STEP 2 - 2018, Q6i): Find all pairs of positive integers (n, p), where p is a prime number, that satisfy n! + 5 = p .

The key observation here is that all numbers below n, by definition, divide n!=nx(n-1)x(n-2)x...x1.So in particular, if n greater or equal to 5, 5 divides n!, and thus is divides n!+5.Hence, if n!+5=p and n greater or equal o 5, 5 must divide p.
But then p must be in fact equal to 5, as prime numbers are divided by only 1 and themselves.Of course this gives n!+5=5 and so n!=0 which clearly can't happen.
OK, so we can't have 5 greater or equal to 5.
We can check "n= 1 or 2 or 3 or 4" manually.
n=1 gives p=6=2x3, not prime.
n=2 gives p=7. prime.
n=3 gives p=11, prime.
n=4 gives p=29, prime.
So the only pairs are: (2,7),(3,11),(4,29).
Remember: if you can solve the problem for every case except from a few, you can later go back fill the missing cases manually.