# Prove that $d\gt 30000$ when $a_1,a_2…a_{15}$ are 15 primes on A.P.

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.

• $d$ must be a multiple of $6$, because all primes greater than $3$ are of the form $6k\pm1$. – user_194421 Jun 2 '18 at 13:18
• Yeah,but it is also a lengthy process before there are $30000/6$=$5000$ possibilities. – Sufaid Saleel Jun 2 '18 at 13:20

$d$ 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 $2\cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13=30030$
• @Piquito: I read the question as "Show that any AP of $15$ primes has $d \gt 30000$. I have no position on whether one exists, but this result reduces the space to search. – Ross Millikan Jun 2 '18 at 14:13
• @Piquito: One can make a handwaving argument that such a progression exists and that there are infinitely many of them. If we take $d=30030$ and let the first number $a$ be huge compared to $d$, the chance that each number in the progression is prime is about $\frac 1{\log a}$, so the chance every number in the progression is prime is about $\frac 1{(\log a)^{15}}$. The sum of this as $a$ goes to infinity diverges. – Ross Millikan Jun 2 '18 at 14:22
• If $a_1=13$, we can still conclude that $13\mid d$ because otherwise $13\mid a_{14}$. In fact, if $a_1\in\{2,3,5,7,11,13\}$, then $a_1\mid a_{1+a_1}$. Hence the condition $a_1>15$ can be dropped. – Hagen von Eitzen Jun 2 '18 at 14:36