1)^{2}-4\left(P^{2}-P-14\right)=m^{2} \\
\left(\frac{m}{2}\right)^{2}=P^{2}+2 P+1-P^{2}+P+14=3 P+15
\end{array}
$$

The possible square numbers are $16,25,36, \ldots$ $3 P+15=16$ (no solution); $3 P+15=25$ (not an integer); $3 P+15=36 \Rightarrow P=7$ The least possible $P=7$