# In what bases is $101$ the only prime in the sequence $1,101,10101,\ldots$?

$$101$$ is the only prime in the sequence $$1,101,10101,\ldots$$ as shown in this Putnam question.

I also know from studying the Collatz conjecture that $$101_2$$ is also the only prime in the same sequence considered as a sequence of base $$2$$ numbers.

In what bases is this true, and is there a general proof for some class of bases?

• There is currently no known general solution to your question. This is since it's solution requires knowledge of primes of the form $101_n=n^2+1$. If the bases can be written $n=2^{2^{k-1}}$, then $2^{2^k}+1$ must be prime. This is called a Fermat prime, and it is currently unknown for what of values of $k$ this number becomes prime, and also if there are finitely/infinitely many of them. See en.m.wikipedia.org/wiki/Fermat_number for more. – Alex S Apr 18 at 2:38

## 1 Answer

The proof in the Putnam question goes through regardless of the base. Replace $$10$$ by $$b$$, $$9$$ by $$b-1$$ and $$11$$ by $$b+1$$. It is true in all bases $$b$$ where $$b^2+1$$ is prime.