TITLE: Show there is no $n>0$ where $n\in \mathbb{Z}$ such that $n$, $n+10$, $n+20$, $n+30$ are all prime
QUESTION [6 upvotes]: Please no solution, just guidance. I've currently worked with a few small primes to see how it holds, with the intention to finding something interesting I can work with, of which I haven't.
Things I understand;

$n$ must be prime.
$n + 10$ is the same as $n+(2\cdot 5)$
$n + 20$ is the same as $n+2(2\cdot 5)$
$n + 30$ is the same as $n+3(2\cdot 5)$
This pattern in particular looks interesting so I've dabbled in it, no prevail.
I've worked with mod $n$, and dabbled there, cannot find anything either.
I've dabbled with congreunce properties, can't find anything there.

I wonder if I could get some assistance with the approach to this kinda problem. Thanks.

REPLY [3 votes]: Since you specifically said "no solution" in bold, I will try to stop short of that.
Consider the number $30n + k$, where $n$ is any integer whatsoever and $k$ satisfies $0 \leq k < 30$. For what values of $k$ could $30n + k$ possibly be prime? Certainly not $k = 4$ or 6. In general, we want $\gcd(30, k) = 1$ (remember that $\gcd(30, 0) = 0$).
This means that if $k$ is one of these numbers: 1, 7, 11, 13, 17, 19, 23, 29, then $30n + k$ might be prime. Notice that 1 and 11 are in there, but 21 is not. 13 and 23 are in there, but 3 is not (of course only $n = 0$ will make $30n + 3$ prime).
I hope this points you in the right direction.