2021 AMC Fall 10A Problem 20

1. When are there two different real solutions? The equation $x^2 + px + q = 0$ has two different real solutions exactly when the discriminant $p^2 - 4q$ is positive. If the discriminant is $0$, there is only one solution (a double root). If it is negative, there are none.
2. Write the conditions. We need neither equation to have two different solutions, so both discriminants must be $0$ or negative:
$$b^2 \le 4c \qquad \text{and} \qquad c^2 \le 4b.$$
3. Bound the numbers. Suppose $b$ is the larger of the two (or they are equal). Then $b^2 \le 4c \le 4b$, so $b^2 \le 4b$, which means $b \le 4$. The same works if $c$ is larger. So both $b$ and $c$ are between $1$ and $4$.
4. Check the pairs. For each $b$, the first condition says $c \ge \frac{b^2}{4}$, and the second says $c^2 \le 4b$:
- $b = 1$: $c \ge \frac14$ and $c^2 \le 4$. So $c = 1$ or $2$.
- $b = 2$: $c \ge 1$ and $c^2 \le 8$. So $c = 1$ or $2$.
- $b = 3$: $c \ge \frac94$ and $c^2 \le 12$. So $c = 3$.
- $b = 4$: $c \ge 4$ and $c^2 \le 16$. So $c = 4$.
5. Count. The pairs are $(1,1)$, $(1,2)$, $(2,1)$, $(2,2)$, $(3,3)$, $(4,4)$: six pairs.

The answer is $\boxed{\textbf{(B)}\ 6}$.

"Two distinct real solutions" is a statement about the discriminant, so the problem turns into two inequalities. Each one limits one number by the other, and using them together gives a hard limit of $4$. After that, only $16$ pairs are possible and a quick check finishes it. Read the words carefully: "two distinct" means a double root is allowed.