+31 How many pairs of positive integers (a,b) satisfy {1}/{a} + {1}/{b}={2}/{17}?
+129850 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 )
+26388 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.
\(\text{The prime factorization of $n^2$ results in all possible decompositions into two unit fractions.} \)
\(n=17\ \text{is odd} \\ n^2 =17*17=17^2*1 \\ \text{All divisors of $n^2=289$ are: $1,\ 17,\ 17^2=289$}\\ \text{Let $n^2=p\times q$ }\)
\(\begin{array}{|r|r|r|c|c|c|c|c|c|c|c| } \hline & p & q & n^2 & s & t & r & k & a & b & \\ & & & =p\cdot q & = \frac{p+q}{2} & = \frac{p-q}{2} & =\frac{t}{2} & = \frac{15+\sqrt{15^2+t^2} }{2} & =k-r & =k+r & \\ \hline 1. & 17^2 & 1 & 289=289 \cdot 1 & 145 & 144 & 72 & 81 & 9 & 153 & \mathbf{\dfrac{2}{17} = \dfrac{1}{9} + \dfrac{1}{153}} \\ \hline 2. & 1 & 17^2 & 289= 1 \cdot 289 & 145 & -144 & -72 & 81 & 153 & 9 & \mathbf{\dfrac{2}{17} = \dfrac{1}{153} + \dfrac{1}{9}} \\ \hline 3. & 17 & 17 & 289= 17 \cdot 17 & 17 & 0 & 0 & 17 & 17 & 17 & \mathbf{\dfrac{2}{17} = \dfrac{1}{17} + \dfrac{1}{17}} \\ \hline \end{array}\)
