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

 Feb 16, 2020
 #1
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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.

 Feb 16, 2020
edited by Hypotenuisance  Feb 16, 2020
edited by Hypotenuisance  Feb 16, 2020
 #3
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Thank you Hypotenuisance! That was the correct answer 😊

Guest Feb 16, 2020
 #5
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@Guest Sure, no problem!laugh

Hypotenuisance  Feb 16, 2020
 #2
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If you want the answer, I believe it is 54 pairs.

 Feb 16, 2020
 #4
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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

laugh  !

 Feb 16, 2020
edited by asinus  Feb 16, 2020
edited by asinus  Feb 17, 2020

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