We see that $$ n=kd$$ for some $$ k\in <!--\; \in \; -->\mathbb{Z}$$, since $$ n\ne <!--\; \ne \; -->0$$ then also we must have $$ k,d\ne <!--\; \ne \; -->0$$ specifically note that $$ |<!--\; |\; -->k|<!--\; |\; -->\ge <!--\; \ge \; -->1$$ so that we have $\[ |<!--\; |\; -->n|<!--\; |\; -->=|<!--\; |\; -->kd|<!--\; |\; -->=|<!--\; |\; -->k|<!--\; |\; -->\cdot <!--\; \cdot \; -->|<!--\; |\; -->d|<!--\; |\; -->\ge <!--\; \ge \; -->1|<!--\; |\; -->d|<!--\; |\; -->=|<!--\; |\; -->d|<!--\; |\; --> \]$ so that $$ |<!--\; |\; -->n|<!--\; |\; -->\ge <!--\; \ge \; -->|<!--\; |\; -->d|<!--\; |\; -->$$

If $$ a=0$$ then because $$ a\mid <!--\; \mid \; -->b$$ then this forces $$ b=0$$ so that $$ a=b=0$$, also if we assumed $$ b=0$$ the same thing occurs.

In the case that neither of $$ a,b=0$$ then recall that if $$ a\mid <!--\; \mid \; -->b$$ we have $$ |<!--\; |\; -->a|<!--\; |\; -->\le <!--\; \le \; -->|<!--\; |\; -->b|<!--\; |\; -->$$ also since $$ b\mid <!--\; \mid \; -->a$$ then also $$ |<!--\; |\; -->a|<!--\; |\; -->\le <!--\; \le \; -->|<!--\; |\; -->b|<!--\; |\; -->$$ so that $$ |<!--\; |\; -->a|<!--\; |\; -->=|<!--\; |\; -->b|<!--\; |\; -->$$.

For the base case of $$ n=2$$ if $$ a\mid <!--\; \mid \; -->b_1$$ and $$ a\mid <!--\; \mid \; -->b_2$$ then we know that there are $$ k_1,k_2\in <!--\; \in \; -->\mathbb{Z}$$ such that $$ b_1=k_{1}a$$ and $$ b_2=k_{2}a$$ so that $$ c_1{b}_1+c_2{b}_2=c_1\left(k_{1}a\right)+c_2\left(k_{2}a\right)=a(c_1{k}_1+c_2{k}_2)$$.

For the induction step, assume the statement holds true for $$ k\in <!--\; \in \; -->{\mathbb{N}}_1$$ and we'll show that it holds true for $$ k+1$$ so suppose that $$ a\mid <!--\; \mid \; -->b_1,\dots <!--\; \dots \; -->b_{k+1}$$ by the induction hypothesis we know that $$ a\mid <!--\; \mid \; -->c_1{b}_1+\cdots <!--\; \cdots \; -->+c_k{b}_k$$ so we get some $$ j^{\mathrm{\prime <!--\; \prime \; -->}}\in <!--\; \in \; -->\mathbb{Z}$$ such that $$ c_1{b}_1+\cdots <!--\; \cdots \; -->+c_k{b}_k=j^{\mathrm{\prime <!--\; \prime \; -->}}a$$ and also some $$ j^{\mathrm{\prime <!--\; \prime \; -->}\mathrm{\prime <!--\; \prime \; -->}}\in <!--\; \in \; -->\mathbb{Z}$$ such that $$ b_{k+1}=j^{\mathrm{\prime <!--\; \prime \; -->}\mathrm{\prime <!--\; \prime \; -->}}a$$ and therefore $\[ c_1{b}_1+\cdots <!--\; \cdots \; -->+c_k{b}_k+c_{k+1}{b}_{k+1}=j^{\mathrm{\prime <!--\; \prime \; -->}}a+j^{\mathrm{\prime <!--\; \prime \; -->}\mathrm{\prime <!--\; \prime \; -->}}a=a(j^{\mathrm{\prime <!--\; \prime \; -->}}+j^{\mathrm{\prime <!--\; \prime \; -->}\mathrm{\prime <!--\; \prime \; -->}}) \]$ so therefore $$ a\mid <!--\; \mid \; -->c_1{b}_1+\cdots <!--\; \cdots \; -->+c_{k+1}{b}_{k+1}$$

