Prime Coincidence

Relevant wiki: Divisibility Rules

Let the prime number $p$ be

$p=\overline{a_1a_2\ldots a_n}$

Where the $a_i$ $i\in\{1,2,\ldots,n\}$ are the digits of $p$

After the process we get $m=\overline{a_1a_2\ldots a_na_n\ldots a_2a_1}$

$m$ is divisible by $11$ if and only if the sum $S$ of its digits with alternated signs is divisible by $11$

$\begin{aligned} S&=\mathbin{\color{#D61F06}a_1} \mathbin{\color{#3D99F6}-a_2}+\ldots \mathbin{\color{#20A900}\pm a_n}\mathbin{\color{#20A900}\mp a_n}\ldots \mathbin{\color{#3D99F6}a_2}\mathbin{\color{#D61F06}-a_1}\\ &=0\equiv 0 \mod{11} \end{aligned}$

Since $m$ is a positive integer divisible by $11$ and it's not $11$ itself (because that would imply $p=1$ , which is not prime), $m$ is always composite

Another nice way of seeing this relationship without relying on divisibility rules is using the fact that $x+1|(x+1)^{2n-1}$ as follows. Let the prime number be

$p=a_n*10^n+a_{n-1}*10^{n-1}+...+a_1*10^1+a_0$

Then the new number is

$n=\sum_{i=0}^{n} a_i \cdot (10^i) (1+10^{2n+1-2i)})$

Since each one of these parts of the sum is divisble by 11, as $1+10|1+10^{2n-1}$ , the whole sum is also divisible by eleven, and since $n$ is not 11 (1 is not prime), $n$ will always be composite.

If you closely observe, each number will be divisible by 11. Hence, there's just no question of the number being a prime.

I checked every prime up to 139 and found no solutions. At some point I had to determine this pattern would likely continue. Though, I don't have a way to write a proof.