I'm trying to prove that if $p,q_1q_2,q_3,q_4,q_5$ are primes and we are given that $q_1q_2q_3q_4q_5+1=p^2$, that then $p$ has to be either $7$, $11$ or $13$.
I succeeded to show that this cannot be true for the three primes smaller than $7$. Also, finding the numbers $q_1q_2,q_3,q_4,q_5$ is easy for $p=7,11,13$. I also showed that for a prime $p>3$, $24=2^3\cdot3|p^2-1$, so if we would find a $p$ larger than $13$ satisfying above conditions we would have $q_1q_2q_3q_4q_5=2^3\cdot3q_5|p^2-1$.
However, from here on I don't really see how to proceed. I looked on Google for this question but I couldn't find anything really helpful.
If $p$ is an odd prime $> 3$, $p^2-1 = (p+1)(p-1)$ is divisible by $24 = 2^3 \cdot 3$, so there is only one more prime available, which can divide only one of $p+1$ and $p-1$. The one it doesn't divide is at most $12$, so $p \le 13$.
EDIT: BTW, if you allow six $q$'s instead of just five, there appear to be infinitely many solutions. See OEIS sequence A106639.