Without evaluating the Legendre symbols, prove the following....

[Math100]

Homework Statement

Without evaluating the Legendre symbols, prove that $1(1|73)+2(2|73)+3(3|73)+\dotsb +72(72|73)=0$. (Hint: As $r$ runs through the numbers $1, 2, ..., 72$, so does $73-r$.)

Relevant Equations

Let $p$ be an odd prime. Then $\sum_{r=1}^{p-1}r(r|p)=0$ if $p\equiv 1\pmod {4}$.

Since $p=73$ in this problem, how should I prove that $\sum_{r=1}^{73-1}r(r|73)=0$? Given that $73=1\pmod {4}$.

 

[fresh_42]

Homework Statement:: Without evaluating the Legendre symbols, prove that $1(1|73)+2(2|73)+3(3|73)+\dotsb +72(72|73)=0$. (Hint: As $r$ runs through the numbers $1, 2, ..., 72$, so does $73-r$.)

Are you sure? We know that $\displaystyle{\sum_{k=1}^{p-1}\left(\dfrac{k}{p}\right)=0}$ for any prime. That would be $(1|73)+(2|73)+(3|73)+\dotsb +(72|73)=0$ in your notation.

Relevant Equations:: Let $p$ be an odd prime. Then $\sum_{r=1}^{p-1}r(r|p)=0$ if $p\equiv 1\pmod {4}$.

If this is really true, then there is nothing to show. $p=73$ is odd and $73\equiv 1\pmod{4}$ so the formula applies to $73.$

Since $p=73$ in this problem, how should I prove that $\sum_{r=1}^{73-1}r(r|73)=0$? Given that $73=1\pmod {4}$.

It is already proven in case the formula is true. Maybe you should prove the formula:
$$
p\equiv 1\pmod{4} \Longrightarrow \sum_{k=1}^{p-1}\left(\dfrac{k}{p}\right)=0
$$
In that case, you should tell us what you already know and what we could use to prove it, e.g. the quadratic residue theorem.

 

[Math100]

Are you sure? We know that $\displaystyle{\sum_{k=1}^{p-1}\left(\dfrac{k}{p}\right)=0}$ for any prime. That would be $(1|73)+(2|73)+(3|73)+\dotsb +(72|73)=0$ in your notation.If this is really true, then there is nothing to show. $p=73$ is odd and $73\equiv 1\pmod{4}$ so the formula applies to $73.$

It is already proven in case the formula is true. Maybe you should prove the formula:
$$
p\equiv 1\pmod{4} \Longrightarrow \sum_{k=1}^{p-1}\left(\dfrac{k}{p}\right)=0
$$
In that case, you should tell us what you already know and what we could use to prove it, e.g. the quadratic residue theorem.

Maybe by substituting $(r|p)=(-1)^{(p-1)/2}(p-r|p)$ into $\sum_{r=1}^{p-1}r(r|p)=0$? But then we have $\sum_{r=1}^{p-1}r(r|p)=\sum_{r=1}^{p-1}r(-1)^{(p-1)/2}(p-r|p)$. How should I proceed from here and simplify to $\sum_{r=1}^{p-1}r(r|p)=0$?

 

[fresh_42]

Smallest test:
\begin{align*}
\left(\dfrac{1}{5}\right)+2\left(\dfrac{2}{5}\right)+3\left(\dfrac{3}{5}\right)+4\left(\dfrac{4}{5}\right)&=
1+2\cdot (-1)+3\cdot (-1)+4\cdot 1=0
\end{align*}
Ok, no counterexample. Let's see what the hint says:
\begin{align*}
\sum_{k=1}^{p-1}k\cdot \left(\dfrac{k}{p}\right)&=\left[1\cdot\left(\dfrac{1}{p}\right)+(p-1)\cdot\left(\dfrac{p-1}{p}\right)\right]+\left[2\cdot\left(\dfrac{2}{p}\right)+(p-2)\cdot\left(\dfrac{p-2}{p}\right)\right]\\
+&\ldots +\left[((p-1)/2)\cdot\left(\dfrac{((p-1)/2)}{p}\right)+((p+1)/2)\cdot\left(\dfrac{((p+1)/2)}{p}\right)\right]
\end{align*}
Can you simplify these terms?
Hint: $\left(\dfrac{k}{p}\right)=\left(\dfrac{p-k}{p}\right)$ if $p\equiv 1\pmod{4}.$ Is the number of brackets $[\cdot]$ even or odd?

