Question

Find the number of bases b \ge 2 such that 100_b - 1_b is prime.

Answer 1

To determine the number of bases \( b \ge 2 \) such that \( 100_b - 1_b \) is prime, we first convert these base \( b \) numbers into decimal.

The number \( 100_b \) in decimal is calculated as:
\[
100_b = 1 \cdot b^2 + 0 \cdot b^1 + 0 \cdot b^0 = b^2.
\]
The number \( 1_b \) in decimal is:
\[
1_b = 1.
\]
Therefore, we have:
\[
100_b - 1_b = b^2 - 1.
\]

Next, we can factor the expression \( b^2 - 1 \):
\[
b^2 - 1 = (b - 1)(b + 1).
\]
For \( b^2 - 1 \) to be prime, one of the factors \( b - 1 \) or \( b + 1 \) must equal 1, as the product of two numbers is prime only if one is 1 and the other is a prime number.

1. **Case \( b - 1 = 1 \)**:
\[
b - 1 = 1 \implies b = 2.
\]
For \( b = 2 \):
\[
b^2 - 1 = 2^2 - 1 = 4 - 1 = 3,
\]
which is prime.

2. **Case \( b + 1 = 1 \)**:
\[
b + 1 = 1 \implies b = 0.
\]
However, this does not meet the requirement \( b \ge 2 \).

Thus, the only valid case that makes \( b^2 - 1 \) prime occurs at \( b = 2 \).

Now, let’s analyze if there are other bases \( b > 2 \). For \( b \ge 3 \):
- Both \( b - 1 \) and \( b + 1 \) are greater than 1.
- When \( b \) is odd, both \( b - 1 \) and \( b + 1 \) are even, hence their product \( (b-1)(b+1) \) cannot be prime.
- When \( b \) is even:
- If \( b = 2k \) for \( k \ge 2 \), then:
\[
b-1 = 2k - 1 \quad \text{(odd)}, \quad b+1 = 2k + 1 \quad \text{(odd)}.
\]
The product \( (b-1)(b+1) \) gives an even number greater than 2, thus not prime.

Therefore, no additional bases greater than or equal to \( 3 \) can make \( b^2 - 1 \) prime.

Finally, the only base \( b \) such that \( 100_b - 1_b \) is prime is \( b = 2 \).

Thus, the total number of bases \( b \ge 2 \) such that \( 100_b - 1_b \) is prime is:
\[
\boxed{1}.
\]