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As it says in the title I want to show that $mathbb Z[frac{2+i}{5}]cap mathbb Q =mathbb Z$.

Set $omega=frac{2+i}{5}$. $mathbb Z[omega]$ is the smallest ring that contains $mathbb Z$ and $omega$. Therefore,

begin{equation}
mathbb Z[omega] = {a_0+a_1 omega+cdots+a_k omega^kmid a_iin mathbb
Z,kin mathbb N}
end{equation}

Let, then, $qinmathbb Z[frac{2+i}{5}]cap mathbb Q$. Since $qinmathbb Q$ we can write $q=frac{m}{n}$, where $m,n$ are relatively prime integers and $mneq 0$. Now, since $qinmathbb Z[frac{2+i}{5}]$ we have that

begin{equation}
q=a_0+a_1 omega+cdots+a_k omega^k
end{equation}

for some $a_iin mathbb Z$. Then,

begin{align}
&frac{m}{n}=a_0+a_1 frac{2+i}{5}+cdots+a_k left(frac{2+i}{5}right)^k \
Longrightarrow &5^km=5^kna_0+5^{k-1}na_1(2+i)+cdots+a_kn(2+i)^k
end{align}

Hence, $nmid 5^km$ and since $gcd(n,m)=1$ we have that $n|5^k$.

Thus, every rational number in $mathbb Z[omega]$ is of the form $frac{m}{5^l}$ where $minmathbb Z$, $l$ is a positive integer and when $lgeq 1$ it holds that $5nmid m$. It suffice then to show that it always hold that $l=0$. I am stuck here. I would very much appreciate some help!

Hint: Look at the equation $5^km = n(5^ka_0+5^{k-1}a_1(2+i)+ cdots + a_k(2+i)^k)$ as an equation between Gaussian integers.
Note that $2+i$ is a prime in the ring of Gaussian integers arising from the factorization of the integer prime $5$, namely $(2+i)(2-i)=5$.
You already noticed that $n$ is a power of $5$. In particular, if $m/n$ is not an integer, you may assume that $m$ is not divisible by $5$.
Hence, viewing it as a Gaussian integer, it should not be divisible by $2+i$ or $2-i$.
Can you see the contradiction from here?