The square of any odd number is $1$ more than a multiple of $8$ I'm taking a single variable calculus course and asked following : 
Say wether the following is a valid proof or not : 

the square of any odd number is 1 more than a multiple of $8$.

Proof : By the division theorem any number can be expressed in one of the forms $4q , 4q+1 , 4q+2 , 4q+3$
Squaring each of these gives : 
$$(4q+1)^2=16q^2+8q+1=8(2q^2+q)+1$$
$$(4q+3)^2=16q^2+24q+9=8(2q^2+3q+1)+1$$
My answer : 
I think this proof is invalid as it does not prove ' the square of any odd number is $1$ more than a multiple of $8$. ' is true for any odd number as it does not prove for the odd number $(2n+1)$ or $(2n-1)$. Is my assertion correct ?
 A: Your proof is basically correct. What you have forgotten is that any odd number can be written as $4q+1$ or $4q+3$ (since the other alternatives according to the division theorem is even and therefore not equal to the odd number in question).
Another alternative is to square the number $2n+1$:
$$(2n+1)^2 = 4n^2+4n+1 = 4(n^2+n) +1 = 4(n+1)n + 1$$
And here either $n+1$ or $n$ is even so $(n+1)n$ is even so  $4(n+1)n$ is a multiple of $8$.
Yet another alternative is by induction. If $(2n+1)^2=8k+1$ you have that $(2(n+1)+1)^2= 4n^2+12n+9 = (2n+1)^2 + 8n+8 = 8(k+n+1) + 1$ and $(2(n-1)+1)^2 = 4n^2 -4n + 1 = (2n+1)^2 - 8n = 8(k-n) + 1$, and trivially it's true fore some integer $n$ so it's therefore true for all integers $n$.
A: The proof is valid and your comment isn't: the proof correctly states that an odd number can be represented as either $4q+1$ or $4q+3$ and argues, in both cases, that its square minus $1$ is divisible by $8$.
A different approach might be with congruences: an odd number is congruent modulo $8$ to $1$, $3$, $5$ or $7$; now
\begin{align}
1^2=1&\equiv 1\pmod{8}\\
3^2=9&\equiv 1\pmod{8}\\
5^2=25&\equiv 1\pmod{8}\\
7^2=49&\equiv 1\pmod{8}
\end{align}
Another approach. Let $n=2q+1$ be odd; then
$$
n^2-1=(n-1)(n+1)=2q(2q+2)=4q(q+1)
$$
and $q(q+1)$ is even, so $4q(q+1)$ is divisible by $8$.
A: The proof is valid.
(1) Since $2n-1=2(n-1)+1$, you don't have to consider $2n-1$ and $2n+1$ separately.
(2) Now consider only $2n+1$. We further divide it into two cases: (i) if $n$ is even, then $n=2q$ for some $q$ and so $2n+1=4q+1$. (ii) if $n$ is odd, then $n=2q+1$ for some $q$ and so $2n+1=4q+3$.
All possible cases are considered.
