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!