[Someone]

> The number you are looking for is e^(sqrt(163) pi) […] It is not an integer of course.

Of course? I’m not aware that we have some theorem other than “we computed it to lots of decimals, and it isn’t an integer” from which that follows.

[less_less]

It's not really "of course", and I don't think we have such a theorem in general. But in this case, I believe the fact that it's not an integer follows from the same theorem that says it's very close to an integer. See eg https://math.stackexchange.com/questions/4544/why-is-e-pi-sq...

Basically e^(sqrt(163)*pi) is the leading term in a Laurent series for an integer, and the other (non-integer) terms are really small but not zero.