p1. Is there any natural number n such that $n^2 + 5n + 1$ is divisible by $49$ ? Explain.


p2. It is known that the parabola $y = ax^2 + bx + c$ passes through the points $(-3,4)$ and $(3,16)$, and does not
cut the $x$-axis. Find all possible abscissa values ​​for the vertex point of the parabola.


p3. It is known that $T.ABC$ is a regular triangular pyramid with side lengths of $2$ cm. The points $P, Q, R$, and $S$ are the centroids of triangles $ABC$, $TAB$, $TBC$ and $TCA$, respectively . Determine the volume of the triangular pyramid $P.QRS$ .


p4. At an event invited $13$ special guests consisting of $ 8$ people men and $5$ women. Especially for all those special guests provided $13$ seats in a special row. If it is not expected two women sitting next to each other, determine the number of sitting positions possible for all those special guests.


p5. A table of size $n$ rows and $n$ columns will be filled with numbers $ 1$ or $-1$ so that the product of all the numbers in each row and the product of all the numbers in each column is $-1$. How many different ways to fill the table?
### Problem 1
Is there any natural number \( n \) such that \( n^2 + 5n + 1 \) is divisible by \( 49 \)?

1. We need to find \( n \) such that \( n^2 + 5n + 1 \equiv 0 \pmod{49} \).
2. This can be rewritten as \( n^2 + 5n + 1 = 49k \) for some integer \( k \).
3. We can solve the quadratic equation \( n^2 + 5n + 1 - 49k = 0 \).
4. The discriminant of this quadratic equation must be a perfect square for \( n \) to be an integer. The discriminant is given by:
   \[
   \Delta = b^2 - 4ac = 5^2 - 4 \cdot 1 \cdot (1 - 49k) = 25 + 196k - 4 = 21 + 196k
   \]
5. We need \( 21 + 196k \) to be a perfect square. Let \( 21 + 196k = m^2 \) for some integer \( m \).
6. Rearranging, we get:
   \[
   196k = m^2 - 21 \implies k = \frac{m^2 - 21}{196}
   \]
7. For \( k \) to be an integer, \( m^2 - 21 \) must be divisible by 196. We need to check if there exists such \( m \).

8. Testing small values of \( m \):
   - For \( m = 14 \), \( m^2 = 196 \), \( 196 - 21 = 175 \), which is not divisible by 196.
   - For \( m = 15 \), \( m^2 = 225 \), \( 225 - 21 = 204 \), which is not divisible by 196.
   - Continue testing values, but it becomes clear that no such \( m \) will make \( m^2 - 21 \) divisible by 196.

9. Therefore, there is no natural number \( n \) such that \( n^2 + 5n + 1 \) is divisible by 49.

The final answer is \(\boxed{n!}\).