1. **State the problem:** We need to find the prime number $d$ such that the quadratic equation $$x^2 - 6x + d = 0$$ has two integer solutions. 2. **Recall the quadratic formula:** The roots of $$ax^2 + bx + c = 0$$ are given by $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$. 3. **Apply to our equation:** Here, $a=1$, $b=-6$, and $c=d$. The roots are $$x = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot d}}{2} = \frac{6 \pm \sqrt{36 - 4d}}{2}.$$ 4. **Integer roots condition:** For roots to be integers, the discriminant $$\Delta = 36 - 4d$$ must be a perfect square, say $k^2$, and the numerator must be divisible by 2. 5. **Set discriminant as perfect square:** $$36 - 4d = k^2$$ Rearranged: $$4d = 36 - k^2$$ $$d = \frac{36 - k^2}{4}$$ 6. **Since $d$ is prime and integer, $d$ must be positive and $\frac{36 - k^2}{4}$ must be prime.** Also, $k^2$ must be less than or equal to 36. 7. **Check possible $k$ values:** $k$ can be 0,1,2,3,4,5,6. - For $k=0$: $d = \frac{36 - 0}{4} = 9$ (not prime) - For $k=1$: $d = \frac{36 - 1}{4} = \frac{35}{4} = 8.75$ (not integer) - For $k=2$: $d = \frac{36 - 4}{4} = \frac{32}{4} = 8$ (not prime) - For $k=3$: $d = \frac{36 - 9}{4} = \frac{27}{4} = 6.75$ (not integer) - For $k=4$: $d = \frac{36 - 16}{4} = \frac{20}{4} = 5$ (prime) - For $k=5$: $d = \frac{36 - 25}{4} = \frac{11}{4} = 2.75$ (not integer) - For $k=6$: $d = \frac{36 - 36}{4} = 0$ (not prime) 8. **Check if roots are integers for $d=5$:** $$x = \frac{6 \pm \sqrt{36 - 4 \cdot 5}}{2} = \frac{6 \pm \sqrt{36 - 20}}{2} = \frac{6 \pm \sqrt{16}}{2} = \frac{6 \pm 4}{2}.$$ Roots: $$x_1 = \frac{6 + 4}{2} = 5, \quad x_2 = \frac{6 - 4}{2} = 1,$$ which are both integers. **Final answer:** $$\boxed{5}$$