Prove that no 5 digit EXTREME PRIMES exist.

Question

I have been trying to come up with my own math problems recently and this is one of my first. It introduces the idea of an extreme prime. I hope that an extreme prime isn't already a thing, because I just used the name to describe a special number. I have a solution to the problem, but I'd like to see smarter solutions and get some feedback on the problem so I can make better ones in the future.

An extreme prime is a number such that every number within the number is prime, expect one-digit numbers, and the number itself is prime. Examples are below for clarity, as I'm bad at explaining.

Examples:

• $617$ is a prime. Also, $61$ is a prime and $17$ is a prime. Therefore $617$ is an extreme prime. Note $6$ is composite: the digits need not be prime.

• $1373$ is prime. Also, $13$ is prime, $37$ is prime, $73$ is prime, $137$ is prime, $373$ is prime. Therefore $1317$ is an extreme prime. Fun fact: $373$ is also the only $3$ digits extreme prime where the digits are prime, so I guess it must be ultra-prime.

The question is to prove that no $5$ digit extreme prime exists. I'm looking forward to some feedback and some ways I can word what an extreme prime is, hope it is fun to solve.

Some other facts I noticed when checking my proof with python (which I do not have a proof for): you may like to try to prove them.

• A $3$ digit extreme prime cannot contain a $2,8$ or $5$.

• A $4$ digit extreme prime cannot contain a $2, 8, 5$ or $4$.

• A $4$ digit extreme prime never starts with $7$.

Quite a few super primes (primes that occupy prime numbered positions in the sequence of all prime numbers) are extreme primes. Can you find them all and create the prime-st number set of all time!

Answer 1

Let $p=d_1d_2\ldots d_n$ be an $n$-digit prime with digits $d_i$. We can define an "extreme" prime recursively as follows: If $n=2$, then $p$ is an extreme prime; if $n\gt2$ then $p$ is extreme if and only if the truncations $d_1d_2\ldots d_{n-1}$ and $d_2\ldots d_n$ are both extreme primes.

It's not hard to see that there only ten $2$-digit primes that can occur in the interior of an $n$-digit extreme prime (with $n\ge4$, so that there is a proper "interior"):

$$11,13,17,19,31,37,71,73,79,97$$

Each of these can be extended on either side to give a $3$-digit extreme prime, but only six of them can be extended on both sides to give a $4$-digit extreme prime. As the OP reports, the $4$-digit extreme primes are

$$1373,3137,3797,6131,6197,9719$$

The only possible extension to a $5$-digit number whose truncations back to four digits both belong to this list is $31373$. (For example, $3797$ cannot be extended on the left because none of the $4$-digit extreme primes are of the form $d_1379$, and it cannot be extended on the right because none of the $4$-digit extreme primes are of the form $797d_4$.) But $31373=137\cdot229$ is not a prime. So there are no $5$-digit extreme primes (hence no extreme primes with more than $5$ digits as well).

Remark: The hardest part of this proof is the part that the OP did, namely identifying the six $4$-digit extreme primes. I don't see any easy way to get to those six without a tedious case-by-case analysis. If anyone can think of a nice way to streamline things, I'd be keen to see it.