The number of distinct pairs of integers $(m, n)$ satisfying $|1 + mn| < |m + n| < 5$ is

We need to find pairs of integers $(m,n)$ where $|1 + mn| < |m + n| < 5$.

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We have two conditions to satisfy:

1. $|m + n| < 5$
2. $|1 + mn| < |m + n|$

From condition 1: $|m + n| < 5$ means $-5 < m + n < 5$

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For condition 2, we'll square both sides since both sides are positive:

$(1 + mn)^2 < (m + n)^2$

Expanding: $1 + 2mn + m^2n^2 < m^2 + 2mn + n^2$

Simplifying: $1 + m^2n^2 < m^2 + n^2$

This gives us: $m^2n^2 - m^2 - n^2 + 1 < 0$

Factoring: $(m^2 - 1)(n^2 - 1) < 0$

Therefore: $(m - 1)(m + 1)(n - 1)(n + 1) < 0$

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For the product $(m - 1)(m + 1)(n - 1)(n + 1) < 0$, we need an odd number of negative factors.

Since $(m + 1) > (m - 1)$ and $(n + 1) > (n - 1)$, the key cases are:

$|m| < 1$ and $|n| > 1$, OR

$|m| > 1$ and $|n| < 1$

Since we need integers, this means:

$m = 0$ and $|n| \geq 2$, OR

$n = 0$ and $|m| \geq 2$

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Case 1: $m = 0$

Constraint becomes $|1| < |n| < 5$, so $1 < |n| < 5$

This gives $n \in \{-4, -3, -2, 2, 3, 4\}$

Valid pairs: $(0, -4), (0, -3), (0, -2), (0, 2), (0, 3), (0, 4)$

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Case 2: $n = 0$

Constraint becomes $|1| < |m| < 5$, so $1 < |m| < 5$

This gives $m \in \{-4, -3, -2, 2, 3, 4\}$

Valid pairs: $(-4, 0), (-3, 0), (-2, 0), (2, 0), (3, 0), (4, 0)$

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Total number of distinct pairs = $6 + 6 = 12$

The key insight is that we need exactly one of $m$ or $n$ to be zero, with the other having absolute value between 2 and 4 (inclusive).