I stumbled across base-3 palindromes of the form for 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 ) and 757 (corresponding to ).

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 which would eliminate a lot of cases in which *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 for 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

Therefore writing we have

Using the facts that

and

we can write

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

**Result 2**. The base-3 palindrome will be divisible by if and only if . Therefore the base-3 palindrome cannot be a Cantor prime if k is not a multiple of 3.

* Proof: *Begin by considering the cyclotomic polynomial

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

and

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

I. for some . In this case the polynomial becomes

Setting we get

so is *not* a root of here.

II. . In this case the polynomial

becomes

Setting we get

so *is* a root of here.

III. . In this case the polynomial

becomes

Setting we get

so *is* a root of here.

There are no other possibilities, so if we set in the polynomials and the result is proved. *QED*

**Result 3**. The base-3 palindrome will be divisible by some number of the form where if k is not a power of 3. Therefore the base-3 palindrome cannot be a Cantor prime if k is not a power of 3.

* Proof:* Consider the polynomial and suppose that k is not a power of 3. Then we can write for some integer such that . Then

where and the result now follows immediately from Result 2 and by setting . *QED*

**Result 4**. Any base-3 palindromic prime of the form will be of the type for , and can therefore be expressed as a sum of two perfect squares where .

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

*QED*