> but it does guarantee that it could be.

in which sense do you mean this? consider for instance

    constexpr int foo(int x) 
    { if(x < 0) throw "error"; return x; }

I would say that "guarantee that it could be" means that as long as the function definition itself "builds" / is validated by the compiler, you can use it in a constexpr context yet this is not the case here:

    constexpr int a = foo(123); // works fine
    constexpr int b = foo(-123); // compile error
    int c = foo(-123); // works fine

so having "constexpr" in the API does not mean that your code will always build ; as soon as you have constexpr the entire implementation of the function is part of the API, thus making the keyword moot like gpderetta says

That's what everybody originally believes (me included)!

   constexpr is_right_answer(int x){
      if (x==42) return true;
      else {
        std::cout <<"try again \n";
        return false;
  }

A function can be constexpr as long as at least one possible invocation can be compile time evaluated. In fact compilers can't generally prove that a function can't be constexpr.

I believe consteval has the semantics you want but I've yet to study it in details.

For more details see: https://www.foonathan.net/2020/10/constexpr-platform/ .