Form of Prime Sierpiński Number of the First Kind

Theorem

Suppose $S_n = n^n + 1$ is a prime Sierpiński number of the first kind.

Then:

$n = 2^{2^k}$

for some integer $k$.

Proof

Aiming for a contradiction, suppose $n$ has an odd divisor $d$.

By Sum of Two Odd Powers:

$\paren {n^{n/d} + 1} \divides \paren {\paren {n^{n/d}}^d + 1^d} = S_n$

thus $S_n$ is composite, which is a contradiction.

Hence $n$ has no odd divisors.

That is, $n$ is a power of $2$.

Write $n = 2^m$.

Aiming for a contradiction, suppose that $m$ has an odd divisor $f$.

By Sum of Two Odd Powers:

$\paren {2^{n m/f} + 1} \divides \paren {\paren {2^{n m/f}}^f + 1^f} = 2^{m n} + 1 = S_n$

thus $S_n$ is composite, which is a contradiction.

Hence $m$ has no odd divisors.

That is, $m$ is a power of $2$.

Therefore we must have:

$n = 2^{2^k}$

for some integer $k$.

$\blacksquare$

Also see

• Prime Sierpiński Numbers of the First Kind