 

[Math100]

Smallest test:
\begin{align*}
\left(\dfrac{1}{5}\right)+2\left(\dfrac{2}{5}\right)+3\left(\dfrac{3}{5}\right)+4\left(\dfrac{4}{5}\right)&=
1+2\cdot (-1)+3\cdot (-1)+4\cdot 1=0
\end{align*}
Ok, no counterexample. Let's see what the hint says:
\begin{align*}
\sum_{k=1}^{p-1}k\cdot \left(\dfrac{k}{p}\right)&=\left[1\cdot\left(\dfrac{1}{p}\right)+(p-1)\cdot\left(\dfrac{p-1}{p}\right)\right]+\left[2\cdot\left(\dfrac{2}{p}\right)+(p-2)\cdot\left(\dfrac{p-2}{p}\right)\right]\\
+&\ldots +\left[((p-1)/2)\cdot\left(\dfrac{((p-1)/2)}{p}\right)+((p+1)/2)\cdot\left(\dfrac{((p+1)/2)}{p}\right)\right]
\end{align*}
Can you simplify these terms?
Hint: $\left(\dfrac{k}{p}\right)=\left(\dfrac{p-k}{p}\right)$ if $p\equiv 1\pmod{4}.$ Is the number of brackets $[\cdot]$ even or odd?

So $\sum_{k=1}^{p-1}k\cdot (\frac{k}{p})=[1+72\cdot (\frac{72}{73})]+[2+71\cdot (\frac{71}{73})]+\dotsb +[36\cdot (\frac{36}{73})+37\cdot (\frac{37}{73})]=73+73+\dotsb +73$?

 

[fresh_42]

So $\sum_{k=1}^{p-1}k\cdot (\frac{k}{p})=[1+72\cdot (\frac{72}{73})]+[2+71\cdot (\frac{71}{73})]+\dotsb +[36\cdot (\frac{36}{73})+37\cdot (\frac{37}{73})]=73+73+\dotsb +73$?

No.

We get from $\left(\dfrac{k}{p}\right)=\left(\dfrac{p-k}{p}\right)$ for $p\equiv 1\pmod{4}$ that
\begin{align*}
\sum_{k=1}^{p-1}k\cdot \left(\dfrac{k}{p}\right)&=\sum_{k=1}^{(p-1)/2}\left[k\cdot\left(\dfrac{k}{p}\right)+(p-k)\cdot\left(\dfrac{p-k}{p}\right)\right]\\
&=\sum_{k=1}^{(p-1)/2}\left[(k+p-k)\cdot \left(\dfrac{k}{p}\right)\right]=p\cdot\sum_{k=1}^{(p-1)/2}\left(\dfrac{k}{p}\right)
\end{align*}

We get from $p\equiv 1\pmod{4}$ that there are an even number of summands, namely $\dfrac{p-1}{2}$ many. This means we have to show that there are equally many Legendre symbols with $+1$ as there are with $-1.$

I assume from the example with $p=5$ that $\left(\dfrac{k}{p}\right)=-\left(\dfrac{k+1}{p}\right)$ if $k< \dfrac{p-1}{2}.$ That would be the easiest way to sort all $\dfrac{p-1}{2}$ summands by pairs of $+1$ and $-1.$ Can you prove this? Or find an argument for why it is true? Or is there another grouping such that
$$
\sum_{k=1}^{(p-1)/2}\left(\dfrac{k}{p}\right)=0\quad ?
$$

 

[Math100]

No.

We get from $\left(\dfrac{k}{p}\right)=\left(\dfrac{p-k}{p}\right)$ for $p\equiv 1\pmod{4}$ that
\begin{align*}
\sum_{k=1}^{p-1}k\cdot \left(\dfrac{k}{p}\right)&=\sum_{k=1}^{(p-1)/2}\left[k\cdot\left(\dfrac{k}{p}\right)+(p-k)\cdot\left(\dfrac{p-k}{p}\right)\right]\\
&=\sum_{k=1}^{(p-1)/2}\left[(k+p-k)\cdot \left(\dfrac{k}{p}\right)\right]=p\cdot\sum_{k=1}^{(p-1)/2}\left(\dfrac{k}{p}\right)
\end{align*}

We get from $p\equiv 1\pmod{4}$ that there are an even number of summands, namely $\dfrac{p-1}{2}$ many. This means we have to show that there are equally many Legendre symbols with $+1$ as there are with $-1.$

