How do I show that except for 5039, there is no prime between 5033 and 5047. I just need a nudge in the right direction, no idea how to start the problem :(
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It is also good to remember that $7! = 5040$. Hence, among the integers from $[7! - 7, 7!+7]$ the only ones that could be prime are $7! \pm 1$. But $7! + 1 = 5041 = 71^2$.
Hence, the only number that can be prime is $7!-1$.
Recall that a number is divisible by $3$ if and only if the sum of the digits is divisible by $3$.
If you write down each number between those two, that is $5034, 5035, \dots , 5046$ you'll notice that each one is either even, divisible by $3$ or divisible by $5$ except 5039 and 5041. Remembering our table of squares, $71^2=5041$, so that completes proof.
Knowing that a composite number will always be a product of primes, let's apply some tests and see the numbers that are factors of each. Note that I may not mention all the factors since one known factor can disprove everything.
Let's begin by scratching out even numbers from the list. Now, for the odd ones.
So, the only prime number now is $5039$ which is good enough to complete the proof.