II_Question6_Pure

(i) Note that m| n! for any m\leqslant n, so, if n\geqslant5 then 5|n!, and 5|5 is always true, whence 5|(n!+5), so n!+5 is not prime for any n\geqslant5. So we can simply try n\in\{1,2,3,4\} by hand, and we see that we get prime numbers for n=2, n=3, and n=4. This means that the pairs are (2,7), (3,11), and (4,29).

(ii) Inspired by the previous part, let’s try to show that there will be no solutions when n gets too large. Looking at the first theorem here, it seems like n\geqslant7 might be possible to rule out, so let’s try that…

Let n\geqslant7. Then if we have some m such that (n,m) satisfies the given equation, the first theorem tells us that m!>(4n)!, so, in particular, m>4n. Now, by the second theorem, there exists some prime p such that 2n<p<4n, but 4n<m, so 2n<p<m. This means that p|m!. But for this to be the case, it must be true that p divides one of the factors of m!=1!\times3!\times\ldots\times(2n-1)!, which can’t be true, since p>2n. So it isn’t possible to have such a pair (n,m) if n\geqslant7. We can pretty easily calculate that (1,1), (2,3), and (3,6) all satisfy the equation, and so we have only to check n\in\{4,5,6\}. Here we can either just write out a list of factorials up to around 14!, which isn’t too much of a hassle, and just check things by hand, or we can use some clever division arguments to find good bounds for m for each given n (for example, when n=5, the left-hand side of the equation has a factor of 5^3, but the smallest m such that m! has a factor of 5^3 is m=15, which is too large). Either way, we can see that the only other pair is (4,10).

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