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 $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ Here, $a=1$, $b=-6$, and $c=d$. 3. **Condition for integer roots:** For roots to be integers, the discriminant must be a perfect square: $$\Delta = b^2 - 4ac = (-6)^2 - 4 \times 1 \times d = 36 - 4d$$ must be a perfect square. 4. **Set the discriminant equal to a perfect square:** Let $k^2 = 36 - 4d$ where $k$ is an integer. 5. **Rewrite to find $d$:** $$d = \frac{36 - k^2}{4}$$ Since $d$ must be prime and an integer, $36 - k^2$ must be divisible by 4 and the result must be prime. 6. **Check possible values of $k$:** Since $k^2 \leq 36$, possible $k$ values are $0,1,2,3,4,5,6$. - 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) 7. **Check if roots are integers for $d=5$:** $$\Delta = 36 - 4 \times 5 = 36 - 20 = 16$$ Roots: $$x = \frac{6 \pm \sqrt{16}}{2} = \frac{6 \pm 4}{2}$$ So roots are: $$x_1 = \frac{6 + 4}{2} = 5, \quad x_2 = \frac{6 - 4}{2} = 1$$ Both are integers. **Final answer:** $$\boxed{5}$$