Quick A: if parameter are not known at compile time, it is like a normal function
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The quoted wording is a little misleading in a sense. If you just take PlusOne in isolation, and observe its logic, and assume that the inputs are known at compile-time, then the calculations therein can also be performed at compile-time. Slapping the constexpr keyword on it ensures that we maintain this lovely state and everything's fine.
But if the input isn't known at compile-time then it's still just a normal function and will be called at runtime.
So the constexpr is a property of the function ("possible to evaluate at compile time" for some input, not for all input) not of your function/input combination in this specific case (so not for this particular input either).
It's a bit like how a function could take a const int& but that doesn't mean the original object had to be const. Here, similarly, constexpr adds constraints onto the function, without adding constraints onto the function's input.
Admittedly it's all a giant, confusing, nebulous mess (C++! Yay!). Just remember, your code describes the meaning of a program! It's not a direct recipe for machine instructions at different phases of compilation.
(To really enforce this you'd have the integer be a template argument.)