TITLE: Prove that $d\gt 30000$ when $a_1,a_2.....a_{15}$ are 15 primes on A.P.
QUESTION [2 upvotes]: Suppose 15 primes $a_1,a_2,......,a_{15}$ are in A.P. If $d$ is the common difference and $a_1\gt 15$ then proved that $d\gt 30000$. $d$ is positive.

It is my problem. There is a few theories and ideas about primes especially the number is primes or the difference between two primes. Can it be done by basic elementary number theory.I tried it to solve by contradiction.
A method is also there . If we check all possible values of $d\lt 30000$(I mean the even values because in this case the value of d is even) and show that there doesn't exist such A.P. then we are done. But it is very lengthy and I believe one or more good ways are there to solve this. Please answer this if you can.

REPLY [4 votes]: $d$ must be even because if $d$ were odd at least one (in fact at least six) of the numbers would be even, greater than $2$ and therefore not a prime.  Similarly, $d$ must be a multiple of $3$ because if it were not at least one of the numbers would be divisible by $3$ and greater than $3$.  The same argument works for $5,7,11,13$ and $2\cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13=30030$
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