Use a simpler formula for Kyber division by q

[randombit]

Simpler approach than #3898

[coveralls]

coverage: 92.083% (-0.005%) from 92.088%
when pulling 8352a22 on jack/simpler-kyber-div-q
into 5f47e7b on master.

[FAlbertDev]

I'm okay with using this more straightforward approach. We can stay with this simple method until we need other CT divisions in the future.
Since we already have a test for our ct_div function in #3898, we may also want to apply it here.

// src/tests/test_kyber.cpp
namespace {

std::vector<Test::Result> test_kyber_utilities() {
   return {
      Botan_Tests::CHECK("constant-time division",
                         [](Test::Result& result) {
                            constexpr auto Q = Botan::KyberConstants::Q;
                            // Check ct_int_div_kyber_q with all possible inputs that may appear in Kyber.
                            for(uint32_t i = 0; i < (1 << 18); ++i) {
                               result.test_is_eq(
                                  Botan::fmt("{}/Q", i), Botan::ct_int_div_kyber_q(i), i / Q);
                            }
                         }),
   };
}

}  // namespace

BOTAN_REGISTER_TEST_FN("kyber", "kyber_utils", test_kyber_utilities);

[randombit]

The function isn't currently exposed anywhere and I'm inclined not to expose it simply to test it, considering it is quite small and should already be very well tested as a side effect of the Kyber tests themselves. I did do an offline test that the expression works for all integers < 2**20.