The Zen Way
x 1 + y 1 = n 1
Let S ( n ) denote the number of ordered pairs ( x , y ) satisfying the equation above for natural numbers x , y , n with n > 1 . Find the value of S ( 6 ) .
The answer is 9.
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2 solutions
Here is a solution similar to yours. x 1 + y 1 = 6 1 6 x + 6 y = x y 6 x + 6 y − x y − 3 6 = − 3 6 ( B y S F F T ) 6 ( x − 6 ) − y ( x − 6 ) = − 3 6 ( 6 − y ) ( x − 6 ) = − 3 6 ⇒ ( y − 6 ) ( x − 6 ) = 3 6 T h e r e f o r e , t h e s o l u t i o n s a r e ( 4 2 , 7 ) , ( 7 , 4 2 ) , ( 2 4 , 8 ) , ( 8 , 2 4 ) , ( 1 8 , 9 ) , ( 9 , 1 8 ) , ( 1 5 , 1 0 ) , ( 1 0 , 1 5 ) a n d ( 6 , 6 ) .
If I were clever or sophisticated I wouldn't do this. However, I wanted to see how the numbers roll out. Both x and y must be greater than or equal to n . I therefore wrote a little code to write out values of these variables that satisfy this equation. Here's what I saw:
Obviously not a proof, but gives me the intuition I always like to have.
rearranging (1/x)+(1/y)=(1/6) we get xy-6y-6x=0. Now we consider simons favourite factorizing trick. We write this as xy-6y-6x+36-36=0. Hence (x-6)(y-6)=36. Now we think combinatorially. The number of distinct pairs of solutions is equal to the number of possible values of x-6 to satisfy the equation since for each distinct (x-6) there cannot be 2 distinct values of (y-6) since it would contradict the equation. Now 36 =(2^2)(3^2). The number of distinct solutions is (2+1)(2+1)=9 by using the fact that if there are n distinct primes appearing x1,x2...xn times respectively the total number of results is (x1+1)(x2+1)...(xn+1)