$$  $$Let $$ p $$$$ 1,p2,p2,p4 $$........$$ pn\text{ be the prime numbers greater than }5\text{ }.\text{if }p1^2+p2^2+p3^2.......pn^2\text{ is divisible by }6\text{ then prove that }n\text{ is divisible by }6 $$

Let $ p_1, p_2, p_3, \ldots, p_n $ be prime numbers greater than 5, and we are asked to prove that if the sum of the squares of these primes is divisible by 6, then $ n $ must be divisible by 6.

### Key Observations:
- **Prime numbers greater than 5**: All prime numbers greater than 5 must be of the form $ 6k \pm 1 $ for some integer $ k $. This is because every integer is either divisible by 6 or of the form $ 6k \pm 1, 6k \pm 2, 6k \pm 3, 6k \pm 4 $. If a number is of the form $ 6k $, it is divisible by 6 and not prime (except for 2 and 3, which are not included since they are less than 5). If a number is of the form $ 6k \pm 2 $, it is divisible by 2 and hence not prime. If a number is of the form $ 6k \pm 3 $, it is divisible by 3 and hence not prime. Thus, all prime numbers greater than 5 must be of the form $ 6k \pm 1 $.

- **Squares of prime numbers of the form $ 6k \pm 1 $**: 
  If a prime number $ p $ is of the form $ 6k \pm 1 $, then:
  \[
  p^2 = (6k \pm 1)^2 = 36k^2 \pm 12k + 1 \equiv 1 \pmod{6}
  \]
  Thus, the square of any prime number greater than 5 is congruent to 1 modulo 6.

### Step-by-Step Proof:

1. Each prime number $ p_i $ (where $ p_i $ is a prime greater than 5) satisfies:
   \[
   p_i^2 \equiv 1 \pmod{6}
   \]
   Hence, for any set of prime numbers $ p_1, p_2, p_3, \dots, p_n $ greater than 5, we have:
   \[
   p_1^2 + p_2^2 + p_3^2 + \dots + p_n^2 \equiv 1 + 1 + 1 + \dots + 1 \pmod{6}
   \]
   where there are $ n $ terms in the sum.

2. Therefore, the sum of the squares of these primes is congruent to $ n \pmod{6} $:
   \[
   p_1^2 + p_2^2 + \dots + p_n^2 \equiv n \pmod{6}
   \]

3. The problem states that the sum of the squares is divisible by 6. This means:
   \[
   n \equiv 0 \pmod{6}
   \]
   Thus, $ n $ must be divisible by 6.

### Conclusion:
We have shown that if the sum of the squares of the primes greater than 5 is divisible by 6, then the number $ n $ of such primes must also be divisible by 6.

### Any questions?
Would you like further clarification or details on any specific part?

Here are five related questions to expand your understanding:

1. What are all the prime numbers of the form $ 6k \pm 1 $ less than 100?
2. How can we prove that the square of any number of the form $ 6k \pm 1 $ is congruent to 1 modulo 6?
3. How would the result change if we included primes less than or equal to 5 in the list?
4. Can you prove that every prime number greater than 5 is of the form $ 6k \pm 1 $?
5. How would the result change if we were working modulo 4 instead of modulo 6?

**Tip**: When working with modular arithmetic, always remember to simplify the expressions early to avoid dealing with large numbers!

Explore the proof that if the sum of squares of prime numbers greater than 5 is divisible by 6, then the number of such primes, n, must also be divisible by 6. This problem delves into prime number properties and modular arithmetic concepts, suitable for advanced high school students.

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