I assume from the example with $p=5$ that $\left(\dfrac{k}{p}\right)=-\left(\dfrac{k+1}{p}\right)$ if $k< \dfrac{p-1}{2}.$ That would be the easiest way to sort all $\dfrac{p-1}{2}$ summands by pairs of $+1$ and $-1.$ Can you prove this? Or find an argument for why it is true? Or is there another grouping such that
$$
\sum_{k=1}^{(p-1)/2}\left(\dfrac{k}{p}\right)=0\quad ?
$$

That seems to be right. Because $73$ has 36 quadratic residues and 36 non-quadratic residues. So they should cancel each other out leaving the answer to $0$. Also, earlier you mentioned that there's another theorem which claims $\sum_{r=1}^{p-1}(r|p)=0$ for any prime, where both the number of quadratic residue and non residue modulo $p$ are exactly $\frac{p-1}{2}$. And if $(\frac{k}{p})=(\frac{p-k}{p})$ for $p\equiv 1\pmod {4}$, then
\begin{align*}
&\sum_{k=1}^{p-1}k(k|p)=\sum_{k=1}^{p-1}(p-k)(p-k|p)\\
&=\sum_{k=1}^{p-1}(p-k)(k|p)\\
&=p\sum_{k=1}^{p-1}(k|p)-\sum_{k=1}^{p-1}k(k|p)\\
&=-\sum_{k=1}^{p-1}k(k|p).\\
\end{align*}
Thus $\sum_{k=1}^{p-1}k(k|p)=0$.

 

[fresh_42]

That seems to be right. Because $73$ has 36 quadratic residues and 36 non-quadratic residues.

Why is this the case?

So they should cancel each other out leaving the answer to $0$. Also, earlier you mentioned that there's another theorem which claims $\sum_{r=1}^{p-1}(r|p)=0$ for any prime, where both the number of quadratic residue and non residue modulo $p$ are exactly $\frac{p-1}{2}$. And if $(\frac{k}{p})=(\frac{p-k}{p})$ for $p\equiv 1\pmod {4}$, then
\begin{align*}
&\sum_{k=1}^{p-1}k(k|p)=\sum_{k=1}^{p-1}(p-k)(p-k|p)\\
&=\sum_{k=1}^{p-1}(p-k)(k|p)\\
&=p\sum_{k=1}^{p-1}(k|p)-\sum_{k=1}^{p-1}k(k|p)\\
&=-\sum_{k=1}^{p-1}k(k|p).\\
\end{align*}
Thus $\sum_{k=1}^{p-1}k(k|p)=0$.

Very nice!

 

[Math100]

Why is this the case?

Very nice!

From $\frac{p-1}{2}$, to find the number of quadratic residues and non-quadratic residues.

 

[fresh_42]

From $\frac{p-1}{2}$, to find the number of quadratic residues and non-quadratic residues.

Yes, but we need equally many among the first half, from $1$ to $\dfrac{p-1}{2}.$ That would be $18$ quadratic residues and $18$ quadratic non-residues among
$$
\left\{\left(\dfrac{1}{73}\right),\left(\dfrac{2}{73}\right),\ldots,\left(\dfrac{36}{73}\right)\right\}
$$

 

[Math100]

Yes, but we need equally many among the first half, from $1$ to $\dfrac{p-1}{2}.$ That would be $18$ quadratic residues and $18$ quadratic non-residues among
$$
\left\{\left(\dfrac{1}{73}\right),\left(\dfrac{2}{73}\right),\ldots,\left(\dfrac{36}{73}\right)\right\}
$$

I was thoughtless.

 

[fresh_42]

Your conclusion in post #7 is fine! (My previous post #10 is only valid if we want to show it directly. So maybe it was me who was thoughtless, not you. I blame it on the time gap.)

Your argument eliminates the coefficients (factors) in front of the Legendre symbols. Unfortunately, the consecutive Legendre symbols $\left(\dfrac{a}{p}\right),\left(\dfrac{a+1}{p}\right)$ do not alternate in general. The case $p=5$ was too small to recognize that. This table shows the distribution: https://en.wikipedia.org/wiki/Legendre_symbol#Table_of_values

So we are left to show that
$$
\sum_{k=1}^{p-1}\left(\dfrac{k}{p}\right)=0.
$$

This is equivalent to what you said: there are equally many quadratic residues as quadratic non-residues among $\{1,2,\ldots,p-1\}.$ Now, why is this the case?

