An arithmetic progression with primes

Question

Here is a math puzzle I thought of a while ago:

Find the longest arithmetic progression that consists only of primes, such that the difference between two consecutive terms is the product of two primes.

Good luck!

Answer 1

The longest such arithmetic progression has length

$5$

Example

$5, 11, 17, 23, 29$

Proof

Let the arithmetic progression be denoted by $a_1, a_2, \ldots$ and consecutive difference be $d = pq$ so that $a_i = (i-1)d + a_1$ and $p$ and $q$ are primes.

If $p$ and $q$ are both odd and $a_1$ is odd then $a_2$ is even and greater than $2$, so $a_2$ is not prime.

If $a_1 = 2$ and $p$ and $q$ are both odd, then $a_3$ is even and the longest possible prime arithmetic progression has length $2$.

Now let us consider $p=2$ and consider everything modulo $3$. If $q \neq 3$, then $pq \not\equiv 0 (\bmod 3)$ and one of $a_2, a_3, a_4$ is divisible by $3$ so the longest prime arithmetic progression in this instance has length $3$ (example $3, 7, 11$)

Our only other case is $pq = 6$.
Considering everything modulo $5$, we find that one of $a_2, a_3, a_4, a_5, a_6$ must be divisible by $5$ so the longest prime arithmetic progression has length $5$ (this is the case for $a_1 = 5$ so that $a_6$ is the next number divisible by $5$).

Answer 2

For arithmetic progression being the numbers prime, we can get only three-term in a series at longest. Surely there will include number 3, cause for any three-term arithmetic series, one of the terms must be divided by 3. And as 3 is prime we can include it. like:

$3, 5, 7$;
$3, 7, 11$;
$3, 17, 31$;
$3, 23, 43$
and so on.