1. **State the problem:** Find all functions $f:\mathbb{R} \to \mathbb{R}$ such that for every $x,y \in \mathbb{R}$, the equation $$1 + f(xy) = f(x + f(y)) + (y - 1)f(x - 1)$$ holds. 2. **Analyze the functional equation:** We want to find $f$ satisfying $$1 + f(xy) = f(x + f(y)) + (y - 1)f(x - 1).$$ 3. **Step 1: Substitute $y=1$** to simplify: $$1 + f(x \cdot 1) = f(x + f(1)) + (1 - 1)f(x - 1) \implies 1 + f(x) = f(x + f(1)) + 0,$$ so $$f(x + f(1)) = 1 + f(x).$$ 4. **Step 2: Define $c = f(1)$**. Then $$f(x + c) = 1 + f(x).$$ This means $f$ shifts by $c$ in the input to increase by 1 in the output. 5. **Step 3: Substitute $x=1$** in the original equation: $$1 + f(1 \cdot y) = f(1 + f(y)) + (y - 1)f(0) \implies 1 + f(y) = f(1 + f(y)) + (y - 1)f(0).$$ 6. **Step 4: Use the shift property from Step 4:** $$f(1 + f(y)) = f(f(y) + 1) = 1 + f(f(y))$$ so the equation becomes $$1 + f(y) = 1 + f(f(y)) + (y - 1)f(0) \implies f(y) = f(f(y)) + (y - 1)f(0).$$ 7. **Step 5: Substitute $y=0$** in the above: $$f(0) = f(f(0)) + (0 - 1)f(0) = f(f(0)) - f(0) \implies f(f(0)) = 2f(0).$$ 8. **Step 6: Use the shift property again:** From Step 4, $f(x + c) = 1 + f(x)$, so applying it to $x = f(0)$: $$f(f(0) + c) = 1 + f(f(0)).$$ But from Step 7, $f(f(0)) = 2f(0)$, so $$f(f(0) + c) = 1 + 2f(0).$$ 9. **Step 7: Try to find $f(0)$ and $c$:** Substitute $x=0$ in the original equation: $$1 + f(0 \cdot y) = f(0 + f(y)) + (y - 1)f(-1) \implies 1 + f(0) = f(f(y)) + (y - 1)f(-1).$$ 10. **Step 8: From Step 6, $f(y) = f(f(y)) + (y - 1)f(0)$, so $$f(f(y)) = f(y) - (y - 1)f(0).$$ Substitute into Step 7: $$1 + f(0) = f(y) - (y - 1)f(0) + (y - 1)f(-1) = f(y) + (y - 1)(f(-1) - f(0)).$$ 11. **Step 9: Rearrange for $f(y)$:** $$f(y) = 1 + f(0) - (y - 1)(f(-1) - f(0)) = 1 + f(0) - (y - 1)f(-1) + (y - 1)f(0).$$ 12. **Step 10: Simplify:** $$f(y) = 1 + f(0) + (y - 1)(f(0) - f(-1)) = 1 + f(0) + y(f(0) - f(-1)) - (f(0) - f(-1)).$$ 13. **Step 11: Combine constants:** $$f(y) = y(f(0) - f(-1)) + 1 + f(0) - f(0) + f(-1) = y(f(0) - f(-1)) + 1 + f(-1).$$ 14. **Step 12: Let $A = f(0) - f(-1)$ and $B = 1 + f(-1)$, so $$f(y) = A y + B,$$ a linear function. 15. **Step 13: Verify linear form in original equation:** Assume $f(x) = A x + B$. Left side: $$1 + f(xy) = 1 + A x y + B.$$ Right side: $$f(x + f(y)) + (y - 1)f(x - 1) = f(x + A y + B) + (y - 1)f(x - 1) = A(x + A y + B) + B + (y - 1)(A(x - 1) + B).$$ Simplify right side: $$A x + A^2 y + A B + B + (y - 1)(A x - A + B) = A x + A^2 y + A B + B + (y - 1)A x - (y - 1)A + (y - 1)B.$$ Group terms: $$A x + (y - 1) A x + A^2 y + A B + B - (y - 1) A + (y - 1) B = A x y + A^2 y + A B + B - A y + A + B y - B.$$ Simplify constants: $$A x y + A^2 y + A B + B - A y + A + B y - B = A x y + y(A^2 - A + B) + (A + A B).$$ 16. **Step 14: Equate left and right sides:** Left side: $1 + A x y + B$ Right side: $A x y + y(A^2 - A + B) + (A + A B)$ Equate: $$1 + A x y + B = A x y + y(A^2 - A + B) + A + A B.$$ Cancel $A x y$ on both sides: $$1 + B = y(A^2 - A + B) + A + A B.$$ 17. **Step 15: Since this holds for all $y$, the coefficient of $y$ must be zero:** $$A^2 - A + B = 0,$$ and the constant terms must satisfy $$1 + B = A + A B.$$ 18. **Step 16: Solve the system:** From $A^2 - A + B = 0$, we get $$B = A - A^2.$$ Substitute into the second equation: $$1 + (A - A^2) = A + A (A - A^2) \\ 1 + A - A^2 = A + A^2 - A^3.$$ Simplify: $$1 + A - A^2 - A - A^2 + A^3 = 0 \\ 1 - 2 A^2 + A^3 = 0.$$ 19. **Step 17: Rearrange:** $$A^3 - 2 A^2 + 1 = 0.$$ Try rational roots: $A=1$: $$1 - 2 + 1 = 0,$$ so $A=1$ is a root. 20. **Step 18: Factor polynomial:** $$(A - 1)(A^2 - A - 1) = 0.$$ The quadratic roots are $$A = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2}.$$ 21. **Step 19: Find corresponding $B$ for each $A$:** - For $A=1$, $$B = 1 - 1^2 = 0.$$ - For $A = \frac{1 + \sqrt{5}}{2}$, $$B = A - A^2 = \frac{1 + \sqrt{5}}{2} - \left(\frac{1 + \sqrt{5}}{2}\right)^2.$$ - For $A = \frac{1 - \sqrt{5}}{2}$, $$B = A - A^2 = \frac{1 - \sqrt{5}}{2} - \left(\frac{1 - \sqrt{5}}{2}\right)^2.$$ 22. **Step 20: Final solutions:** $$f(x) = x,$$ $$f(x) = \frac{1 + \sqrt{5}}{2} x + \frac{1 + \sqrt{5}}{2} - \left(\frac{1 + \sqrt{5}}{2}\right)^2,$$ $$f(x) = \frac{1 - \sqrt{5}}{2} x + \frac{1 - \sqrt{5}}{2} - \left(\frac{1 - \sqrt{5}}{2}\right)^2.$$