Is there a base-4 prime repunit for n > 2?

Surely it is easy, as you suspected, to prove that (1/3)(4^n - 1) is composite for all odd n >= 3.

Let n = 2p + 1 with p >= 1.

(1/3)(4^(2p + 1) - 1) = (1/3)(2^(4p + 2) - 1) = (1/3)[2^(2p + 1) - 1]*[2^(2p + 1) + 1]

Now it is easy to prove that one or other of the final brackets is divisible by 3 (do you want this proof?) and that leaves the product of a pair of integers both greater than 1 so the result is not prime.

ALWAYS COMPOSITE (provided n > 2)

[second added note]

Just factor 4^n - 1.

a_n = (4^n - 1) / 3 = (2^2n - 1) / 3 =

(2^n + 1)(2^n - 1) / 3

For n > 2 the above shows a_n will have 2 factors both > 1,

so it is always composite.

Note however that a_2 = 5 is indeed prime.

=====================================

a_n is composite for any composite n -

a_n is represented as a string of n ones (base 4). Just group the 1's into groups of k, where k is any factor of n. Then it's clear that a_n is a multiple of 111..11 (k ones, base 4).

Therefore if any a_n is prime, it is for prime n.

=====================================

Added later:

My number theory is weak, but I couldn't help playing around

with it. For what it's worth, I got this result.

For any prime p > 2, a corollary to Fermat's Little Theorem

gives 4^p = 4 (mod p) where "=" means congruent here.

Then 4^p -1 = 3 (mod p)

hence (4^p -1) / 3 = 1 (mod p)

which may be interesting, but this says nothing about whether (4^p - 1)/3 is prime.

We can narrow down the list of potential numbers a little.

As you note, every other number, the ones corresponding to n being even, are divisible by 5.

At n = 3 we have 21, which is divisible by 3. Since the pattern follows like this:

a_(n+1) = 4*a_n + 5

If we apply this pattern to itself we get results like

a_(n+2) = 4*[4*a_n + 5] + 5 = 16*a_n + 25

a_(n+3) = 4*[16*a_n + 25] + 5 = 64*a_n + 125

etc.

If you do this 6 times, then you will find that

a_(n+6) = 4096*a_n + 6825. [note that 6825 is divisible by 3]

--> If a_n is divisible by 3, as is the case with n = 2 --> 21, then every 6th number in the sequence is also divisible by 3. Saying "every 6th number" is the same as saying "when n = an odd multiple of 3". This lets us rule out more number.

The result will be composite when n is even or for odd multiple of 3. When n is an even multiple of three then the result is also divisible by 5, as noted.

No, not a proof, but a way to further limit the options of n that might make the repunit prime. So far, every one I have checked have been composite, even when n is a prime number.

5 = (4^2-1)/3 is the only one. Look at the bottom section of the link.

Your last line was a good guess =)