 

[Math100]

Your conclusion in post #7 is fine! (My previous post #10 is only valid if we want to show it directly. So maybe it was me who was thoughtless, not you. I blame it on the time gap.)

Your argument eliminates the coefficients (factors) in front of the Legendre symbols. Unfortunately, the consecutive Legendre symbols $\left(\dfrac{a}{p}\right),\left(\dfrac{a+1}{p}\right)$ do not alternate in general. The case $p=5$ was too small to recognize that. This table shows the distribution: https://en.wikipedia.org/wiki/Legendre_symbol#Table_of_values

So we are left to show that
$$
\sum_{k=1}^{p-1}\left(\dfrac{k}{p}\right)=0.
$$

This is equivalent to what you said: there are equally many quadratic residues as quadratic non-residues among $\{1,2,\ldots,p-1\}.$ Now, why is this the case?

Let $a$ be a primitive root of $p$.
Then the integers $a^{1}, a^{2}, ..., a^{p-1}$ form a reduced residue system modulo $p$
such that $\varphi(p)=p-1$, where $r\in\left \{ 1, 2, ..., p-1 \right \}$.
This implies $r\equiv a^{k}\pmod {p}$ for $1\leq k\leq p-1$.
By Euler's Criterion, we have that
$(\frac{r}{p})=(\frac{a^{k}}{p})\equiv (a^{k})^{\frac{p-1}{2}}\equiv (a^{\frac{p-1}{2}})^{k}\equiv (-1)^{k}\pmod {p}$.
Since $gcd(a, p)=1$, it follows that $a^{\frac{p-1}{2}}\equiv -1\pmod {p}$.
Note that both $(\frac{r}{p})$ and $(-1)^{k}$ are equal to either $1$ or $-1$.
Thus $\sum_{r=1}^{p-1}\frac{r}{p}=\sum_{k=1}^{p-1}(-1)^{k}=0$.

 

[fresh_42]

Let $a$ be a primitive root of $p$.
Then the integers $a^{1}, a^{2}, ..., a^{p-1}$ form a reduced residue system modulo $p$
such that $\varphi(p)=p-1$, where $r\in\left \{ 1, 2, ..., p-1 \right \}$.
This implies $r\equiv a^{k}\pmod {p}$ for $1\leq k\leq p-1$.
By Euler's Criterion, we have that
$(\frac{r}{p})=(\frac{a^{k}}{p})\equiv (a^{k})^{\frac{p-1}{2}}\equiv (a^{\frac{p-1}{2}})^{k}\equiv (-1)^{k}\pmod {p}$.
Since $gcd(a, p)=1$, it follows that $a^{\frac{p-1}{2}}\equiv -1\pmod {p}$.
Note that both $(\frac{r}{p})$ and $(-1)^{k}$ are equal to either $1$ or $-1$.
Thus $\sum_{r=1}^{p-1}\frac{r}{p}=\sum_{k=1}^{p-1}(-1)^{k}=0$.

Sorry, but this sounds rather confusing. E.g. you end a sentence with a specification of $r$ that was never mentioned before. Let me explain the ideas behind the statement.

We are looking for remainders modulo a prime $p$. These form the additive group $\mathbb{Z}_p=\{0,1,2,\ldots,p-1\}.$ If $a,b\in \mathbb{Z}_p$ then $a+b \pmod{p}$ is again in $\mathbb{Z}_p$. We have a zero, addition in associative and we have inverse elements: $a+(p-a) \equiv 0 \pmod{p}.$ It is even a commutative group since $a+b\equiv b+a\pmod{p}.$ These conditions altogether make $\mathbb{Z}_p$ and abelian, additive group.

We have even a field $\mathbb{Z}_p$ because we can distributively muliply in $\mathbb{Z}_p$ and all elements
$$
G:=\mathbb{Z}_p^* = \{a\in \mathbb{Z}_p\,|\,\exists \,b\in \mathbb{Z}_p\, : \,a\cdot b=1\}=\{a\in \mathbb{Z}_p\,|\,\operatorname{gcd}(a,p)=1\}=\{1,2,\ldots,p-1\}=\mathbb{Z}_p -\{0\}
$$
except for the zero form a multiplicative group with $\varphi (p)=p-1$ elements. Since all elements in $\mathbb{Z}_p$ are of order $p$ and all elements of $\mathbb{Z}_p^*$ are coprime to $p$, they all generate $G.$ This means $G=\{a^k\,|\,0< k< p\}$ for every $a\in G.$

