In this article, we do a slightly more recent problem. Still number theory, but this time involving factorials. A relatively more relaxed and simple problem compared to other problems posted, but may still require some thinking! Before looking at the solution, give the problem a go!

We first introduce some notation to make our lives easier. We say that a divides b if b/a is an integer and for this we write a | b.

Now, suppose that we have some divisor of k!, call it d. Then we have that since d | k!m!, then d | (k! + m!). However, we already know that d | k!, and therefore d|m! as well!

What does this mean for us? Well we can indeed set d = k! since we chose d arbitrarily. Therefore we have that k! | m!, meaning that we can write j(k!) = m! for some integer j since k! must be a multiple of m!. Rewriting, we yield:

Now, we can cancel out the m!s and we just get the following, simple equation:

Finally, rearranging we have that

The only way for this to happen is by having j = 1 and n! — 1 =1 In other words, we precisely have k = m and n! = 2 meaning n = 2. Therefore, our solutions are given by:

However, there is one key thing to notice. In fact, 0! = 1! and thus we also have solutions given by:

Note that the factorial function, even when analytically extended is NOT defined for negative integers, thus we did not need to worry about them in the first place.

There is one set of solutions we have missed, however. Note that we can also have (n!-2, n!-1, n) for n ≥ 3 since we can factor the left hand side and it works out nicely. The reader should ask themselves why there are no more solutions left.

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