#1**+1 **

The first step to solving this problem is to figure out sample values of m. \(m\) can be 1, 2, 3, or 4. 5 would be too large as 5 squared is 25 and 22-25 is a negative number, it says ONLY positive integers. 0 isn't a positive number. So there are more than 4 pairs of positive integers (m, n) that satisfy the equation \(m^2 +n <22\). Then figure out n for each m.

Hypotenuisance Feb 16, 2020

#4**+1 **

How many pairs of positive integers (m,n) satisfy \(m^2 + n < 22\) ?

**Hello Guest!**

\(\{m,n\} \subset\mathbb N\\ \mathbb L=m^2+n\\ \mathbb L\in \mathbb \{\mathbb R<22\}\)

\(n=1\to m\in \{1,2,3,4 \}\\ n=2\to m\in \{1,2,3,4 \}\\ n=3\to m\in \{1,2,3,4 \}\\ n=4\to m\in \{1,2,3,4 \}\\ n=5\to m\in \{1,2,3,4 \}\)

\(n=6\to m\in \{1,2,3 \}\\ n=7\to m\in \{1,2,3 \}\\ n=8\to m\in \{1,2,3 \}\\ n=9\to m\in \{1,2,3 \}\\ n=10\to m\in \{1,2,3 \}\\ n=11\to m\in \{1,2,3\}\)

\(n=12\to m\in \{1,2,3 \}\\ n=13\to m\in \{1,2 \}\\ n=14\to m\in \{1,2 \}\\ n=15\to m\in \{1,2 \}\\ n=16\to m\in \{1,2 \} \)

\(n=17\to m\in \{1,2 \}\\ n=18\to m=1\\ n=19\to m=1 \\ n=20\to m=1\\ \)

53 pairs of positive integers (m,n) satisfy m2 + n < 22

** !**

asinus Feb 16, 2020