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.
(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.
