We know that an arithmetic progression has to have a composite number since there are arbitrarily large gaps between primes. But I was wondering whether the following construction is possible:
Can there exist an $m$ number of arithmetic progressions say $\{a_1^n\},\{a_2^n\},....,\{a_m^n\}$(here $\{a_1^n\},\{a_2^n\}.$ represents an arithmetic progression)such that for each $n$ the set $A_n=\{a_1(n),....,a_m(n)\}$($a_m(n)$ represents the $n$ term of the AP $\{a_m^n\}$) has atleast one prime.
So I have heard that the if we have an arithmetic progression say $a+nq$ then the number of primes less than $x$ if $(a,q)=1$ is $\frac{cx}{\ln x}$. So if there are $m$ arithmetic progressions then in $n$ terms for $n$ sufficiently large we can all AP's combined have atmost have $\frac{cmn}{\ln n}$ but if the above where possible then we have $n$ primes atleast. This is a contradiction. I am not sure whether my argument here is correct so if anyone can point if something is wrong it would be great. Also I am having to use very strong theorems so I am also looking for an easy way to solve it. Thanks.
No, this is impossible. Write your APs as $b_i + d_i n$, and assume $b_i>1$ (else shift the index $n$ until it is true).
Now set $N = \prod_{j=1 }^m b_j$.
Then $a_i(N)= b_i + d_i \prod_{j=1 }^m b_j$. This is divisible by $b_i$ for each $i$. So $a_i(N)$ is not prime for all $i$.