Arithmetic sequence where every term is prime?

Question

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.

Answer 1

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 positive $k$. Then the element is $$kb^2 +(k+a)b + a = (kb+a)(b+1)$$ which is composite.

Answer 2

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

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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.

Answer 3

This is for your information:

There is an arithmetic progression which has ten prime terms and its first term is 199 and common difference is 210:

$199, 409, 829,1039,1249, 1459, 1669,1879, 2089$

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!

Answer 4

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 $n$.

Answer 5

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.