How many pairs $(m, n)$ of positive integers satisfy the equation $m^2 + 105 = n^2$?

We need to find how many pairs $(m, n)$ of positive integers satisfy $m^2 + 105 = n^2$.

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Starting with $m^2 + 105 = n^2$:

$n^2 - m^2 = 105$

The left side is a difference of squares, so we can factor:

$(n-m)(n+m) = 105$

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We need to find the prime factorization of 105:

$105 = 3 \times 5 \times 7$

When we have $105 = 3^1 \times 5^1 \times 7^1$, the total number of factors is:

$(1+1)(1+1)(1+1) = 8$ factors

The 8 factors of 105 are: 1, 3, 5, 7, 15, 21, 35, 105

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We need all ways to write 105 as a product of two positive integers.

Since $m$ and $n$ are positive integers:

$n + m > 0$ (always true for positive integers)

Then, $n - m > 0$ otherwise the product would become negative. (so $n > m$)

$n + m > n - m$ (since $m > 0$)

We need factor pairs $(a,b)$ where the product is 105:

$(1, 105)$

$(3, 35)$

$(5, 21)$

$(7, 15)$

Here, $a = n-m$ and $b = n+m$

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Now, we can find the exact value of $n$ and $m$

$a+b = n-m + n+m = 2n$

Therefore, $n = \dfrac{a+b}{2}$

$b-a = n+m - (n-m) = 2m$

Therefore, $m = \dfrac{b-a}{2}$

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Using the above equations, the pairs are: $(52, 53)$, $(16, 19)$, $(8, 13)$, and $(4, 11)$.

Hence, there are 4 pairs $(m, n)$ of positive integers that satisfy the equation.