Why can the squares of the first $k$ odd numbers never cover all residue classes for a prime? Suppose I'm considering a prime number $p$ and the set of squares of the first $k$ odd numbers. Then I'm trying to show that this set of squares can never cover all the residue classes mod $p$. Since the number of classes is at most $k$, the result is clear if $p>k$. But why does it still hold for any prime $p\leq k$?
For example, if $p=3$, then no number from the set {$1^2,3^2,5^2,...,(2k-1)^2$} is congruent to 2 $mod$ 3, if $p=5$, then no number from this set is congruent to 1 $mod$ 5 etc. This fact seems so easy to prove, but am I missing something very easy here?  
 A: The residue classes that are covered are quadratic residues, with the possible inclusion of zero (often not treated as a quadratic residue, though it obviously is the square of zero) should $p$ be itself one of the first $k$ odd numbers.
It is well-known that quadratic residues constitute only half of the nonzero residue classes for an odd prime.  For the case $p=2$ we observe that odd numbers squared are congruent always to one mod $2$.
A: Given a prime number $p$ there are $p-1$ non-zero residue classes  modulo $p$.
In the case $k>p$, try to divide it by $p$ with remainder.  Write $k=qp+r$ with $r< p$. Now $k^2=(qp+r)^2=q^2p^2+2qpr + r^2\equiv r^2\pmod p$.
So it is clear that by considering $k$ bigger than $p$, their squares cannot produce what  you cannot produce as a square of elements less than $p$.
Now for a number $b$ less than $p$ note that we also have $p-b<p$, and $(p-b)^2=p^2-2bp+b^2\equiv b^2 \pmod p$. That $b$ and $p-b$ have the same squares mod $p$.  That is, squaring is a 2-to-1 function. So squares cannot hit all the residue classes, they can hit exactly $(p-1)/2$ residue classes. 
