[b]i.)[/b] Let $g(x) = x^5 + x^4 + x^3 + x^2 + x + 1.$ What is the remainder when the polynomial $g(x^{12}$ is divided by the polynomial $g(x)$?

[b]ii.)[/b] If $k$ is a positive number and $f$ is a function such that, for every positive number  $x, f(x^2 + 1 )^{\sqrt{x}} = k.$ Find the value of 

\[ f( \frac{9 +y^2}{y^2})^{\sqrt{ \frac{12}{y} }} \] for every positive number $y.$

[b]iii.)[/b] The function $f$ satisfies the functional equation $f(x) + f(y) = f(x+y) - x \cdot y - 1$ for every pair $x,y$ of real numbers. If $f(1) = 1,$ then find the numbers of integers $n,$ for which $f(n) = n.$
### Part i)
1. Given the polynomial \( g(x) = x^5 + x^4 + x^3 + x^2 + x + 1 \).
2. We need to find the remainder when \( g(x^{12}) \) is divided by \( g(x) \).
3. Notice that \( g(x) \) can be rewritten using the formula for the sum of a geometric series:
   \[
   g(x) = \frac{x^6 - 1}{x - 1}
   \]
   because \( x^6 - 1 = (x - 1)(x^5 + x^4 + x^3 + x^2 + x + 1) \).
4. Therefore, \( g(x) \) has roots at the 6th roots of unity, excluding 1.
5. Since \( g(x) \) has roots at \( \omega, \omega^2, \omega^3, \omega^4, \omega^5 \) where \( \omega = e^{2\pi i / 6} \), we need to consider \( g(x^{12}) \) at these roots.
6. Notice that \( x^{12} = (x^6)^2 \), and since \( \omega^6 = 1 \), we have \( \omega^{12} = 1 \).
7. Therefore, \( g(x^{12}) \) evaluated at any root of \( g(x) \) will be \( g(1) = 1 + 1 + 1 + 1 + 1 + 1 = 6 \).
8. Since \( g(x) \) divides \( g(x^{12}) - 6 \), the remainder when \( g(x^{12}) \) is divided by \( g(x) \) is 6.

\[
\boxed{6}
\]

### Part ii)
1. Given \( f(x^2 + 1)^{\sqrt{x}} = k \) for every positive number \( x \).
2. We need to find the value of \( f\left( \frac{9 + y^2}{y^2} \right)^{\sqrt{\frac{12}{y}}} \) for every positive number \( y \).
3. Let \( x = \frac{3}{y} \). Then \( x^2 = \left( \frac{3}{y} \right)^2 = \frac{9}{y^2} \).
4. Therefore, \( x^2 + 1 = \frac{9}{y^2} + 1 = \frac{9 + y^2}{y^2} \).
5. Substituting \( x = \frac{3}{y} \) into the given functional equation, we get:
   \[
   f\left( \frac{9 + y^2}{y^2} \right)^{\sqrt{\frac{3}{y}}} = k
   \]
6. We need to find \( f\left( \frac{9 + y^2}{y^2} \right)^{\sqrt{\frac{12}{y}}} \).
7. Notice that \( \sqrt{\frac{12}{y}} = 2 \cdot \sqrt{\frac{3}{y}} \).
8. Therefore, we have:
   \[
   f\left( \frac{9 + y^2}{y^2} \right)^{\sqrt{\frac{12}{y}}} = \left( f\left( \frac{9 + y^2}{y^2} \right)^{\sqrt{\frac{3}{y}}} \right)^2 = k^2
   \]

\[
\boxed{k^2}
\]

### Part iii)
1. Given the functional equation \( f(x) + f(y) = f(x+y) - xy - 1 \) for every pair \( x, y \) of real numbers.
2. We know \( f(1) = 1 \).
3. Set \( x = 0 \) in the functional equation:
   \[
   f(0) + f(y) = f(y) - 1 \implies f(0) = -1
   \]
4. Set \( y = 1 \) in the functional equation:
   \[
   f(x) + f(1) = f(x+1) - x - 1 \implies f(x) + 1 = f(x+1) - x - 1 \implies f(x+1) = f(x) + x + 2
   \]
5. Using the recurrence relation \( f(x+1) = f(x) + x + 2 \), we can find \( f(n) \) for any integer \( n \).
6. Calculate \( f(n) \) for positive integers:
   \[
   f(2) = f(1) + 1 + 2 = 1 + 3 = 4
   \]
   \[
   f(3) = f(2) + 2 + 2 = 4 + 4 = 8
   \]
   \[
   f(4) = f(3) + 3 + 2 = 8 + 5 = 13
   \]
   \[
   \vdots
   \]
   \[
   f(n) = \frac{(n+1)(n+2)}{2} - 2
   \]
7. We need to find \( n \) such that \( f(n) = n \):
   \[
   \frac{(n+1)(n+2)}{2} - 2 = n
   \]
   \[
   (n+1)(n+2) = 2n + 4
   \]
   \[
   n^2 + 3n + 2 = 2n + 4
   \]
   \[
   n^2 + n - 2 = 0
   \]
   \[
   (n+2)(n-1) = 0
   \]
   \[
   n = -2 \text{ or } n = 1
   \]
8. Since \( n \) must be an integer, the only solutions are \( n = -2 \) and \( n = 1 \).

\[
\boxed{2}
\]