Let S(n) denote the number of ordered pairs (x,y) satisyfing \frac{1}{x}+\frac{1}{y}=\frac{1}{n}
where n > 1,, and x,y,n \in N
(1) FInd the value of S(6)
(2)Show that if n is prime , then S(n) = 3 always
23My approach
\frac{1}{x}+\frac{1}{y}=\frac{k}{kn}=\frac{t+k-t}{kn}=\frac{t}{kn}+\frac{k-t}{kn}
where, k>t
so, x=\frac{kn}{t} \; \;\;\; and \;\;y = \frac{kn}{k-t}
and then putting n = 6 , and k=2,3,4,....... and putting t accordingly , we get suitable values of x,y .
Is this approach good enough to solve the problem ??
Pleasse giv general method for dis sum
1yes there is a general method
after some manipulation u will get
(x-n)(y-n)=n^2 \\ \\\texttt{the no.of ways of breaking n in two factors } \\ \\\texttt{prime factorise }6^2 =2^2.3^2 \\ \\\texttt{this leads to solving this quesion} \\ a+b=2 \\ c+d=2 \\ ^3C_1^2=9
a prime square will be of form p^2 can be resolved in two factors in exactly three ways
p,p
p^2,1
1,p^2