Returning const value from arithmetic operator overload with move assignment

Let's say I have the following minimal example class:

#include <iostream>

class Foo {
public:
    Foo() = default;

    Foo(const Foo&) = default;
    Foo(Foo&&) noexcept = default;

    Foo& operator=(const Foo& rhs) {
        std::cout << "copy\n";

        return *this;
    }

    Foo& operator=(Foo&& rhs) noexcept {
        std::cout << "move\n";

        return *this;
    }

    Foo operator+(const Foo& rhs) const {
        Foo x; // with some calculation

        return x;
    }
};

int main() {
    Foo a, b, c;

    a = b + c;
}

This prints move as expected. Now according to Effective C++ Item 3, I should return const Foo from operator+ to avoid construct like a + b = c, i.e.:

// To avoid a + b = c
const Foo operator+(const Foo& rhs) const {}

Unfortunately, this suddenly starts calling copy assignment instead of move assignment operator. [I'm using gcc 4.8.4 on Ubuntu, but it is probably nothing related to compiler]

How can I ensure that a + b = c fails to compile and in the same time move assignment is called for a = b + c? Or with the introduction of move semantics, is there no way to achieve both of them in the same time?