# Math Puzzle

#### eddybob123

##### Active member
1) Find all solutions (a,b) to the equation

$$\displaystyle \frac{1}{a}+\frac{1}{b}=\frac{1}{8}$$.

2) Generalize to

$$\displaystyle \frac{1}{a}+\frac{1}{b}=\frac{1}{n}$$.

If no one gets the answer in long time I'll post hints.

#### agentmulder

##### Active member
Hi there eddybob123 , welcome to MHB !

-agentredlum

#### caffeinemachine

##### Well-known member
MHB Math Scholar
1) Find all solutions (a,b) to the equation

$$\displaystyle \frac{1}{a}+\frac{1}{b}=\frac{1}{8}$$.

2) Generalize to

$$\displaystyle \frac{1}{a}+\frac{1}{b}=\frac{1}{n}$$.

If no one gets the answer in long time I'll post hints.
For the generalized part put $a=n-x$ and $b=n-y$. One gets $n^2=xy$.

#### Albert

##### Well-known member
1) Find all solutions (a,b) to the equation

$$\displaystyle \frac{1}{a}+\frac{1}{b}=\frac{1}{8}$$.

2) Generalize to

$$\displaystyle \frac{1}{a}+\frac{1}{b}=\frac{1}{n}$$.

If no one gets the answer in long time I'll post hints.
$$\displaystyle \frac{1}{a}+\frac{1}{b}=\frac{1}{n}$$.
$\dfrac{1}{n}=\dfrac {k}{nk}=\dfrac {1+(k-1)}{nk}=\dfrac {1}{nk}+\dfrac {k-1}{nk}$ (k=2,3,----)
here nk must be a multiple of (k-1)
take n=8
$\dfrac {1}{8}=\dfrac {1}{16}+\dfrac{1}{16}=\dfrac {1}{24}+\dfrac{1}{12}= \dfrac {1}{40}+\dfrac{1}{10}=\dfrac {1}{72}+\dfrac{1}{9}$
I found four pairs of (a,b)

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#### Albert

##### Well-known member
$n=xy ,\,\, x\geq 1,\,\, y\geq 1$
$\dfrac {1}{n}=\dfrac {k}{nk}=\dfrac {1+(k-1)}{nk}=\dfrac{1}{nk}+\dfrac{k-1}{nk}$
here nk mod (k-1)=0-----(1)
$k=x,y,(x+y),---$
$if \,\, n=8=1\times 8=2\times 4$
$\therefore k=2,4,(2+4),(1+8),(2+4+1+8)----(2)$
compare (1) and (2) and choose suitable k,and get corresponding a and b

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