Let's define $b:= a^{226} +4a +1$. We know that $b$ is odd because $54$ is even and the gcd is odd. But if $a$ were odd, $b$ would be even; so $a$ is also even.
We also know that $3\nmid a$ (since $3\nmid b$, if it did divide $a$, it would be a divisor of $1$, which is absurd.) Applying Fermat's theorem, we have
$$a^{226} + 4a + 1 = (a^2)^{113} + 4a + 1 \equiv 4a + 2 \equiv 0 \pmod 3$$
This means that $a \equiv 1 \pmod 3$. We infer the following congruences:
$$a \equiv 0 \pmod2 \\ a \equiv 1 \pmod 3$$
If I had a $\pmod 9$ instead of a $\pmod 3$ in the last congruence, I'd be able to aply the Chinese Remainder Theorem.
How can I bound the values of $r_{9}(a)$, given that $r_{3}(a)=1$?