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