# Question involving prime numbers, *brothers* numbers.

I thought about the following problem, probably it already appears in mathematical literature.

Definition 1:

Operator $\unrhd$, is binary operation, defined for natural numbers as follows: To every $n,m$ naturals, $n\unrhd m$ is first comes $n$ and then comes $m$.

Example: if $n=56$, $m=67$, then $n\unrhd m = 5667$

Definition 2:

Primes $p_k$ and $p_s$ will called brothers, if $p_k \unrhd p_s$ or $p_s \unrhd p_k$ are primes.

Example: $p_1=2$ and $p_2=3$ are brothers, as $2 \unrhd 3=23$ is prime.

Question:

Prove that there is infinite brothers numbers.

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I suspect this is beyond reach. If we could prove there are infinitely many primes $p$ such that $10p+3$ is prime, we'd have it, but this is beyond what's currently known. – Gerry Myerson Feb 15 '13 at 2:18
The first thing I thought was to attack this problem (mow an open one) by contradiction: suppose that there are finite brothers numbers. Let $A$ be such set. Now, there is a maximal length of brothers number in $A$, suppose that $\text{length$\{p_i \unrhd p_j\}$}=k$. Now - a problem. Sorry, if it sounds stupid. :) – Salech Alhasov Feb 15 '13 at 2:53
@GerryMyerson I agree this seems just out of reach, but we can prove that for infinitely many prime pairs $(p,q)$, the numbers $10p+q$, $100p+q$, $1000p+q$, $10000p+q$ are simultaneously prime. The catch, of course, is that we don't know how to prove this for bounded $q$! – Erick Wong Feb 15 '13 at 4:32

It would be most interesting to find a pair of primes that are brothers in one base but not in another. Among $2$, $3$, $5$, and $7$, I think the brothers in base ten are also brothers in binary, although not always in the same order. – Todd Wilcox Feb 15 '13 at 4:00