, 2020

We first show that $b$ is $a$-good iff $b$ is even and $p \mid a$ for all primes $p \le b$. Suppose there is a prime $p \le b$ with $p \nmid a$, take $t = v_p(b!)$. Then there exist positive integers $c$ such that $ac \equiv 1 \pmod{p^{t+1}}$. Take large enough $c$ (so that $an \ge b$) and let $n = (p-1)c$, then $an \equiv p-1 \pmod{p^{t+1}}$. Let $X = an(an-1)(an-2)\cdots(an-b+1)$. Since $p \le b$, $(an - (p-1))$ divides $X$, so $v_p(X) \ge t+1$, hence $v_p\left(\binom{an}{b}\right) = v_p(X) - v_p(b!) \ge 1$. Since $p \mid an+1$, $an+1$ does not divide $\binom{an}{b}-1$, thus $b$ is not $a$-good.
On the other hand, if $p \mid a$ for all prime $p \le b$, then $b!$ is coprime to $an+1$. $an+1$ divides $\binom{an}{b}-1 = \frac{X}{b!}-1$ iff $X \equiv b! \pmod{an+1}$. Easy to check that $X \equiv (-1)^b b! \pmod{an+1}$. Thus we know that $b$ is $a$-good iff $b$ is even.
This completes the claim.

Back to the problem, suppose $b$ is $a$-good, then $b$ is even and $p \mid a$ for all primes $p \le b$. Since $b+2$ is not $a$-good, we must have $b+1$ is prime.

First, we show the only if part of the claim in Solution 1. Suppose that we have a prime $p \le b$ with $p \nmid a$. Suppose the $p^t$ digit of the expansion of $b$ in base $p$ is nonzero. Since $p \nmid a$, can find $n$ such that the $p^0$ digit of $an$ is $p-1$ and the $p^t$ digit of $an$ is 0. Thus, $p \mid an+1$ and $p \mid \binom{an}{b}$ by Lucas' theorem. Thus, $b$ is not $a$-good.
Now we show that if $b$ is $a$-good but $b+2$ is not, then there must be a prime dividing $an+1$ for some $n$, which also divides $(b+1)(b+2)$. Indeed, the ratio between $\binom{an}{b+2}$ and $\binom{an}{b}$ is $\frac{(b+1)(b+2)}{(an-b)(an-b-1)}$. By $a$-good property, choose $n$ so that $\binom{an}{b+2} \not\equiv 1 \pmod{an+1}$ and $\binom{an}{b} \equiv 1 \pmod{an+1}$, which means that the given ratio must not be $1 \pmod{an+1}$. So $(b+1)(b+2)$ cannot be coprime to $an+1$. Since $p \nmid a$ for all primes $p \le b$, these condition force that either $(b+1)$ or $(b+2)$ to be a prime.
By choosing a prime $an+1$ (by Dirichlet's theorem), easy to check that $b$ must be even. So $b+2$ cannot be prime.