Find the numbes of a prime arithmetic sequence I need to find five prime numbers that form an arithmetic sequence with a common difference of 6. I tried to use that 3 will be divisor of the common difference and congruence, but I need to show it in a way that an High school student can understand it. All help will be appreciate.
 A: This is a question you need to think about modulo $30$, but there really are just six cases to consider, because it all wraps around:


*

*Obviously $p_1 \not \equiv 0$, nor $6, 12, 18, 24$.

*If $p_1 \equiv 1 \pmod{30}$ then the other primes would be $7, 13, 19, 25$, but a prime can't be $25 \pmod{30}$.

*$p_1 \not \equiv 2 \pmod{30}$ should be obvious.

*If $p_1 \equiv 3 \pmod{30}$ then the other primes would be $9$, nope, this won't work.

*$p_1 \not \equiv 4 \pmod{30}$ should be obvious.

*This leaves us with $p_1 \equiv 5 \pmod{30}$ as the only possibility that works. But $p_1$ must be actual $5$, and not $35, 65, 95, \ldots$

A: It is distressing how innumerate some high school students are. John's answer with congruence relations might fly over their heads, even though it seems perfectly basic to us. If they at least can understand variables, we can restate John's answer and spell out some of the details he glossed over because he assumes the high school students will understand (I do not share his optimism).
First we label the smallest prime in the progression $p_1$. Then we want $p_2 = p_1 + 6$, $p_3 = p_1 + 12$, and so on and so forth to $p_5 = p_1 + 24$, such that $p_1, p_2, p_3, p_4, p_5$ are all prime.
So essentially we want to solve $p_1 = 6k + c$, with $0 \leq c < 6$. Clearly, $c = 0$ will not work because then $p_1$ is a multiple of 6 and thus not prime. $c = 2$ works if $k = 0$ but then $p_2$ is also even and thus obviously composite. $c = 3$ also works if $k = 0$ but then $p_2$ is a multiple of 3 and obviously composite. There is no need to even consider $c = 4$.
This leaves only two viable options: $c = 1$ or $c = 5$. You should already know that 1 is not a prime number, regardless of whether or not you know the best explanation for that. So for $c = 1$ to work, we require $k > 0$. At this point it makes sense for us to switch from $6k$ to $30k$. We have then $p_1 = 30k + 7$. But then $p_4 = 30k + 25$, which is obviously divisible by 5 and greater than 5, and thus composite. We run into a similar problem if we try $p_1 = 30k + 13$ or $30k + 19$ or $30k + 1$.
Finally this leaves us with $c = 5$ for $p_1 = 6k + c$. If $p_1 = 30k + 11$, then $p_5 = 30k + 35 = 30(k + 1) + 5$, yada, yada, yada, a multiple of 5 is unavoidable.
But there is one positive multiple of 5 that is prime, and that is 5 itself. The only possible positive solution is $p_1 = 5$. But if you accept negative numbers, there is just one other solution: $p_1 = -29$.
A: Hint: A good place to start is guess and check to try and find a small example. Starting with $2$ won't work as $2+6$ is composite. A similar problem arises when starting with $3$. What about starting with $5$?
