+0

# I’m confused

0
421
4

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

#2
+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 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
+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)   Jun 20, 2019
#2
+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 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
+1

Thanks, heureka......I like that method  !!!!!   CPhill  Jun 21, 2019
#4
+2

Thank you, CPhill !

heureka  Jun 22, 2019