Are there infinitely many integers $n$ such that both $27n+23$ and $169n+144$ are simultaneously prime ?

Question

ganeshie8 · Answer

http://mathworld.wolfram.com/DirichletsTheorem.html

ikram002p · Answer

so just showing that gcd( 27n+23,169n+144 )=1 ?

isaidavila · Answer

Let's find out if 27 and 23 are relatively primes 
(27;23)=1 they are
Now 144 and 169 and they are
So now from dirichlet theorem there are infinite n numbers for this series contain infinite primes

kainui · Answer

My way that I think works is find a,b,x,y that make this possible for any integer k,

$$27(ak+b)+23=169(xk+y)+144$$

It ends up leading to basically solving this:

$$27b=169y+121$$
which has as one possible solution
$$b=17$$$$y=2$$
Anyways point is, now I can just pick 

$n=ak+b=xk+y$

and by Dirichlet's theorem this should be just as good as any other solution...? I just gotta make sure stuff is relatively prime or something huh... I dunno I am too hyper from eating candy today to check my work sorry lol

ikram002p · Answer

@Kainui  use Euclid algorithm to show they are relatively prime =D