Show there is no $n>0$ where $n\in \mathbb{Z}$ such that $n$, $n+10$, $n+20$, $n+30$ are all prime

Question

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.

Answer 1

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.

Answer 2

You're working in base $10$, right? If the sum of digits of a number is divisible by $3$, then the number itself is also divisible by $3$. For example, the digits of $489$ add up to $21$, which means that $489$ is divisible by $3$.

If we add $10$ to that, we get $499$, and its digits add up to $22$. Add another $10$ to get $509$, so the digits only add up to $14$, but notice that $2 + 2 = 4$ and $1 + 4 = 5$. Add another $10$ to get $519$, with digits adding up to $15$, that's divisible by $3$ just like $489$ is.

Answer 3

$0, \pm 1$ is a complete residue class modulo $3$ (i.e. all integers, $z \mod 3$, are equivalent to either $0, 1$ or $-1$).

If $n\equiv 0 \mod 3$ then $n + 30\equiv 0 \mod 3$ and $3|n+30$ and is not prime. ($n$ itself could be $3$ and be prime.)

If $n\equiv 1\mod3$ then $n + 20\equiv 1+20\equiv 0 \mod 3$ and $3|n + 20$.

And if $n \equiv -1 \mod 3$ then $n+1 \equiv -1+10\equiv 0 \mod 3$ and $3|n+10$.

So one of $n+10; n+20;$ or $n+30$ is divisible by $3$ and not prime.