Arithmetic sequence where every term is prime?

I conjecture that there does not exist an arithmetic sequence where every number is prime. Put another way, the set $$S_{a,b} = \{a + \delta b \ \vert \delta \in \mathbb{N}\}$$ Where $a, b \in \mathbb{N}$

contains some composite number.

I have no idea how to approach such a question. Googling let me to theorems about densities of primes in arithmetic sequences (Green-Tao and the like), but nothing that answers this elementary question.

• What happens if $\delta=a$? – quasi Apr 17 '18 at 8:00
• @quasi That takes care of $a≠1$ but if $a=1$ it's not always sufficient. Fortunately it's not hard to see that $b\delta+1$ contains arbitrarily large multiples of $b+1$. – MJD Apr 17 '18 at 8:32
• If $a,b\in \Bbb N$ and $\gcd (a,b)=1$ then there are infinitely many $d\in \Bbb N$ such that $a+bd$ is prime. This is called Dirichlet's Theorem. It is not easy or elementary. – DanielWainfleet Apr 17 '18 at 9:49
• Would 3-5-7 or 7-37-67 be a case? – usiro Apr 21 '18 at 9:16

As quasi points out in a comment, when $\delta=a$ the element is $a(b+1)$ which is composite unless possibly $a=1$.

More generally just take $\delta = kb+k+a$ for any $k$ whatever. Then the element is $$kb^2 +(k+a)b + a = (kb+a)(b+1)$$ which is composite.

• This post brought to you by the P.D.Q. Bernoulli Institute of Lower Mathematics. – MJD Apr 17 '18 at 9:00
• Much appreciated :) I came up with the $a(b + 1)$ right after posting the question, but didn't notice the corner case of $a = 1$. – Siddharth Bhat Apr 17 '18 at 9:00
• Is this an "insight" proof, is there a method to come up with this? – Siddharth Bhat Apr 17 '18 at 9:01
• Quasi's observation takes care of $a≠1$. For $a=1$ I just looked at an example, picking $4b+1$ arbitrarily. Because $\gcd(4,5)=1$ this was obviously going to hit a multiple of 5 every 5 terms. The generalization was immediate: $1+\delta b$ contains infinitely many multiples of $b+1$. – MJD Apr 17 '18 at 9:06
• @siddharth I wrote up my thought process in more detail: blog.plover.com/math/arithmetic-sequences.html – MJD Apr 20 '18 at 6:53

I answered similar questions here and here.

"An Introduction To The Theory Of Numbers" by Hardy, Theorem 21, page 18.

THEOREM 21. No polynomial $f(n)$ with integral coefficients, not a constant, can be prime for all $n$, or for all sufficiently large $n$.

In this case $f(\delta)=a+\delta b$. It's a more powerful result, but the proof is very simple, as you will see.

This is quasi-immediate: $a+ab$ is composite.

Update:

There is a special case which invalidates the claim: $a=1$.

Using MJD's method, $a+(a+b+1)b=(a+b)(b+1)$ is always composite.

• "quasi-immediate" is funny! I didn't get it until just now. – MJD Apr 19 '18 at 13:02
• @MJD: though you found a harder version of it. – Yves Daoust Apr 19 '18 at 13:06
• I meant your pun: it was in fact "quasi-immediate" because quasi had immediately posted a comment pointing it out. – MJD Apr 19 '18 at 13:15
• @MJD: unintentional, I didn't get it either ! :) – Yves Daoust Apr 19 '18 at 13:27

We can in fact prove that the sequence contains infinitely many multiples of every number $n$ that is relatively prime to $b$. (Like $n = b + 1$ in MJD's answer.)

Pick any number $n$ such that $\gcd(b, n) = 1$. This means there exist integers $x, y$ such that $nx - by = 1$. (The nice thing about $n=b+1$ is that we can write down $x$ and $y$ explicitly, as $x=y=1$.)

Now take $\delta = nk + ay$ for any $k$.

Then, $a + \delta b = a + (nk + ay)b = a + nkb + a(nx - 1) = n(kb + ax)$ which is a multiple of $n$.

$199, 409, 829,1039,1249, 1459, 1669,1879, 2089$
Also due to Sierpinski theorem the longest Arithmetic progression with prime numbers has 13 terms, first number is $4943$ and the common difference is $60060$.
There is also a theorem that claims there is infinitely many Arithmetic progressions with 13 numbers and common difference $30030$ with all numbers prime, but there has not been found even one yet!