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How many pairs of positive integers (a,b) satisfy {1}/{a} + {1}/{b}={2}/{17}?

 May 22, 2019
 #1
avatar+102417 
+3

1/a + 1/b  =  2/17

 

One obvious solution is    a  = 17    and b = 17

 

Another can be found using the "Egyptian Fraction" method

 

(1) Take the ceiling of  [ 17/2]  = ceiling [ 8.5]  =   9

(2)  Take the reciprocal of this   =  1/9

(3)  Subtract this fraction from 2/17

 

2/17  - 1/9   =

 

18/153 -  17/153  =  1/153

 

So....two other possibilities  are  either

 

1/9 + 1/153    or    1/153  + 1/9

 

So....the possible positive pairs of (a, b)  =

 

(17,17)

(9, 153)  and

(153, 9 )

 

cool cool cool

 May 22, 2019
edited by CPhill  May 22, 2019
edited by CPhill  May 22, 2019
 #2
avatar+22896 
+2

How many pairs of positive integers (a,b) satisfy \(\dfrac{ {1}}{{a}} + \dfrac{{1}}{{b}}=\dfrac{{2}}{{17}}\) ?

 

\(\text{For odd $n>2$ there is always at least one decomposition into exactly two unit fractions: $\dfrac{2}{n} = \dfrac{1}{a} + \dfrac{1}{b}$ }\)

Finding all possibilities.

\(\text{The prime factorization of $n^2$ results in all possible decompositions into two unit fractions.} \)

 

\(n=17\ \text{is odd} \\ n^2 =17*17=17^2*1 \\ \text{All divisors of $n^2=289$ are: $1,\ 17,\ 17^2=289$}\\ \text{Let $n^2=p\times q$ }\)

 

\(\begin{array}{|r|r|r|c|c|c|c|c|c|c|c| } \hline & p & q & n^2 & s & t & r & k & a & b & \\ & & & =p\cdot q & = \frac{p+q}{2} & = \frac{p-q}{2} & =\frac{t}{2} & = \frac{15+\sqrt{15^2+t^2} }{2} & =k-r & =k+r & \\ \hline 1. & 17^2 & 1 & 289=289 \cdot 1 & 145 & 144 & 72 & 81 & 9 & 153 & \mathbf{\dfrac{2}{17} = \dfrac{1}{9} + \dfrac{1}{153}} \\ \hline 2. & 1 & 17^2 & 289= 1 \cdot 289 & 145 & -144 & -72 & 81 & 153 & 9 & \mathbf{\dfrac{2}{17} = \dfrac{1}{153} + \dfrac{1}{9}} \\ \hline 3. & 17 & 17 & 289= 17 \cdot 17 & 17 & 0 & 0 & 17 & 17 & 17 & \mathbf{\dfrac{2}{17} = \dfrac{1}{17} + \dfrac{1}{17}} \\ \hline \end{array}\)

 

laugh

 May 22, 2019
edited by heureka  May 22, 2019

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