Drop the base

$a^2 - b^2 = 5^{10101}$

$(a+b)(a-b) = 5^{10101}$

Now, we have to consider which pairs of factors of $5^{10101}$ are able to equal $(a+b)(a-b)$ . For example, we can have:

$5^{10101} \times 5^{0}$ ,

$5^{10100} \times 5^{1}$ ,

$5^{10099} \times 5^{2}$ ,

$...$

$5^{5052} \times 5^{5049}$ ,

$5^{5051} \times 5^{5050}$ .

In total, there are $5051$ possible pairs of factors. Since $a$ and $b$ are positive, $(a+b)$ must be the larger factor. In all of these cases, due to the parity of $a$ and $b$ , $(a+b)(a-b)$ can be made, meaning that there are $\boxed{5051}$ distinct pairs of $(a,b)$ .

If we generalise for $5^n$ , since $5$ is prime, there will be a determinable number of pairs of factors. When $n$ is odd, there are $\frac{n+1}{2}$ possible pairs. However, if $n$ is even, e.g. $5^{2k}$ , we get a pair of factors of $5^k \times 5^k$ . This would make $b = 0$ which is impossible, therefore this pair must be discounted. Therefore, an even $n$ has the same result as the odd number $n-1$ . In general, we can write this out in a couple of ways:

$\frac{n + (n\mod{2})}{2}$

or

$\lceil \frac{n}{2} \rceil$