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:
\[ \boxed{0} \]