Welpp

I looked at some cases, and it always works out to 2.

No matter what n is, a = 7

7 mod 9 = 7

Let $n$ be a positive integer. If $a\equiv (3^{2n}+4)^{-1}\pmod{9}$,
what is the remainder when $a$ is divided by $9$?

$\begin{array}{|rcll|} \hline \mathbf{a} &\equiv& \mathbf{ (3^{2n}+4)^{-1}\pmod{9} } \\ && \boxed{ 3^{2n}+4 = \left(3^2\right)^n+4 = 9^n+4 } \qquad \gcd(9^n+4,9) = \gcd(4,9)=1!\\ a &\equiv& (9^n+4)^{-1}\pmod{9} \\ && \boxed{(9^n+4)^{-1}\pmod{9} = (9^n+4)^{\phi(9)-1} \quad | \quad \phi(9)=9*\left( 1-\dfrac{1}{3}\right)=6 \\ =(9^n+4)^{5} \\ \mathbf{(9^n+4)^{-1}\pmod{9} =(9^n+4)^{5}}\pmod{9} } \\ a &\equiv& (9^n+4)^{5} \pmod{9} \\\\ a &\equiv& \underbrace{\binom50 9^{5n}+\binom51 9^{4n}*4+\binom52 9^{3n}*4^2+\binom53 9^{2n}*4^3+\binom54 9^{n}*4^4}_{=0\pmod{9} }+ \binom55 4^5 \pmod{9} \\\\ a &\equiv& \binom55 4^5 \pmod{9} \\\\ a &\equiv& 4^5 \pmod{9} \\\\ a &\equiv& 1024 \pmod{9} \\\\ a &\equiv& 1024-113*9 \pmod{9} \\\\ \mathbf{a} &\equiv& \mathbf{7 \pmod{9}} \\ \hline \end{array}$