A class for rational number (p/q) with overloading + and << operator

I wanted to add two rational numbers and display them in the form of p/q using overloading the operators + and <<. I'm using friend function, because the function for addition and display are taking multiple and different types of parameters. Inside the addition function I'm performing a normal addition of fractions, like how we do in real life. But when i run the code i get an error that can't convert Rational to Rational(),
Error: Rational.cpp: In function 'int main()': Rational.cpp:51:15: error: assignment of function 'Rational R3()' R3 = R1 + R2; Rational.cpp:51:15: error: cannot convert 'Rational' to 'Rational()' in assignment*

i have no idea why it's saying that .... ??

C++

#include <iostream>
using namespace std;

class Rational
{
    private:
        int P;
        int Q;
    public:
        Rational(int p = 1, int q = 1)
        {
            P = p;
            Q = q;
        }
    
    friend Rational operator+(Rational r1, Rational r2);

    friend ostream & operator<<(ostream &out, Rational r3);
};

Rational operator+(Rational r1, Rational r2)
    {
        Rational temp;
        if(r1.Q == r2.Q)
        {
            temp.P = r1.P + r2.P;
            temp.Q = r1.Q;
        }
        else
        {
            temp.P = ((r1.P) * (r2.Q)) + ((r2.P) * (r1.Q));
            temp.Q = (r1.Q) * (r2.Q);
        }

        return temp;
    }

ostream & operator<<(ostream &out, Rational r3)
{
    out<<r3.P<<"/"<<r3.Q<<endl;
    return out;
}
int main()
{
    Rational R1(3,4);
    Rational R2(5,6);
    Rational R3();

    R3 = R1 + R2;
    cout<<R3;

}

This

Rational R3();

declares a function called R3 that returns a Rational and takes no parameters. It does not define R3 to be a default constructed Rational. Change the line to any of the below

 Rational R3; 
 Rational R3{};
 auto R3 = Rational();
 auto R3 = Rational{};