Is number $1$ the only natural number with this property?

Let us call a number $b \in \mathbb N$ gcd-friendly if there exists $n_0(b) \in \mathbb N$ and $a_0(b) \in \mathbb N$ such that for every $n \geq n_0(b)$ and $a \geq a_0(b)$ we have $\gcd (\sum_{i=0}^{n}a^i,b)=1$.

The number $1$ is obviously gcd-friendly because $1$ is relatively prime to every other natural number.

Even numbers obviously are not gcd-friendly because if we suppose they are and choose some odd number $a$ that is greater than $a_0(b)$ and some odd number $n$ that is greater than $n_0(b)$ we would have that $\sum_{i=0}^{n}a^i$ is even so we would have $\gcd (\sum_{i=0}^{n}a^i,b)\geq2\neq1$.

The question is:

Is the number $1$ the only natural number which is gcd-friendly?

• If you take $b \geqslant 2$, that you suppose that $a_0$ and $n_0$ exist, then you could take $a = b\cdot a_0 + 1$ and $n = b\cdot (n_0 + 1) - 1$ (and then $b \mid \sum \limits_{k=0}^n a^k$) – charmd Apr 23 '16 at 16:52
• @charMD: I think that comment should be an answer. – joriki Apr 23 '16 at 16:54

If you take $b \geqslant 2$, that you suppose that $a_0$ and $n_0$ exist, then you could take $a = b\cdot a_0 + 1$ and $n = b\cdot (n_0 + 1) - 1$, and then $\sum \limits_{k=0}^n a^k \equiv \sum \limits_{k=0}^n 1^k \equiv n+1 \equiv b(n_0 +1) \equiv 0 \pmod{b}$ ; so $b \mid \sum \limits_{k=0}^n a^k$.
Hence $1$ is the only gcd-friendly number.