For $1leq aleq p-1$ and $5leq p$, show that $sum_{(a/p)=1}^{} a equiv 0 pmod{p}$
$begingroup$
For $1leq aleq p-1$ and $5leq p$, show that $$sum_{(a/p)=1} a equiv 0 pmod{p}$$ where $(a/p)$ is the Legende symbol.
I know that there are as many quadratic residues as quadratic nonresidues, but I have no idea of how they are distributed.
Hints or complete solutions would be appreciated.
number-theory
edited Jan 4 at 22:05
Maria Mazur
47.1k1260120
asked Jan 4 at 22:05
John John
394
$endgroup$
add a comment |
$begingroup$
For $1leq aleq p-1$ and $5leq p$, show that $$sum_{(a/p)=1} a equiv 0 pmod{p}$$ where $(a/p)$ is the Legende symbol.
I know that there are as many quadratic residues as quadratic nonresidues, but I have no idea of how they are distributed.
Hints or complete solutions would be appreciated.
number-theory
edited Jan 4 at 22:05
Maria Mazur
47.1k1260120
asked Jan 4 at 22:05
John John
394
$endgroup$
2
$begingroup$
Hint: If $r$ is a primitive root of $p$ then the quadratic residues are congruent mod $p$ to the even powers of $r$.
$endgroup$
– MisterRiemann
Jan 4 at 22:09
add a comment |
$begingroup$
For $1leq aleq p-1$ and $5leq p$, show that $$sum_{(a/p)=1} a equiv 0 pmod{p}$$ where $(a/p)$ is the Legende symbol.
I know that there are as many quadratic residues as quadratic nonresidues, but I have no idea of how they are distributed.
Hints or complete solutions would be appreciated.
number-theory
edited Jan 4 at 22:05
Maria Mazur
47.1k1260120
asked Jan 4 at 22:05
John John
394
$endgroup$
For $1leq aleq p-1$ and $5leq p$, show that $$sum_{(a/p)=1} a equiv 0 pmod{p}$$ where $(a/p)$ is the Legende symbol.
I know that there are as many quadratic residues as quadratic nonresidues, but I have no idea of how they are distributed.
Hints or complete solutions would be appreciated.
number-theory
number-theory
edited Jan 4 at 22:05
Maria Mazur
47.1k1260120
asked Jan 4 at 22:05
John John
394
edited Jan 4 at 22:05
Maria Mazur
47.1k1260120
asked Jan 4 at 22:05
John John
394
edited Jan 4 at 22:05
Maria Mazur
47.1k1260120
edited Jan 4 at 22:05
Maria Mazur
47.1k1260120
edited Jan 4 at 22:05
Maria Mazur
47.1k1260120
47.1k1260120
asked Jan 4 at 22:05
John John
394
asked Jan 4 at 22:05
John John
394
asked Jan 4 at 22:05
John John
394
394
2
$begingroup$
Hint: If $r$ is a primitive root of $p$ then the quadratic residues are congruent mod $p$ to the even powers of $r$.
$endgroup$
– MisterRiemann
Jan 4 at 22:09
add a comment |
2
$begingroup$
Hint: If $r$ is a primitive root of $p$ then the quadratic residues are congruent mod $p$ to the even powers of $r$.
$endgroup$
– MisterRiemann
Jan 4 at 22:09
2
2
$begingroup$
Hint: If $r$ is a primitive root of $p$ then the quadratic residues are congruent mod $p$ to the even powers of $r$.
$endgroup$
– MisterRiemann
Jan 4 at 22:09
$begingroup$
Hint: If $r$ is a primitive root of $p$ then the quadratic residues are congruent mod $p$ to the even powers of $r$.
$endgroup$
– MisterRiemann
Jan 4 at 22:09
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Note that, quadratic residues are precisely members of the set, ${1^2,2^2,dots,left(frac{p-1}{2}right)^2}$, where I assume the rest of the arithmetic to be done modulo $p$.
Now,
$$
sum_{k=1}^{frac{p-1}{2}}k^2 equiv frac{frac{p-1}{2}cdot frac{p+1}{2}p}{6} pmod{p}.
$$
Since $pneq 2,3$, the result follows immediately.
answered Jan 4 at 22:14
AaronAaron
1,912415
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
Note that, quadratic residues are precisely members of the set, ${1^2,2^2,dots,left(frac{p-1}{2}right)^2}$, where I assume the rest of the arithmetic to be done modulo $p$.
Now,
$$
sum_{k=1}^{frac{p-1}{2}}k^2 equiv frac{frac{p-1}{2}cdot frac{p+1}{2}p}{6} pmod{p}.
$$
Since $pneq 2,3$, the result follows immediately.
answered Jan 4 at 22:14
AaronAaron
1,912415
$endgroup$
add a comment |
$begingroup$
Note that, quadratic residues are precisely members of the set, ${1^2,2^2,dots,left(frac{p-1}{2}right)^2}$, where I assume the rest of the arithmetic to be done modulo $p$.
Now,
$$
sum_{k=1}^{frac{p-1}{2}}k^2 equiv frac{frac{p-1}{2}cdot frac{p+1}{2}p}{6} pmod{p}.
$$
Since $pneq 2,3$, the result follows immediately.
answered Jan 4 at 22:14
AaronAaron
1,912415
$endgroup$
add a comment |
$begingroup$
Note that, quadratic residues are precisely members of the set, ${1^2,2^2,dots,left(frac{p-1}{2}right)^2}$, where I assume the rest of the arithmetic to be done modulo $p$.
Now,
$$
sum_{k=1}^{frac{p-1}{2}}k^2 equiv frac{frac{p-1}{2}cdot frac{p+1}{2}p}{6} pmod{p}.
$$
Since $pneq 2,3$, the result follows immediately.
answered Jan 4 at 22:14
AaronAaron
1,912415
$endgroup$
Note that, quadratic residues are precisely members of the set, ${1^2,2^2,dots,left(frac{p-1}{2}right)^2}$, where I assume the rest of the arithmetic to be done modulo $p$.
Now,
$$
sum_{k=1}^{frac{p-1}{2}}k^2 equiv frac{frac{p-1}{2}cdot frac{p+1}{2}p}{6} pmod{p}.
$$
Since $pneq 2,3$, the result follows immediately.
answered Jan 4 at 22:14
AaronAaron
1,912415
answered Jan 4 at 22:14
AaronAaron
1,912415
answered Jan 4 at 22:14
AaronAaron
1,912415
answered Jan 4 at 22:14
AaronAaron
1,912415
1,912415
add a comment |
add a comment |
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Hint: If $r$ is a primitive root of $p$ then the quadratic residues are congruent mod $p$ to the even powers of $r$.
$endgroup$
– MisterRiemann
Jan 4 at 22:09