[b]p1.[/b] Show that $9x + 5y$ is a multiple of$ 17$ whenever $2x + 3y$ is a multiple of $17$.


[b]p2.[/b] Express the five distinct fifth roots of $1$ in terms of radicals.


[b]p3.[/b] Prove that the three perpendiculars dropped to the three sides of an equilateral triangle from any point inside the triangle have a constant sum.


[b]p4.[/b] Find the volume of a sphere which circumscribes a regular tetrahedron of edge $a$.


[b]p5.[/b] For any integer $n$ greater than $1$, show that $n^2-2n + 1$ is a factor at $n^{n-1}-1$.


PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].
### Problem 1
Show that \( 9x + 5y \) is a multiple of \( 17 \) whenever \( 2x + 3y \) is a multiple of \( 17 \).

1. We start with the given condition: \( 2x + 3y \equiv 0 \pmod{17} \).
2. We need to show that \( 9x + 5y \equiv 0 \pmod{17} \) under this condition.
3. Notice that:
   \[
   9x + 5y \equiv 9x + 5y \pmod{17}
   \]
4. We can express \( 9x + 5y \) in terms of \( 2x + 3y \):
   \[
   9x + 5y \equiv 9x + 5y \equiv 9x + 5y - 4(2x + 3y) \pmod{17}
   \]
5. Simplify the expression:
   \[
   9x + 5y - 4(2x + 3y) = 9x + 5y - 8x - 12y = x - 7y
   \]
6. Since \( 2x + 3y \equiv 0 \pmod{17} \), we have:
   \[
   2x + 3y = 17k \quad \text{for some integer } k
   \]
7. Substitute \( x \) and \( y \) from the above equation:
   \[
   x = \frac{17k - 3y}{2}
   \]
8. Substitute \( x \) back into \( x - 7y \):
   \[
   \frac{17k - 3y}{2} - 7y = \frac{17k - 3y - 14y}{2} = \frac{17k - 17y}{2} = \frac{17(k - y)}{2}
   \]
9. Since \( 17 \) is a multiple of \( 17 \), the expression is zero modulo \( 17 \):
   \[
   \frac{17(k - y)}{2} \equiv 0 \pmod{17}
   \]
10. Therefore, \( 9x + 5y \equiv 0 \pmod{17} \).

\(\blacksquare\)

### Problem 2
Express the five distinct fifth roots of \( 1 \) in terms of radicals.

1. The fifth roots of unity are the solutions to the equation \( z^5 = 1 \).
2. These roots are given by:
   \[
   z_k = e^{2\pi i k / 5} \quad \text{for } k = 0, 1, 2, 3, 4
   \]
3. Expressing these in terms of radicals:
   \[
   z_0 = 1
   \]
   \[
   z_1 = e^{2\pi i / 5} = \cos\left(\frac{2\pi}{5}\right) + i\sin\left(\frac{2\pi}{5}\right)
   \]
   \[
   z_2 = e^{4\pi i / 5} = \cos\left(\frac{4\pi}{5}\right) + i\sin\left(\frac{4\pi}{5}\right)
   \]
   \[
   z_3 = e^{6\pi i / 5} = \cos\left(\frac{6\pi}{5}\right) + i\sin\left(\frac{6\pi}{5}\right)
   \]
   \[
   z_4 = e^{8\pi i / 5} = \cos\left(\frac{8\pi}{5}\right) + i\sin\left(\frac{8\pi}{5}\right)
   \]

The final answer is \( \boxed{ 1, e^{2\pi i / 5}, e^{4\pi i / 5}, e^{6\pi i / 5}, e^{8\pi i / 5} } \).