# Smallest 10 Primes in Arithmetic Sequence

## Theorem

The smallest $10$ primes in arithmetic sequence are:

$199 + 210 n$

for $n = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9$.

These are also the smallest $8$ and $9$ primes in arithmetic sequence.

## Proof

 $\ds 199 + 0 \times 210$ $=$ $\ds 199$ which is the $46$th prime $\ds 199 + 1 \times 210$ $=$ $\ds 409$ which is the $80$th prime $\ds 199 + 2 \times 210$ $=$ $\ds 619$ which is the $114$th prime $\ds 199 + 3 \times 210$ $=$ $\ds 829$ which is the $145$th prime $\ds 199 + 4 \times 210$ $=$ $\ds 1039$ which is the $175$th prime $\ds 199 + 5 \times 210$ $=$ $\ds 1249$ which is the $204$th prime $\ds 199 + 6 \times 210$ $=$ $\ds 1459$ which is the $232$nd prime $\ds 199 + 7 \times 210$ $=$ $\ds 1669$ which is the $263$rd prime $\ds 199 + 8 \times 210$ $=$ $\ds 1879$ which is the $289$th prime $\ds 199 + 9 \times 210$ $=$ $\ds 2089$ which is the $316$th prime

But note that $199 + 10 \times 210 = 2299 = 11^2 \times 19$ and so is not prime.

Now we show that this is the smallest $10$ primes in arithmetic sequence.

By Divisibility of Common Difference of Arithmetic Sequence of Primes, the common difference $d$ of any $10$ primes in arithmetic sequence must be divisible by all primes less that $10$.

That is, the common difference is a multiple of $2 \times 3 \times 5 \times 7 = 210$.

Suppose $d = 210$.

Then the first term $p_1$ of the arithmetic sequence cannot be $11$, as $11 + 210 = 221 = 13 \times 17$.

Note that $210 \equiv 1 \pmod {11}$.

Hence if the $p_1$ gives a remainder $r > 1$ upon divison by $11$, the $\paren {12 - r}^{th}$ term of the sequence is equivalent to:

$r + \paren {12 - r - 1} \times 210 \equiv 0 \pmod {11}$

and is divisible by $11$, which is impossible as all terms are primes.

Therefore $p_1 \equiv 1 \pmod {11}$.

The first few primes of this form are:

$23, 67, 89, 199, \dots$

and we eliminate the first three candidates because:

$23 + 5 \times 210 = 1073 = 29 \times 37$
$67 + 3 \times 210 = 697 = 17 \times 41$
$89 + 1 \times 210 = 229 = 13 \times 23$

which are not primes, and we have checked that $199 + 210 n$ are indeed $10$ primes in arithmetic sequence.

Now suppose $d > 210$.

Then $d \ge 420$ and the last term is greater than $420 \times 9 = 3780$.

This shows that the one we found here is the smallest $10$ primes in arithmetic sequence.

$\Box$

### Smallest $9$ primes in arithmetic sequence

Most of the analysis above applies, but we need to also consider $p_1 \equiv 2 \pmod {11}$.

This gives the first few primes:

$13, 79, 101, 167, 211, \dots$

and we have:

$13 + 6 \times 210 = 1273 = 19 \times 67$
$79 + 1 \times 210 = 289 = 17^2$
$101 + 3 \times 210 = 731 = 17 \times 43$
$167 + 1 \times 210 = 377 = 13 \times 29$
$221 > 199$

so there are no smaller $9$ primes in arithmetic sequence.

$\Box$

### Smallest $8$ primes in arithmetic sequence

Most of the analysis above applies, but we need to also consider $p_1 \equiv 3 \pmod {11}$.

This gives the first few primes:

$3, 47, 113, 157, 179, 223, \dots$

and we have:

$3 + 1 \times 210 = 213 = 3 \times 71$
$47 + 7 \times 210 = 1517 = 37 \times 41$
$113 + 1 \times 210 = 323 = 17 * 19$
$157 + 5 \times 210 = 1207 = 17 * 71$
$179 + 7 \times 210 = 1649 = 17 * 97$
$223 > 199$

so there are no smaller $8$ primes in arithmetic sequence.

$\blacksquare$