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How many distinct ordered pairs of positive integers $(m,n)$ are there so that the sum of the reciprocals of $m$ and $n$ is $\frac14$?

 Jun 20, 2019

Best Answer 

 #2
avatar+25226 
+2

How many distinct ordered pairs of positive integers \((m,n)\) are there
so that the sum of the reciprocals of \(m\) and \(n\) is \(\dfrac14\)?

 

\(\text{From the relationship} \\ \dfrac{1}{z} = \dfrac{1}{m} + \dfrac{1}{n} \\ \text{follows immediately that $m>z$ and $n> z$ must be.}\\ \text{You can write $m=z+a$ and $n=z+b$ }\\ \text{Now the result:}\\ \dfrac{1}{z} = \dfrac{1}{z+a} + \dfrac{1}{z+b} \\ \)

 

\(\begin{array}{|rcll|} \hline \dfrac{1}{z} &=& \dfrac{1}{z+a} + \dfrac{1}{z+b} \\\\ \dfrac{1}{z} &=& \dfrac{2z+a+b}{z^2+za+zb+ab} \\\\ z^2+za+zb+ab &=& z(2z+a+b) \\ z^2+za+zb+ab &=& 2z^2+za+zb \\ z^2+za+zb+{\color{red}ab} &=& z^2+za+zb + {\color{red}z^2} \quad & \quad \text{by comparison follows } \boxed{z^2=ab} \\ \hline \end{array} \)

 

\(\text{Each pair $(a, b)=$ (divider, co-divider) of $n^2$ gives a solution }\\ \text{ from $\dfrac{1}{z} = \dfrac{1}{z+a} + \dfrac{1}{z+b} $.}\)

 

\(\text{if z = 4:}\\ \text{The divisors of $z^2=16$ are $1, 2, 4, 8, 16$ ($5$ divisors) }\)

 

\(\text{So there are $ \mathbf{5}$ distinct ordered pairs of positive integers $(m,n)$ }\)

 

\(\begin{array}{|c|c|c|c|c|} \hline 4^2 & divider & co-divider & \\ = 16 & a & b & ab & \dfrac{1}{4} = \dfrac{1}{4+a} + \dfrac{1}{4+b} \\ \hline & 1 & 16 & 1\cdot 16 = 16& \mathbf{\dfrac{1}{4} =} \dfrac{1}{4+1} + \dfrac{1}{4+16} = \mathbf{\dfrac{1}{5} + \dfrac{1}{20}} \\ \hline & 2 & 8 & 2\cdot 8 = 16& \mathbf{\dfrac{1}{4} =} \dfrac{1}{4+2} + \dfrac{1}{4+8}= \mathbf{\dfrac{1}{6} + \dfrac{1}{12}} \\ \hline & 4 & 4 & 4\cdot 4 = 16& \mathbf{\dfrac{1}{4} =} \dfrac{1}{4+4} + \dfrac{1}{4+4}= \mathbf{\dfrac{1}{8} + \dfrac{1}{8}} \\ \hline & 8 & 2 & 8\cdot 2 = 16& \mathbf{\dfrac{1}{4} =} \dfrac{1}{4+8} + \dfrac{1}{4+2}= \mathbf{\dfrac{1}{12} + \dfrac{1}{6}} \\ \hline & 16 & 1 & 16\cdot 1 = 16& \mathbf{\dfrac{1}{4} =} \dfrac{1}{4+16} + \dfrac{1}{4+1}= \mathbf{\dfrac{1}{20} + \dfrac{1}{5}} \\ \hline \end{array}\)

 

The distinct ordered pairs of positive integers \((m,n) = \sigma_0(z^2)\)

http://oeis.org/A000005

 

laugh

 Jun 21, 2019
edited by heureka  Jun 21, 2019
edited by heureka  Jun 21, 2019
edited by heureka  Jun 21, 2019
edited by heureka  Jun 21, 2019
 #1
avatar+111389 
+2

1/m  + 1/n  = 1/4

 

[ m + n]  / mn  = 1/4         which implies that

 

mn / [ m + n ]  = 4

 

mn  =  4 [ m + n]

 

mn  =  4m + 4n

 

mn - 4n  = 4m

 

n [ m - 4 ]  =  4m

 

n  =       4m

            _____

             m - 4

 

When

 

m = 5   n  =  20

m = 6   n  =  12

m = 8   n  = 8

m = 12  n  = 6

m = 20  n = 5

 

So   (m, n)  = (5, 20)  (6,12) (8, 8)  (12, 6)  and (20, 5)

 

 

cool cool cool

 Jun 20, 2019
 #2
avatar+25226 
+2
Best Answer

How many distinct ordered pairs of positive integers \((m,n)\) are there
so that the sum of the reciprocals of \(m\) and \(n\) is \(\dfrac14\)?

