How many distinct ordered pairs of positive integers $(m,n)$ are there so that the sum of the reciprocals of $m$ and $n$ is $\frac14$?
+26367 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 \((m,n) = \sigma_0(z^2)\)
+128406 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)
+26367 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 \((m,n) = \sigma_0(z^2)\)
