Pairs of integers satisfying absolute inequalities

Squaring $|1+mn| < |m+n|$: $(1+mn)^2 < (m+n)^2 \implies 1 + 2mn + m^2n^2 < m^2 + 2mn + n^2$ $m^2n^2 - m^2 - n^2 + 1 < 0 \implies (m^2 - 1)(n^2 - 1) < 0$

For integers, this requires one of $m^2 - 1$ or $n^2 - 1$ to be negative (i.e. equal to $-1$, meaning $m=0$ or $n=0$).

If $m = 0$: $|1 + 0| < |0 + n| < 5 \implies 1 < |n| < 5$ $n$ can be $\{-4, -3, -2, 2, 3, 4\}$ (6 choices).

If $n = 0$: $1 < |m| < 5 \implies m \in \{-4, -3, -2, 2, 3, 4\}$ (6 choices).

Total pairs $= 6 + 6 = 12$.