Question: I have been interested in tetration for a while and I have this question: what is the tetration-root of $2$? ($x^x=2$). In tetration form, it can be written as 2^^x. In general, I am just curious on how to find the tetration-root of any number. i.e. $x^x=a$.
Is there is an infinite formula, product, or Taylor series for this type of equation?

Answer: 
*Hint:* To prove that $G \subseteq H$, we need to solve
$$
(m,n) = a(2,1) + b(1,3)
$$
for $a,b \in \mathbb Z$, given that
$$
2m+n=5t
$$
Express the first equation in matrix terms and use the second equation when you invert the matrix.
First, realize that $(5,0)=3\*(2,1)-(1,3)$, so $(5,0)\in H$. Similarly, realize that $(0,5)=2\*(1,3)-(2,1)$, so $(0,5)\in H$

To prove $G\subseteq H$, consider the opposite.

Let $g\in G,g=(m,n)$ be the element with the least nonnegative $m$ such that $g\notin H$. For $m<0$, consider $-g$.

If $m\geq 5$, consider $g=(m-5,n)+(5,0)$. Since $m-5\equiv m \pmod 5$, $(m-5,n)$ must also be in $G$, and since $g\notin H$ and $(5,0)\in H$, $(m-5,n)$ must not be in $H$. But since $m>5$, $m-5$ is positive, so $g$ is not the element with the least positive $m$ which is not in $H$.

Otherwise, $m<5$. Since $\forall g\in G,g=(m,n),2m+n\equiv 0\pmod 5$, so $n\equiv -2m\equiv 3m \pmod 5$, so $g=(m,n)\equiv (m,3m)\equiv m\*(1,3) \pmod 5$. Since $3m\equiv n \pmod 5$, $3m=5m'+n$. $g=m\*(1,3)-m'\*(0,5)$, and since $(1,3),(0,5)\in H$, a linear combination of $(1,3)$ and $(0,5)$ is also in $H$, so $g\in H$, which is a contradiction.