Help proving half the integers in a complete residue system are quadratic residues The following theorem is what I need to prove:

Let $p$ be an odd prime. Then half the numbers not congruent to $0$ in any complete residue system modulo p are perfect squares modulo $p$ and half are not.

I have done examples that help demonstrate this property:

Consider the complete residue system $B=\{1,2,3,…12\}$ where for each value $B$ in we perform $f(b)=b^2 \space (\text{mod} \space 13)$
$f(1)=(1)^2 \space (\text{mod} \space  13)=1$,   $f(2)=(2)^2 \space (\text{mod} \space 13)=4$, $f(3)=(3)^2 \space (\text{mod} \space 13)=4$,  $f(5)=12$, $f(6)=10$,  $f(7)=10$, $f(8)=12$,   $f(9)=3$, $f(10)=9$,   $f(11)=4$,   $f(12)=1$
We see that $f$ maps all the values in the set $B$ to $F(B)=\{1,4,9,12,10\}$. This set $F(B)$ is our quadratic residues $a_i$. Now observe that only half the values in B exist in $F(B)$. This demonstrates what the theorem was discussing; that only half the integers in a complete residue system represent valid quadratic residues.

I can recognize the pattern but I'm at a complete loss for how to prove this theorem. Before anyone tries to close my question, yes, I have read other answer such as: Show that exactly half of the integers in the set are quadratic residues. But I don't understand the logic of the hint. I don't know how $x^2 \equiv (p-x)^2 \pmod p$ was formulated nor do I see how to use it to prove the theorem. Could someone help guide me through a full proof that deliberatly explains the logic of each step?
Please note I have absolutely no knowledge of Group Theory and I'd prefer to focus on a proof that doesn't use Primitive Roots (because I suck at primitive roots lol). Thank you for any help!
I think that the best explanatory approach is to consider my comments, following your question, against the specific example of $(p=11).$
First consider the values $\{1^2, 2^2, 3^2, 4^2, 5^2\}.$  Is it possible that two of those 5 values have the same residue, $\pmod{11}$?
Suppose that $a,b \in \{1,2,3,4,5\} ~: ~a \neq b.$
Further suppose that $a^2 \equiv_{11} b^2.$  
Then, $(a-b)(a+b) = (a^2 - b^2) \equiv_{11} 0.$
This implies that $11 ~| ~(a^2 - b^2) = (a-b)(a+b).$
However, by assumption, $a\neq b \implies 11 ~\not| ~(a-b).$ 
Further, by assumption, $1 \leq a,b \leq 5 \implies (a+b) &lt; 11 \implies 11 ~\not| ~(a+b).$
Therefore, given that 11 is prime, the assumption that there exists $a,b \in \{1,2,3,4,5\} ~: ~a \neq b$ such that $a^2 \equiv_{11} b^2$ has led to a contradiction.
That is, you have that $11 ~| ~(a-b)(a+b)$, yet at the same time, it is impossible for $11$ to divide either $(a-b)$ or $(a+b)$.  Because $11$ is a prime, you know that whenever $11$ divides $(r \times s)$, then either $11 | r$ or $11 | s.$
As a result of this contradiction, I am free to conclude that $\{1^2, 2^2, 3^2, 4^2, 5^2\}$ represent $5$ distinct residues, $\pmod{11}.$

Addendum
The intent of the above analysis is to show that in general, $\{1^2, 2^2, \cdots, [(p-1)/2]^2\}$ must represent $[(p-1)/2]$ distinct residues, $\pmod{11}$.
The next step is to show, continuing to use $(p=11)$ as an example, that the exact number of quadratic residues must be $[(p-1)/2].$
Consider $6^2 = (11 - 5)^2 = 11^2 - [(2) \times (11) \times (5)] + 5^2.$ 
Clearly, from the above equation, you have that $6^2 \equiv_{11} 5^2.$ 
This implies that $6^2$ and $5^2$ must represent the exact same quadratic residue, $\pmod{11}.$
Now consider $7^2 = (11 - 4)^2 = 11^2 - [(2) \times (11) \times (4)] + 4^2.$ 
From the above equation, you have that $7^2 \equiv_{11} 4^2.$ 
This implies that $7^2$ and $4^2$ must represent the exact same quadratic residue, $\pmod{11}.$
Now consider $8^2 = (11 - 3)^2 = 11^2 - [(2) \times (11) \times (3)] + 3^2.$ 
From the above equation, you have that $8^2 \equiv_{11} 3^2.$ 
This implies that $8^2$ and $3^2$ must represent the exact same quadratic residue, $\pmod{11}.$
Now consider $9^2 = (11 - 2)^2 = 11^2 - [(2) \times (11) \times (2)] + 2^2.$ 
From the above equation, you have that $9^2 \equiv_{11} 2^2.$ 
This implies that $9^2$ and $2^2$ must represent the exact same quadratic residue, $\pmod{11}.$
Now consider $(10)^2 = (11 - 1)^2 = 11^2 - [(2) \times (11) \times (1)] + 1^2.$ 
From the above equation, you have that $(10)^2 \equiv_{11} 1^2.$ 
This implies that $(10)^2$ and $1^2$ must represent the exact same quadratic residue, $\pmod{11}.$
Therefore, you have that the quadratic residues represented by $\{6^2, 7^2, 8^2, 9^2, (10)^2\}$ must be the exact same $5$ quadratic residues as the residues represented by $\{1^2, 2^2, 3^2, 4^2, 5^2\}.$
Therefore, for $(p=11)$ there are exactly $5$ quadratic residues, among $\{1^2, 2^2, \cdots, (10)^2\}.$
This analysis easily generalizes to any odd prime $p$.
The foundation of this analysis is that if $a$ is any element in $\{1,2,\cdots, [(p-1)/2]\}$, then the following two things will be true.

$(p-a)$ must be an element in $\{[(p-1)/2] + 1, \cdots, (p-1)\}.$

$(p-a)^2 = p^2 - (2)(p)(a) + a^2 \equiv_{p} a^2.$


Therefore the $[(p-1)/2]$ quadratic residues represented by 
$\{\langle[(p-1)/2] + 1\rangle^2, \cdots, (p-1)^2\}$ 
must be the exact same quadratic residues as
$\{1^2, 2^2, \cdots, [(p-1)/2]^2\}.$
Therefore, for any odd prime $p$, there will be exactly $[(p-1)/2]$ distinct quadratic residues.

Note
Focusing on $(p=11)$, the quadratic residues are not represented by $\{1,2,3,4,5\}.$
Instead, the quadratic residues are represented by
$\{1^2, 2^2, 3^2, 4^2, 5^2\}.$
Let me answer your queries in different manner.
$S_p = \{ 0, 1, \dots p - 1 \}$ set of complete residues.

$p - 1 \equiv - 1 \; (\text{mod } p)$
$p - 2 \equiv -2 \; (\text{mod } p)$
$\vdots$
$p - \frac{p -1}{2} \equiv - \frac{p-1}{2} \; (\text{mod } p)$

So we can rewrite our set of complete residues:
$- \frac{p-1}{2}, \dots, -1, 0, 1, \dots, \frac{p-1}{2}$

Check that the new set also contains $p$ many elements.


Now, any square number can be thought as a bucket. Only $0$ goes to a bucket alone and others goes in pair $x$ and $-x$.