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 ?

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 $8$.
