[Number Theory]
Problem: $\quad$ Suppose that $p$ and $q$ are two-digit prime numbers such that $p^{2}-q^{2}=2 p+6 q+8$. Compute the largest possible value of $p+q$.

Solution: Subtract from both sides and regroup to obtain $p^{2}-2 p-\left(q^{2}+6 q\right)=8$. Completing both squares yields $(p-1)^{2}-(q+3)^{2}=0$. The left side is a difference of two squares; factor to obtain $((p-1)+(q+3))((p-1)-(q+3))=0$, whence $(p+q+2)(p-q-4)=0$. For positive primes $p$ and $q$, the first factor $p+q+2$ must also be positive. Therefore the second factor $p-q-4$ must be zero, hence $p-4=q$. Now look for primes starting with 97 and working downward. If $p=97$, then $q=93$, which is not prime; if $p=89$, then $q=85$, which is also not prime. But if $p=83$, then $q=79$, which is prime. Thus the largest possible value of $p+q$ is $83+79=\mathbf{1 6 2}$.

Final answer: 162