Let S(n) denotes the number of ordered pairs (x,y) satisfying `1/x+1/y=1/n, " where " n gt 1 " and " x,y,n in N`.
`" "` (i) Find the value of S(6).
`" "` (ii) Show that, if n is prime, then S(n)=3, always.

Answer

To solve the problem, we will tackle each part step by step.

### Part (i): Find the value of S(6)

1. **Start with the equation**: We have the equation \( \frac{1}{x} + \frac{1}{y} = \frac{1}{n} \). For \( n = 6 \), the equation becomes:
\[ \frac{1}{x} + \frac{1}{y} = \frac{1}{6} \]

2. **Rearranging the equation**: We can rewrite the equation as:
\[ \frac{x + y}{xy} = \frac{1}{6} \]
By cross-multiplying, we get:
\[ 6(x + y) = xy \]

3. **Rearranging further**: Rearranging gives us:
\[ xy - 6x - 6y = 0 \]
Adding \( 36 \) to both sides, we can factor:
\[ xy - 6x - 6y + 36 = 36 \]
This can be factored as:
\[ (x - 6)(y - 6) = 36 \]

4. **Finding pairs (a, b)**: Let \( a = x - 6 \) and \( b = y - 6 \). Then we need to find the pairs \( (a, b) \) such that:
\[ ab = 36 \]
The natural number pairs \( (a, b) \) that satisfy this are:
- \( (1, 36) \)
- \( (2, 18) \)
- \( (3, 12) \)
- \( (4, 9) \)
- \( (6, 6) \)
- \( (9, 4) \)
- \( (12, 3) \)
- \( (18, 2) \)
- \( (36, 1) \)

5. **Counting ordered pairs**: Each pair \( (a, b) \) corresponds to an ordered pair \( (x, y) \) since \( x = a + 6 \) and \( y = b + 6 \). Therefore, we have:
- \( (7, 42) \)
- \( (8, 24) \)
- \( (9, 18) \)
- \( (10, 15) \)
- \( (12, 12) \)
- \( (15, 10) \)
- \( (18, 9) \)
- \( (24, 8) \)
- \( (42, 7) \)

This gives us a total of **9 ordered pairs**.

Thus, \( S(6) = 9 \).

### Part (ii): Show that if n is prime, then S(n) = 3

1. **Start with the equation**: For a prime number \( n \), we have:
\[ \frac{1}{x} + \frac{1}{y} = \frac{1}{n} \]

2. **Rearranging the equation**: As before, we can rewrite it as:
\[ xy - nx - ny = 0 \]
Adding \( n^2 \) gives:
\[ xy - nx - ny + n^2 = n^2 \]
This can be factored as:
\[ (x - n)(y - n) = n^2 \]

3. **Finding pairs (a, b)**: Let \( a = x - n \) and \( b = y - n \). We need to find pairs \( (a, b) \) such that:
\[ ab = n^2 \]
The factors of \( n^2 \) are \( (1, n^2) \), \( (n, n) \), and \( (n^2, 1) \).

4. **Counting ordered pairs**: The pairs \( (a, b) \) correspond to:
- \( (1, n^2) \) gives \( (n + 1, n^2 + n) \)
- \( (n, n) \) gives \( (2n, 2n) \)
- \( (n^2, 1) \) gives \( (n^2 + n, n + 1) \)

This results in exactly **3 ordered pairs**.

Thus, for any prime number \( n \), \( S(n) = 3 \).

### Summary of Solutions:
- (i) \( S(6) = 9 \)
- (ii) If \( n \) is prime, then \( S(n) = 3 \).