Infinite Number of Even Fermat Pseudoprimes

Theorem
Despite their relative rarity, there exist an infinite number of even Fermat pseudoprimes.

Proof
In the context of Wells, Fermat pseudoprime probably means Fermat pseudoprimes to the base $2$.

Consider the equation:
 * $2^m - 2 \equiv 0 \pmod m$

Any $m$ satisfying the above equation is a Fermat pseudoprime.

We show that for each even $m$ satisfying the above equation, there exists a prime $p$ such that $m p$ also satisfies the equation.

Write $m = 2 n$.

Then $2^{2 n - 1} - 1 \equiv 0 \pmod n$.

By Zsigmondy's Theorem, there exists a prime $p$ such that:
 * $p \divides 2^{2 n - 1} - 1$
 * $p \nmid 2^k - 1$ for all $k < 2 n - 1$

By this we have $2^{2 n - 1} \equiv 1 \pmod p$.

By Fermat's Little Theorem $2^{p - 1} \equiv 1 \pmod p$.

Let $d = \gcd \set {2 n - 1, p - 1}$.

We have $d \le 2 n - 1$ and $2^d \equiv 1 \pmod p$.

This gives $d = 2 n - 1$ and thus $2 n - 1 \divides p - 1$.

Thus there exists $x \in \N$ such that:
 * $p = \paren {2 n - 1} x + 1 > n$

This shows that $p \perp n$.

From $2 p n - 1 = 2 n - 1 + 2 n \paren {p - 1} = \paren {2 n - 1} \paren {1 + 2 n x}$ we have:
 * $2^{m p - 1} = 2^{\paren {2 n - 1} \paren {1 + 2 n x}} \equiv 1 \pmod {n, p}$

By Chinese Remainder Theorem:
 * $2^{m p - 1} \equiv 1 \pmod {n p}$

By Congruence by Product of Moduli:
 * $2^{m p} \equiv 2 \pmod {m p}$

showing that $m p$ is also an even Fermat pseudoprime.

The first even Fermat pseudoprime is $161 \, 038$.

We can use the above method repeatedly to construct infinitely many even Fermat pseudoprimes, though it is impractical since these numbers grow exponentially.