The number of distinct pairs of integers $(m,n)$ satisfying $|1+mn|<| m+n|<5$ is? $|1+mn|<5$
$\Rightarrow -6 < mn < 4$
$|m+n|<5$
$\Rightarrow -5<m+n<5$
Now as $m$ and $n$ are integers , I tried to make cases such the product of $mn$ is equal to $-5,-4,-3,-2,-1,0,1,2$ and $3$ as $ -6 < mn < 4$ but I end up getting too many pairs and also I am confused what to do in the case where $mn=0$.
 A: The official solution is right: There are 12 solutions:

*

*(-4, 0)

*(-3, 0)

*(-2, 0)

*(0, -4)

*(0, -3)

*(0, -2)

*(0, 2)

*(0, 3)

*(0, 4)

*(2, 0)

*(3, 0)

*(4, 0)

A: In what follows, $M$ and $N$ denote $|m|$ and $|n|$ respectively (otherwise we get an unreadable mess of vertical bars).

*

*If $m,n$ are both non-zero and have the same sign, then $|1+mn|=1+MN$ and $|m+n|=M+N$. So we have
$$1+MN < M+N$$
$$\therefore MN-M-N+1<0$$
$$\therefore (M-1)(N-1)<0$$
So either $M-1$ or $N-1$ must be $<0$, which is impossible if $M>0$ and $N>0$.

*If $m$ and $n$ are both non-zero and have different signs, we may suppose without loss of generality that $M>N$. So $|1+mn|=MN-1$, and $|m+n|=M-N$, and we have
$$MN-1<M-N$$
$$\therefore MN-M+N-1<0$$
$$\therefore (M+1)(N-1)<0$$
which is again impossible if $M>0$ and $N>0$.

Therefore one of $m,n$ must be zero!
A: Another way to get rid of the absolute values is to square:
$|1+mn|^2<|m+n|^2<5^2\implies (1+mn)^2<(m+n)^2<25$ $\implies$ $1<m^2+n^2-m^2n^2\implies0<(n^2-1)(1-m^2)$. Now if $n^2\ge 2$ then we should have $m=0\implies n^2<25\implies n=|2|,|3|,|4|$. Due to symmetricality we also have $n=0$ and $m=|2|,|3|,|4|$. Last we can check that these 12 pairs of integers are really solutions.
