### Instruction: How many pairs of positive integers $(m,n)$ satisfy $m^2 + n < 22?$.### Response: I want to find all the pairs of positive integers that satisfy the given inequality.
One way to approach this problem is to try different values of $m$ and see how they affect the possible values of $n$.
If $m = 1$, then $m^2 = 1$ and $n < 22 - 1 = 21$.
Since $n$ has to be a positive integer, the possible values of $n$ are $1, 2, 3, \dots, 20$.
That's 20 pairs of the form $(1,n)$ that work.
If $m = 2$, then $m^2 = 4$ and $n < 22 - 4 = 18$.
The possible values of $n$ are $1, 2, 3, \dots, 17$.
That's 17 pairs of the form $(2,n)$ that work.
If $m = 3$, then $m^2 = 9$ and $n < 22 - 9 = 13$.
The possible values of $n$ are $1, 2, 3, \dots, 12$.
That's 12 pairs of the form $(3,n)$ that work.
If $m = 4$, then $m^2 = 16$ and $n < 22 - 16 = 6$.
The possible values of $n$ are $1, 2, 3, 4, 5$.
That's 5 pairs of the form $(4,n)$ that work.
If $m = 5$, then $m^2 = 25$ and $n < 22 - 25 = -3$.
There are no positive integers that are less than -3, so there are no pairs of the form $(5,n)$ that work.
Similarly, if $m > 5$, then $m^2 > 25$ and $n < 22 - m^2 < 0$.
So there are no more pairs of positive integers that satisfy the inequality.
Therefore, the total number of pairs is $20 + 17 + 12 + 5 = 54$.