A character from $G$ is a group homomorphism $\alpha \, : \,G\longrightarrow \mathbb{C}-\{0\}$ such that $\alpha (a)\cdot\alpha (b)=\alpha (a\cdot b)$ where $a,b\in G$ and multiplication in $G$ is modulo $p.$ The characters of $G$ form again a multiplicative group by setting $(\alpha \cdot \beta )(a):=\alpha (a)\cdot \beta (a).$ Say this group of characters is $M.$ Let us fix an (primitive) element $a\in G.$ Then every character is defined by its value on $a.$ Say $\alpha(a)=z.$ Then $\alpha (a^k)=\alpha(a)^k=z^k$ and all elements of $G$ are of the form $a^k.$ The neutral element is the principal (trivial) character $\chi_0$ which maps all elements on $1$. Then $\alpha^{-1}(a^k):=z^{-1}$ defines the inverse character: $(\alpha \cdot \alpha^{-1})(a)=\alpha (a^k)\cdot \alpha^{-1}(a^k)=z^k\cdot (z^{-1})^k=1^k=1=\chi_0(a^k).$ Furthermore, $\alpha^{p-1}(a)=\alpha(a^{p-1})=\alpha(1)=1=\chi_0(a),$ so every character is of finite order and $z$ has to be a $p-1$-th root of unity. This means we can identify the group of characters of $G$ with $G=\{1,2,\ldots,p-1\}$ itself

The Legendre symbol for $p\equiv 1\pmod{4}$ is quadratic since $\left(\dfrac{a}{p}\right)\cdot\left(\dfrac{a}{p}\right)=\left(\dfrac{a^2}{p}\right)=1.$ We define the subgroup $G^2=\{a^2\,|\,a\in G\}< G$ of all squares modulo $p$ in $G.$

Let $a\in G$ be a primitive element again. Then $a$ is of even order $p-1$ and $2n\not\equiv 1\pmod{p-1}.$ (Proof: If $2n\equiv 1\pmod{p-1}$ then $2\,|\,(p-1)\,|\,(2n-1)$ which is impossible.) Thus $(a^n)^2=a^{2n}\neq a$ for all $n\in \mathbb{N}.$ This means that $a$ is no square and $aG^2\neq G^2.$ But $aG^2$ is of order $2$ in $G/G^2=\{G^2,aG^2\}.$ It exists therefore exactly one non-trivial homomorphism $G/G^2\longrightarrow \mathbb{C}-\{0\}$ given by $G^2\longmapsto \chi_0$ and $aG^2\longmapsto \left(\dfrac{\cdot}{p}\right)$ the Legendre symbol. Since $G^2$ and $aG^2$ and $G=G^2\cup aG^2,$ we have equally many elements in $G^2$ and $aG^2$, i.e. we have equally many squares in $G$ as non-squares. Finally,
$$
\sum_{k=1}^{p-1}\left(\dfrac{k}{p}\right)=\sum_{k\in aG^2}\left(\dfrac{k}{p}\right)+\sum_{k\in G^2}\left(\dfrac{k}{p}\right)=\left|aG^2\right|\cdot (-1)+\left|G^2\right|\cdot (+1)=\left|G^2\right|\cdot (-1+1)=0
$$
and

\begin{align*}
&\sum_{k=1}^{p-1}k(k|p)=\sum_{k=1}^{p-1}(p-k)(p-k|p)\\
&=\sum_{k=1}^{p-1}(p-k)(k|p)\\
&=p\sum_{k=1}^{p-1}(k|p)-\sum_{k=1}^{p-1}k(k|p)\\
&=-\sum_{k=1}^{p-1}k(k|p).\\
\end{align*}
Thus $\sum_{k=1}^{p-1}k(k|p)=0$.

 

[Math100]

Sorry, but this sounds rather confusing. E.g. you end a sentence with a specification of $r$ that was never mentioned before. Let me explain the ideas behind the statement.

We are looking for remainders modulo a prime $p$. These form the additive group $\mathbb{Z}_p=\{0,1,2,\ldots,p-1\}.$ If $a,b\in \mathbb{Z}_p$ then $a+b \pmod{p}$ is again in $\mathbb{Z}_p$. We have a zero, addition in associative and we have inverse elements: $a+(p-a) \equiv 0 \pmod{p}.$ It is even a commutative group since $a+b\equiv b+a\pmod{p}.$ These conditions altogether make $\mathbb{Z}_p$ and abelian, additive group.

