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 $b-1=1$: $$b-1=1⟹b=2.$$ For $b=2$: $$b^2-1=2^2-1=4-1=3,$$ which is prime.

2. Case $b+1=1$: $$b+1=1⟹b=0.$$ 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: $$b-1=2k-1\text{(odd)},b+1=2k+1\text{(odd)}.$$ 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: $$\boxed{{\displaystyle 1}}.$$