11 is Only Palindromic Prime with Even Number of Digits

Theorem
$11$ is the only palindromic prime with an even number of digits when expressed in decimal notation.

Proof
Let $P$ be a palindromic number with $2 n$ digits:
 * $P = \left[{a_{2 n - 1} a_{2 n - 2} \ldots a_2 a_1 a_0}\right]_{10}$

Thus:
 * $P = \displaystyle \sum_{j \mathop = 0}^{n - 1} a_j + 10^{2 n - 1 - j}$

Consider the summation:
 * $S = \displaystyle \sum_{k \mathop = 0}^{2 n - 1} \left({-1}\right)^k a_k$

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

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

Thus:
 * $\left({-1}\right)^k = - \left({-1}\right)^{2 n - 1 - k}$

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.