[b]p1.[/b] Arrange the whole numbers $1$ through $15$ in a row so that the sum of any two adjacent numbers is a perfect square. In how many ways this can be done?


[b]p2.[/b] Prove that if $p$ and $q$ are prime numbers which are greater than $3$ then $p^2 - q^2$ is divisible by $24$.


[b]p3.[/b] If a polyleg has even number of legs he always tells truth. If he has an odd number of legs he always lies. 
Once a green polyleg told a dark-blue polyleg ”- I have $8$ legs. And you have only $6$ legs!” 
The offended dark-blue polyleg replied ”-It is me who has $8$ legs, and you have only $7$ legs!” 
A violet polyleg added ”-The dark-blue polyleg indeed has $8$ legs. But I have $9$ legs!” 
Then a stripped polyleg started ”None of you has $8$ legs. Only I have $8$ legs!” 
Which polyleg has exactly $8$ legs?


[b][b]p4.[/b][/b] There is a small puncture (a point) in the wall (as shown in the figure below to the right). The housekeeper has a small flag of the following form (see the figure left). Show on the figure all the points of the wall where you can hammer in a nail such that if you hang the flag it will close up the puncture.
[img]https://cdn.artofproblemsolving.com/attachments/a/f/8bb55a3fdfb0aff8e62bc4cf20a2d3436f5d90.png[/img]


[b]p5.[/b] Assume $ a, b, c$ are odd integers. Show that the quadratic equation $ax^2 + bx + c = 0$ has no rational solutions. 


PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].
1. **Step 1: Understanding the problem**

   We need to prove that if \( p \) and \( q \) are prime numbers greater than 3, then \( p^2 - q^2 \) is divisible by 24.

2. **Step 2: Analyzing primes greater than 3**

   Any prime number greater than 3 must be of the form \( 6k \pm 1 \). This is because:
   - If a prime number \( p \) were of the form \( 6k \), it would be divisible by 6 and hence not prime.
   - If \( p \) were of the form \( 6k + 2 \) or \( 6k + 4 \), it would be even and greater than 2, hence not prime.
   - If \( p \) were of the form \( 6k + 3 \), it would be divisible by 3 and greater than 3, hence not prime.

   Therefore, \( p \) and \( q \) can be written as \( p = 6m \pm 1 \) and \( q = 6n \pm 1 \) for some integers \( m \) and \( n \).

3. **Step 3: Squaring the primes**

   Let's square \( p \) and \( q \):
   \[
   p^2 = (6m \pm 1)^2 = 36m^2 \pm 12m + 1
   \]
   \[
   q^2 = (6n \pm 1)^2 = 36n^2 \pm 12n + 1
   \]

4. **Step 4: Subtracting the squares**

   Now, consider \( p^2 - q^2 \):
   \[
   p^2 - q^2 = (36m^2 \pm 12m + 1) - (36n^2 \pm 12n + 1)
   \]
   \[
   p^2 - q^2 = 36m^2 \pm 12m + 1 - 36n^2 \mp 12n - 1
   \]
   \[
   p^2 - q^2 = 36(m^2 - n^2) \pm 12(m \mp n)
   \]

5. **Step 5: Modulo 6 analysis**

   Since \( p \) and \( q \) are of the form \( 6k \pm 1 \), their squares modulo 6 are:
   \[
   (6k \pm 1)^2 \equiv 1 \mod 6
   \]
   Therefore:
   \[
   p^2 \equiv 1 \mod 6 \quad \text{and} \quad q^2 \equiv 1 \mod 6
   \]
   \[
   p^2 - q^2 \equiv 1 - 1 \equiv 0 \mod 6
   \]

6. **Step 6: Modulo 4 analysis**

   Since \( p \) and \( q \) are odd primes, they can be written as \( 4k \pm 1 \). Their squares modulo 4 are:
   \[
   (4k \pm 1)^2 \equiv 1 \mod 4
   \]
   Therefore:
   \[
   p^2 \equiv 1 \mod 4 \quad \text{and} \quad q^2 \equiv 1 \mod 4
   \]
   \[
   p^2 - q^2 \equiv 1 - 1 \equiv 0 \mod 4
   \]

7. **Step 7: Combining results**

   Since \( p^2 - q^2 \) is divisible by both 6 and 4, it must be divisible by their least common multiple, which is 24.

\[
\boxed{p^2 - q^2 \text{ is divisible by 24}}
\]