For today’s interesting problem, we are looking to find all positive integer pairs n and m which satisfy the equation above, when n and m are distinct. If we play around with small n and m we can quickly see that 2⁴ = 16 = 4² , so the pair 2 and 4 are certainly one solution. Actually, it is the only solution.

I noticed this interesting fact during my secondary school education. Additionally you can use this to prove that e is the point where a change in the exponent causes a larger result than a change in the base. E.g. (x+d)^y > x^(y+d) when y >= e for any real value d.

Only assuming y is also less than x.

Thanks for sharing. Are there any other posts on your blog you highly recommend?

1 and 1 are also a solution ?

Yeah OP updated the title of the article to include "distinct".

the original article says "distinct, positive." It was mis-transcribed to HN.

other way around - op updated title after submitting to hn

The problem requires that n and m are distinct integers.

Is this why ... 42?

1^2 = 1 * 1 = 1

2^1 = 2

I shouldn't comment when I'm half asleep...

1^2 != 2^1