How many pairs of perfect squares between and inclusive differ by a prime number?

To find how many pairs of perfect squares between $a$ and $b$ (inclusive) differ by a prime number, we can follow these steps:

1. Break Down the Problem

• Identify the range of perfect squares between $a$ and $b$.
• For each perfect square, find pairs with another perfect square that differ by a prime number.

2. Relevant Concepts

• A perfect square can be represented as $n^2$, where $n$ is a positive integer.
• The difference between two perfect squares, $n^2$ and $m^2$, can be expressed as: $|n^2 - m^2| = |(n - m)(n + m)|$
• The absolute difference must be a prime number.

3. Analysis and Detail

• Let $p = n^2 - m^2$ be a prime number.
• We need to check pairs $(n^2, m^2)$ such that $n^2 < m^2$ and: $p = m^2 - n^2$

4. Verify and Summarize

• Iterate through all integers $n$ and $m$ such that $n^2$ and $m^2$ are in the range from $a$ to $b$.
• For each pair, check if the difference $m^2 - n^2$ is a prime number.

Final Answer

The total count of pairs of perfect squares that differ by a prime number will depend on the specific values of $a$ and $b$. If you provide the specific range, I can calculate and find the number of such pairs.