If $11 \mid n$, $a(n) \equiv 5^n \not\equiv 0 \pmod{11}$ and $b(n) \equiv 1 \not\equiv 0 \pmod{11}$, so the statement holds. Otherwise, $11 \mid n^5+5^n \iff 11 \mid (n^5+5^n)^2=n^{10}+2n^55^n+1 \iff 11 \mid 2n^55^n+2 \iff 11 \mid n^55^n+1$.

$n^5 \equiv 0,1,10 \pmod {11}$ and $5^n \equiv 1,3,4,5,9 \pmod{11}$ So $11|5^n+n^5 \iff 5^n \equiv 1, n^5 \equiv 10 \pmod{11} \iff 11|n^5*5^n+1$

This should work, right? Clearly, $gcd(11,n)=1$ $$ n^{10} \equiv 1 \pmod{11} \implies 11 | n^5+5^n+5^n n^{10} - 5^n \implies 11 | n^5 (n^5 5^n+1) \implies 11| n^5 5^n +1$$ And, if $$ 11 | n^5 5^n+1 \implies 11 | n^{10} 5^n +n^5 \implies 11|n^5+5^n$$

Clearly, $ 11 \nmid n $ Proving "If" direction : We are considering $ 11 \mid n^{5} 5^{n} + 1 $ to be true! By Fermat's Little Theorem, We know $n^{10} \equiv 1 \pmod{11}$ This tells us, \[ n^{5} 5^{n} + 1 \equiv n^{5} ( n^{5} 5^{n} + 1) \equiv n^{10} 5^{n} + n^{5} \equiv 5^{n} + n^{5} \equiv 0 \pmod{11} \]$\blacksquare$ Proving "Only If" Direction : Here, We are considering $ 11 \mid n^{5} + 5^{n} $ to be true! \[ n^{5} + 5^{n} \equiv ( n^{5} + 5^{n} )^2 \equiv n^{10} + 5^{2n} + 2 n^{5} 5^{n} \equiv 2( 1+ n^{5} 5^{n}) \equiv n^{5} 5^{n} + 1 \equiv 0 \pmod{11} \]$\blacksquare$ Since we have shown both directions—that $a(n) \equiv 0 \pmod{11} \Rightarrow b(n) \equiv 0 \pmod{11}$ and $ b(n) \equiv 0 \pmod{11} \Rightarrow a(n) \equiv 0 \pmod{11}, $ we conclude that $a(n)$ is divisible by $11$ if and only if $ b(n)$ is divisible by $ 11$. Q.E.D