Base case, set $$ m=0$$ then $$ n^m=1$$ which is odd. For the induction step assume it holds true for $$ k\in <!--\; \in \; -->{\mathbb{N}}_0$$ so that $$ n^k$$ is odd, which means that there is some $$ j\in <!--\; \in \; -->\mathbb{Z}$$ such that $$ n^k=2j+1$$ therefore $$ n^{k+1}=(2j+1)n$$ but $$ n$$ is odd so there is some $$ j^{\mathrm{\prime <!--\; \prime \; -->}}\in <!--\; \in \; -->\mathbb{Z}$$ such that $$ n=2j^{\mathrm{\prime <!--\; \prime \; -->}}+1$$ and thus $\[ n^{k+1}=(2j+1)(2j^{\mathrm{\prime <!--\; \prime \; -->}}+1)=4j{j}^{\mathrm{\prime <!--\; \prime \; -->}}+2j+2j^{\mathrm{\prime <!--\; \prime \; -->}}+1=2(2j{j}^{\mathrm{\prime <!--\; \prime \; -->}}+j+j^{\mathrm{\prime <!--\; \prime \; -->}})+1 \]$ hence $$ n^{k+1}$$ is prime so the statement follows from the principle of induction.

On the other hand if we know that $$ b\le <!--\; \le \; -->\sqrt{a}$$ and that $$ c\le <!--\; \le \; -->\sqrt{a}$$, since $$ b\ne <!--\; \ne \; -->a$$ then we know that $$ b\ne <!--\; \ne \; -->\sqrt{a}$$, therefore more specifically we see that $$ b<<!--\; <\; -->\sqrt{a}$$ this implies that $$ a=bc<<!--\; <\; -->\sqrt{a}\sqrt{a}=a$$ which is a contradiction. In the case of $$ c\ge <!--\; \ge \; -->\sqrt{a}$$ and $$ b\ge <!--\; \ge \; -->\sqrt{a}$$ it follows in a symmetrical manner.

We have that $$ n=mk$$ for some $$ k\in <!--\; \in \; -->\mathbb{Z}$$ and we want to prove that $$ a^{mk}-<!--\; -\; -->1=(a^m-<!--\; -\; -->1)j$$ for some $$ j\in <!--\; \in \; -->\mathbb{Z}$$, a good guess would be that $$ j=a^{(m-<!--\; -\; -->1)k}$$ which produces $$ (a^m-<!--\; -\; -->1)\left(a^{(m-<!--\; -\; -->1)k}\right)=a^{mk}-<!--\; -\; -->a^{(m-<!--\; -\; -->1)k}$$ and so we have to get rid of the $$ a^{(m-<!--\; -\; -->1)k}$$ term, which inspires us to try $$ j=a^{(m-<!--\; -\; -->1)k}+a^{(m-<!--\; -\; -->2)k}$$ which yields $\[ (a^m-<!--\; -\; -->1)(a^{(m-<!--\; -\; -->1)k}+a^{(m-<!--\; -\; -->2)k})=a^{km}+a^{(m-<!--\; -\; -->1)k}-<!--\; -\; -->(a^{(m-<!--\; -\; -->1)k}+a^{(m-<!--\; -\; -->2)k})=a^{km}-<!--\; -\; -->a^{(m-<!--\; -\; -->2)k} \]$

As we've just run into the same problem as last time we see that adding $$ a^{(m-<!--\; -\; -->3)k}$$ would remove the most recent residue term, but then leave the new residue term $$ a^{(m-<!--\; -\; -->3)k}$$ . Therefore formally using induction we can show that $\[ (a^m-<!--\; -\; -->1)\left(\underset{i=0}{\overset{m-<!--\; -\; -->1}{\sum <!--\; \sum \; -->}}{a}^{ik}\right)=a^{mk}-<!--\; -\; -->1=a^n-<!--\; -\; -->1 \]$ where $$ 1$$ is really the residue term $$ a^{\left(0\right)k}$$