Question

Find the number of bases b \ge 2 such that 144_b is prime.

Answer 1

To find the number of bases \( b \ge 2 \) such that \( 144_b \) is prime, we first need to convert \( 144_b \) from base \( b \) to base 10. The representation \( 144_b \) can be expanded as follows:

\[
144_b = 1 \cdot b^2 + 4 \cdot b^1 + 4 \cdot b^0 = b^2 + 4b + 4
\]

We can simplify this expression:

\[
144_b = b^2 + 4b + 4 = (b + 2)^2
\]

Now, we need to analyze whether \( (b + 2)^2 \) can be prime. A prime number is defined as a natural number greater than 1 that has no positive divisors other than 1 and itself.

Since \( (b + 2)^2 \) represents a square of an integer, it is equal to the product of \( (b + 2) \) by itself. The only situation in which a square \( n^2 \) can be prime is when \( n = 1 \) because:

\[
n^2 = p \implies n = p \text{ (which is prime) or } n = 1 \text{ (thus } p = 1)
\]

Here, we have:

\[
(b + 2)^2 = p
\]

If \( b + 2 \) equals 1, then:

\[
b + 2 = 1 \implies b = -1
\]

However, this is not valid as \( b \) must be at least 2.

Since \( b + 2 \) being any integer greater than 1 results in \( (b + 2)^2 \) being composite (as it has at least the factors \( 1, b + 2, (b + 2)^2 \)), we conclude that there are no base values \( b \geq 2 \) such that \( 144_b \) can be prime.

Thus, the number of bases \( b \ge 2 \) such that \( 144_b \) is prime is:

\[
\boxed{0}
\]