# A note on finding more base-3 palindromic primes of the form 1 + 3^k + (3^k)^2

I stumbled across base-3 palindromes of the form $1 + 3^k + (3^k)^2$ for $k \in \mathbb{N}$ and became especially interested in them when I realised the reciprocal of any such number must belong to the middle-third Cantor set. In particular, primes of this form will be Cantor primes which I have explored before. The only examples of primes of this form I am aware of are 13 (corresponding to $k = 1$) and 757 (corresponding to $k = 3$).

I am very interested in finding more primes of this base-3 palindromic form, if there are any, or if not, I would like to see a mathematical argument which shows there cannot be any more. (Since the question of whether or not there are infinitely many palindromic primes in general is a major open problem, it would be a major breakthrough to prove there are infinitely many primes of this particular palindromic form). At the very least, it would be nice to establish some conditions on $k$ which would eliminate a lot of cases in which $1 + 3^k + (3^k)^2$ cannot be prime.

I have noticed the following four properties about them, which I will document here:

Result 1. All base-3 palindromes of the form $1 + 3^k + (3^k)^2$ for $k \in \mathbb{N}$ are such that their reciprocal belongs to the middle-third Cantor set. Therefore, in particular, all primes of this form are Cantor primes.

Proof: We have

$1 + 3^k + (3^k)^2 = \frac{3^{3k} - 1}{3^k - 1}$

Therefore writing $p = 1 + 3^k + (3^k)^2$ we have

$\frac{1}{p} = \frac{3^k - 1}{3^{3k} - 1}$

Using the facts that

$\frac{3^k - 1}{2}$ $= 1 + 3 + 3^2 + \cdots + 3^{k-1}$

and

$\frac{1}{3^{3k} - 1}$ $= \frac{1}{3^{3k}} + \frac{1}{(3^{3k})^2} + \cdots$

we can write

$\frac{1}{p}$ $= 2(1 + 3 + 3^2 + \cdots + 3^{k-1})$ $\{\frac{1}{3^{3k}} + \frac{1}{(3^{3k})^2} + \cdots\}$

This is an expression for $1/p$ which corresponds to a base-3 representation involving only the digits 0 and 2, and therefore $1/p$ must belong to the middle-third Cantor set. QED

Result 2. The base-3 palindrome $1 + 3^k + (3^k)^2$ will be divisible by $1 + 3 + 3^2 = 13$ if and only if $gcd(k, 3) = 1$. Therefore the base-3 palindrome $1 + 3^k + (3^k)^2$ cannot be a Cantor prime if k is not a multiple of 3.

Proof: Begin by considering the cyclotomic polynomial

$\Phi_3(x) = 1 + x + x^2$

This has as its roots the two primitive cube roots of unity

$w = exp(i2\pi/3) = (-1 + \sqrt{3}i)/2$

and

$w^2 = exp(i4\pi/3) = (-1 - \sqrt{3}i)/2$

If $w$ is a root of $1 + x^k + (x^k)^2$ then so is $w^2$ (since roots always occur in conjugate pairs), and in this case it must be that $1 + x + x^2$ is a factor of $1 + x^k + (x^k)^2$. Now there are three possibilities for k:

I. $k = 3s$ for some $s \in \{0, 1, 2, \ldots \}$. In this case the polynomial $1 + x^k + (x^k)^2$ becomes

$1 + x^{3s} + x^{6s}$

Setting $x = w$ we get

$1 + w^{3s} + w^{6s}$ $= 1 + 1 + 1 = 3$

so $w$ is not a root of $1 + x^k + (x^k)^2$ here.

II. $k = 3s + 1$. In this case the polynomial

$1 + x^k + (x^k)^2$

becomes

$1 + x^{3s+1} + x^{6s+2}$

Setting $x = w$ we get

$1 + w + w^2 = 0$

so $w$ is a root of $1 + x^k + (x^k)^2$ here.

III. $k = 3s + 2$. In this case the polynomial

$1 + x^k + (x^k)^2$

becomes

$1 + x^{3s+2} + x^{6s+4}$

Setting $x = w$ we get

$1 + w^2 + w^4$ $= 1 + w^2 + w = 0$

so $w$ is a root of $1 + x^k + (x^k)^2$ here.

There are no other possibilities, so if we set $x = 3$ in the polynomials $1 + x + x^2$ and $1 + x^k + (x^k)^2$ the result is proved. QED

Result 3. The base-3 palindrome $1 + 3^k + (3^k)^2$ will be divisible by some number of the form $1 + 3^{3^r} + (3^{3^r})^2$ where $r \in \{0, 1, 2, \ldots\}$ if k is not a power of 3. Therefore the base-3 palindrome $1 + 3^k + (3^k)^2$ cannot be a Cantor prime if k is not a power of 3.

Proof: Consider the polynomial  $1 + x^k + (x^k)^2$ and suppose that k is not a power of 3. Then we can write $k = 3^r \cdot s$ for some integer $s > 1$ such that $gcd(s, 3) = 1$. Then

$1 + x^k + (x^k)^2 = 1 + y^s + (y^s)^2$

where $y = x^{3^r}$ and the result now follows immediately from Result 2 and by setting $x = 3$. QED

Result 4. Any base-3 palindromic prime of the form $1 + 3^k + (3^k)^2$ will be of the type $4q + 1$ for $q \in \mathbb{N}$, and can therefore be expressed as a sum of two perfect squares $m^2 + n^2$ where $m, n \in \mathbb{N}$.

Proof: If $1 + 3^k + (3^k)^2$ is prime then k must be a power of 3 and hence odd. Odd powers of 3 are always of the form $4q + 3$ and even powers of 3 are always of the form $4q + 1$. Therefore $1 + 3^k + (3^k)^2$ will be a sum of each of these two forms, plus the number 1, and this must always give a number of the form $4q + 1$. It is a well known theorem of Fermat that all primes of the form $4q + 1$ can be expressed as a sum of two perfect squares. QED