# Proof

• Assume, $$d\ge 0$$ is a fixed integer.
• Case $$d=0$$:
• Let $$n\in \mathbb Z_0$$.
• By definition of $Z_0$ we have $$\gcd(n,0)=1$$ only for $$n=1$$, it follows that $$\mathbb Z_0=\{1\}$$.
• Trivially, $$\mathbb Z_0$$ is divisor-closed.
• Case $$d = 1$$:
• Let $$n\in \mathbb Z_1$$.
• Since we have $$\gcd(n,1)=1$$ for all $$n\in \mathbb Z$$, it follows $$\mathbb Z_1=\mathbb Z$$.
• Trivially, $$\mathbb Z$$ is divisor-closed.
• Case $$d > 1$$:
• There is a unique factorization $$|d|=p_1^{e_1}\cdot\ldots\cdot p_k^{e_k}$$ with $$e_i\ge 0$$ for $$i=1,\ldots,k$$ by the fundamental theorem of arithmetic.
• Let $$n\in \mathbb Z_d$$ and let $$t\mid n$$. We have to show that $$t\in\mathbb Z_d$$.
• If $$t=1$$, then $$\gcd(t,d)=1$$. Thus, $$t\in \mathbb Z_d$$.
• If $$t > 1$$, then also $$n > 1$$ and we have two unique factorizations: * $$n=q_1^{f_1}\cdot\ldots\cdot q_m^{f_m}$$ be with $$0\le f_j$$ for $$j=1,\ldots,m$$ and * $$t=q_1^{g_1}\cdot\ldots\cdot q_m^{g_m}$$ with $$0\le g_j\le f_j$$ for $$j=1,\ldots,m$$, since $$t\mid n$$.
• Because $$n\in \mathbb Z_d$$, we have $$\gcd(n,d)=1$$.
• Thus, there is no such index $$l$$, $$l=1,\ldots,\min(m,k)$$, for which both exponents $$e_l > 0$$ and $$g_l > 0$$.
• Otherwise, $$\gcd(q_l^{e_l},q_l^{g_l})=q_l^{\min(e_l,g_l)} > 1$$, because $$\min(e_l,g_l)>1$$, and we would have found a proper common divisor of $$n$$ and $$d$$, contradicting $$n\in\mathbb Z_d$$.
• For the same reason, $t$ and$$d$$ are co-prime.
• Thus, $$t\in\mathbb Z_d$$.

Github: ### References

#### Bibliography

1. Piotrowski, Andreas: Own Research, 2014