11 is Only Palindromic Prime with Even Number of Digits

Theorem

Let $P$ be a palindromic number with $2 n$ digits:

$P = \sqbrk {a_{2 n - 1} a_{2 n - 2} \ldots a_2 a_1 a_0}_{10}$

Thus:

$P = \ds \sum_{j \mathop = 0}^{n - 1} a_j + 10^{2 n - 1 - j}$

Consider the summation:

$S = \ds \sum_{k \mathop = 0}^{2 n - 1} \paren {-1}^k a_k$

As $a_k = a_{2 n - 1 - k}$ we have:

$\ds S$

$=$

$\ds \sum_{k \mathop = 0}^{n - 1} \paren {-1}^k a_k + \paren {-1}^{2 n - 1 - k} a_k$

$\ds $

$=$

$\ds \sum_{k \mathop = 0}^{n - 1} a_k \paren {\paren {-1}^k + \paren {-1}^{2 n - 1 - k} }$

As $2 n - 1$ is odd, it follows that $k$ and $2 n - 1 - k$ are of opposite parity.

Thus:

$\paren {-1}^k = -\paren {-1}^{2 n - 1 - k}$

and it follows that:

$S = 0$

It follows by Divisibility by 11 that $P$ is divisible by $11$.

Thus, except for $11$ itself, a palindromic number with an even number of digits cannot be prime.

The result follows.

$\blacksquare$