We have even a field $\mathbb{Z}_p$ because we can distributively muliply in $\mathbb{Z}_p$ and all elements
$$
G:=\mathbb{Z}_p^* = \{a\in \mathbb{Z}_p\,|\,\exists \,b\in \mathbb{Z}_p\, : \,a\cdot b=1\}=\{a\in \mathbb{Z}_p\,|\,\operatorname{gcd}(a,p)=1\}=\{1,2,\ldots,p-1\}=\mathbb{Z}_p -\{0\}
$$
except for the zero form a multiplicative group with $\varphi (p)=p-1$ elements. Since all elements in $\mathbb{Z}_p$ are of order $p$ and all elements of $\mathbb{Z}_p^*$ are coprime to $p$, they all generate $G.$ This means $G=\{a^k\,|\,0< k< p\}$ for every $a\in G.$

A character from $G$ is a group homomorphism $\alpha \, : \,G\longrightarrow \mathbb{C}-\{0\}$ such that $\alpha (a)\cdot\alpha (b)=\alpha (a\cdot b)$ where $a,b\in G$ and multiplication in $G$ is modulo $p.$ The characters of $G$ form again a multiplicative group by setting $(\alpha \cdot \beta )(a):=\alpha (a)\cdot \beta (a).$ Say this group of characters is $M.$ Let us fix an (primitive) element $a\in G.$ Then every character is defined by its value on $a.$ Say $\alpha(a)=z.$ Then $\alpha (a^k)=\alpha(a)^k=z^k$ and all elements of $G$ are of the form $a^k.$ The neutral element is the principal (trivial) character $\chi_0$ which maps all elements on $1$. Then $\alpha^{-1}(a^k):=z^{-1}$ defines the inverse character: $(\alpha \cdot \alpha^{-1})(a)=\alpha (a^k)\cdot \alpha^{-1}(a^k)=z^k\cdot (z^{-1})^k=1^k=1=\chi_0(a^k).$ Furthermore, $\alpha^{p-1}(a)=\alpha(a^{p-1})=\alpha(1)=1=\chi_0(a),$ so every character is of finite order and $z$ has to be a $p-1$-th root of unity. This means we can identify the group of characters of $G$ with $G=\{1,2,\ldots,p-1\}$ itself

The Legendre symbol for $p\equiv 1\pmod{4}$ is quadratic since $\left(\dfrac{a}{p}\right)\cdot\left(\dfrac{a}{p}\right)=\left(\dfrac{a^2}{p}\right)=1.$ We define the subgroup $G^2=\{a^2\,|\,a\in G\}< G$ of all squares modulo $p$ in $G.$

Let $a\in G$ be a primitive element again. Then $a$ is of even order $p-1$ and $2n\not\equiv 1\pmod{p-1}.$ (Proof: If $2n\equiv 1\pmod{p-1}$ then $2\,|\,(p-1)\,|\,(2n-1)$ which is impossible.) Thus $(a^n)^2=a^{2n}\neq a$ for all $n\in \mathbb{N}.$ This means that $a$ is no square and $aG^2\neq G^2.$ But $aG^2$ is of order $2$ in $G/G^2=\{G^2,aG^2\}.$ It exists therefore exactly one non-trivial homomorphism $G/G^2\longrightarrow \mathbb{C}-\{0\}$ given by $G^2\longmapsto \chi_0$ and $aG^2\longmapsto \left(\dfrac{\cdot}{p}\right)$ the Legendre symbol. Since $G^2$ and $aG^2$ and $G=G^2\cup aG^2,$ we have equally many elements in $G^2$ and $aG^2$, i.e. we have equally many squares in $G$ as non-squares. Finally,
$$
\sum_{k=1}^{p-1}\left(\dfrac{k}{p}\right)=\sum_{k\in aG^2}\left(\dfrac{k}{p}\right)+\sum_{k\in G^2}\left(\dfrac{k}{p}\right)=\left|aG^2\right|\cdot (-1)+\left|G^2\right|\cdot (+1)=\left|G^2\right|\cdot (-1+1)=0
$$
and

What a long proof. But what does $aG^{2} \mapsto ({\frac{\cdot }{p}})$ mean/indicate? I've never seen this before.