Prove $n^2 − 4n = 8t + 5$ for some integer t. I am stuck on this proof.  Any help is appreciated.
I am trying to prove the following statement.
Let n be any odd integer. Prove $n^2 − 4n = 8t + 5$ for some integer t.
 A: $n \equiv 1,3,5,7$ $(\operatorname{mod} 8)$
$\implies n^2\equiv 1$ $(\operatorname{mod} 8)$
$\implies n^2-4n\equiv 5$ $(\operatorname{mod} 8)$
A: If $n$ is an odd integer, it can be written as $n=2m+1$, with $m$ an integer. Substituting this gives:
$$n^2-4n = (2m+1)^2-4(2m+1) = 4m^2-4m-3.$$
To show that this can be written as $8t+5$ for some integer $t$, we can subtract 5 from $n^2-4n$ and show that it is divisible by 8. Subtracting 5 gives:
$$(n^2-4n)-5 = 4(m-2)(m+1).$$
Clearly, we can divide by 4, so we are left to show that:
$$(m-2)(m+1)$$
is divisible by 2. This is easy: if $m$ is odd, than $(m+1)$ is even and therefore divisible by 2. If $m$ is even, than $(m-2)$ is also even and therefore divisible by 2.
Thus, we can divide $n^2-4n-5$ by 8, so $n^2-4n$ can be written as $8t+5$ for some integer $t$.
A: Since integer $n$ is odd we have $n=2m+1$ for some integer $m$. That leads to:$$n^2-4n-5=4m^2-4m-8=4m(m-1)+8$$
One of the consecutive integers $m-1$ and $m$ is even so that $8$ divides $4m(m-1)$ and consequently divides $4m(m-1)+8$. 
Writing $4m(m-1)+8=8t$ we end up with:$$n^2-4n=8t+5$$
If you are familiar with 'mod' then the answer of user170039 is more efficient.