 

\(\text{From the relationship} \\ \dfrac{1}{z} = \dfrac{1}{m} + \dfrac{1}{n} \\ \text{follows immediately that $m>z$ and $n> z$ must be.}\\ \text{You can write $m=z+a$ and $n=z+b$ }\\ \text{Now the result:}\\ \dfrac{1}{z} = \dfrac{1}{z+a} + \dfrac{1}{z+b} \\ \)

 

\(\begin{array}{|rcll|} \hline \dfrac{1}{z} &=& \dfrac{1}{z+a} + \dfrac{1}{z+b} \\\\ \dfrac{1}{z} &=& \dfrac{2z+a+b}{z^2+za+zb+ab} \\\\ z^2+za+zb+ab &=& z(2z+a+b) \\ z^2+za+zb+ab &=& 2z^2+za+zb \\ z^2+za+zb+{\color{red}ab} &=& z^2+za+zb + {\color{red}z^2} \quad & \quad \text{by comparison follows } \boxed{z^2=ab} \\ \hline \end{array} \)

 

\(\text{Each pair $(a, b)=$ (divider, co-divider) of $n^2$ gives a solution }\\ \text{ from $\dfrac{1}{z} = \dfrac{1}{z+a} + \dfrac{1}{z+b} $.}\)

 

\(\text{if z = 4:}\\ \text{The divisors of $z^2=16$ are $1, 2, 4, 8, 16$ ($5$ divisors) }\)

 

\(\text{So there are $ \mathbf{5}$ distinct ordered pairs of positive integers $(m,n)$ }\)

 

\(\begin{array}{|c|c|c|c|c|} \hline 4^2 & divider & co-divider & \\ = 16 & a & b & ab & \dfrac{1}{4} = \dfrac{1}{4+a} + \dfrac{1}{4+b} \\ \hline & 1 & 16 & 1\cdot 16 = 16& \mathbf{\dfrac{1}{4} =} \dfrac{1}{4+1} + \dfrac{1}{4+16} = \mathbf{\dfrac{1}{5} + \dfrac{1}{20}} \\ \hline & 2 & 8 & 2\cdot 8 = 16& \mathbf{\dfrac{1}{4} =} \dfrac{1}{4+2} + \dfrac{1}{4+8}= \mathbf{\dfrac{1}{6} + \dfrac{1}{12}} \\ \hline & 4 & 4 & 4\cdot 4 = 16& \mathbf{\dfrac{1}{4} =} \dfrac{1}{4+4} + \dfrac{1}{4+4}= \mathbf{\dfrac{1}{8} + \dfrac{1}{8}} \\ \hline & 8 & 2 & 8\cdot 2 = 16& \mathbf{\dfrac{1}{4} =} \dfrac{1}{4+8} + \dfrac{1}{4+2}= \mathbf{\dfrac{1}{12} + \dfrac{1}{6}} \\ \hline & 16 & 1 & 16\cdot 1 = 16& \mathbf{\dfrac{1}{4} =} \dfrac{1}{4+16} + \dfrac{1}{4+1}= \mathbf{\dfrac{1}{20} + \dfrac{1}{5}} \\ \hline \end{array}\)

 

The distinct ordered pairs of positive integers \((m,n) = \sigma_0(z^2)\)

http://oeis.org/A000005

 

laugh

heureka Jun 21, 2019
edited by heureka  Jun 21, 2019
edited by heureka  Jun 21, 2019
edited by heureka  Jun 21, 2019
edited by heureka  Jun 21, 2019
 #3
avatar+111389 
+1

Thanks, heureka......I like that method  !!!!!

 

cool cool cool

CPhill  Jun 21, 2019
 #4
avatar+25226 
+2

Thank you, CPhill !

heureka  Jun 22, 2019

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