Proving the Division Algorithm using induction Let $n \in \mathbb{N}$. For every $m \in \mathbb{Z}$, there exist unique $q, r \in \mathbb{Z}$ such that $ m = qn+r$ and $0 \le r \le n-1$. We call $q$ the quotient and $r$ the remainder when dividing into $m$.
I'm having trouble proving this with induction. I believe the idea is to first prove for all $m \in \mathbb{Z} _{\ge 0}$ by induction, then prove for all $m \in \mathbb{Z}$ but $m \notin \mathbb{Z} _{\ge 0}$ using the induction proof on $-m$, since $-m \in \mathbb{N}$.
I'd appreciate any help, thank you.

Uniqueness doesn't need induction. Suppose $m=qn+r=q'n+r'$, where $0\le r\le n-1$. It's not restrictive to assume $r\le r'$, so we have $0\le r'-r\le n-1$; but $r'-r=n(q-q')$, so
$$
0\le n(q-q')<n
$$
As $n>0$, this implies
$$
0\le q-q'<1
$$
so $q=q'$ and therefore $r'=r$.

The proof of existence can be conveniently split into the cases $m\ge0$ and $m<0$.
The first case is done by induction. The case $m=0$ is obvious: take $q=0$ and $r=0$. Assume you know $m=qn+r$, with $0\le r<n$; then
$$
m+1=qn+r+1
$$
If $r+1=n$, then $m+1=q(n+1)+0$, otherwise $r+1<n$ (using the hypothesis that $r\le n-1$, so $r+1\le n$) and the assert is true.
Now let's prove the case $m<0$. From the first case we get
$$
-m=qn+r
$$
with $0\le r<n$. If $r=0$, then $m=(-q)n+0$ and we're done. Otherwise $0<r<n$ and
$$
m=(-q)n-r=(-q)n-n+n-r=(-q-1)n+(n-r)
$$
where $0<n-r<n$ and we're done.
