properties of Quadratic Gauss sum
$begingroup$
Proof that $$ G(n,p^k)=pG(n,p^{k-2}) if k ≥ 2 and p is an odd prime number or if k ≥ 4 and p = 2. $$
Where $$ G(n,m)=sum_{x=0}^{m-1}eleft(frac{nx^2}{m}right) , for gcd(n,m)=1$$
My idea is to split up the sum and consider , say $$
G(n,2^k)=sum_{x=0}^{2^k-1}eleft(frac{nx^2}{2^k}right)=
sum_{x=0}^{2^{k-2}-1}eleft(frac{frac{1}{4}nx^2}{2^{k-2}}right)+sum_{x=2^{k-2}}^{2^{k-1}-1}eleft(frac{frac{1}{4}nx^2}{2^{k-2}}right)+sum_{x=2^{k-1}}^{3*2^{k-2}-1}eleft(frac{frac{1}{4}nx^2}{2^{k-2}}right)+sum_{x=3*2^{k-2}}^{2^{k}-1}eleft(frac{frac{1}{4}nx^2}{2^{k-2}}right)$$
Due to the periodicity this is equal to $$ 4sum_{x=0}^{2^{k-2}-1}eleft(frac{frac{1}{4}nx^2}{2^{k-2}}right) $$
but this does not bring me further .
Thanks for helping .
number-theory quadratic-forms
edited Dec 13 '18 at 17:23
Larry
2,41331129
asked Dec 13 '18 at 16:52
MatilloMatillo
107
$endgroup$
add a comment |
$begingroup$
Proof that $$ G(n,p^k)=pG(n,p^{k-2}) if k ≥ 2 and p is an odd prime number or if k ≥ 4 and p = 2. $$
Where $$ G(n,m)=sum_{x=0}^{m-1}eleft(frac{nx^2}{m}right) , for gcd(n,m)=1$$
My idea is to split up the sum and consider , say $$
G(n,2^k)=sum_{x=0}^{2^k-1}eleft(frac{nx^2}{2^k}right)=
sum_{x=0}^{2^{k-2}-1}eleft(frac{frac{1}{4}nx^2}{2^{k-2}}right)+sum_{x=2^{k-2}}^{2^{k-1}-1}eleft(frac{frac{1}{4}nx^2}{2^{k-2}}right)+sum_{x=2^{k-1}}^{3*2^{k-2}-1}eleft(frac{frac{1}{4}nx^2}{2^{k-2}}right)+sum_{x=3*2^{k-2}}^{2^{k}-1}eleft(frac{frac{1}{4}nx^2}{2^{k-2}}right)$$
Due to the periodicity this is equal to $$ 4sum_{x=0}^{2^{k-2}-1}eleft(frac{frac{1}{4}nx^2}{2^{k-2}}right) $$
but this does not bring me further .
Thanks for helping .
number-theory quadratic-forms
edited Dec 13 '18 at 17:23
Larry
2,41331129
asked Dec 13 '18 at 16:52
MatilloMatillo
107
$endgroup$
$begingroup$
For $p$ odd, $y$ is an inversible square $bmod p^k$ iff it is square non-zero $bmod p$. So $sum_{x=1}^{p^k} e(frac{nx^2}{p^k}) =2 sum_{l=1}^{k/2} sum_{a=1}^{p-1}sum_{r=1}^{p^{k-2l-1}} e(frac{n p^{2l}(a^2+r p)}{p^{k}})$
$endgroup$
– reuns
Dec 13 '18 at 21:03
add a comment |
$begingroup$
Proof that $$ G(n,p^k)=pG(n,p^{k-2}) if k ≥ 2 and p is an odd prime number or if k ≥ 4 and p = 2. $$
Where $$ G(n,m)=sum_{x=0}^{m-1}eleft(frac{nx^2}{m}right) , for gcd(n,m)=1$$
My idea is to split up the sum and consider , say $$
G(n,2^k)=sum_{x=0}^{2^k-1}eleft(frac{nx^2}{2^k}right)=
sum_{x=0}^{2^{k-2}-1}eleft(frac{frac{1}{4}nx^2}{2^{k-2}}right)+sum_{x=2^{k-2}}^{2^{k-1}-1}eleft(frac{frac{1}{4}nx^2}{2^{k-2}}right)+sum_{x=2^{k-1}}^{3*2^{k-2}-1}eleft(frac{frac{1}{4}nx^2}{2^{k-2}}right)+sum_{x=3*2^{k-2}}^{2^{k}-1}eleft(frac{frac{1}{4}nx^2}{2^{k-2}}right)$$
Due to the periodicity this is equal to $$ 4sum_{x=0}^{2^{k-2}-1}eleft(frac{frac{1}{4}nx^2}{2^{k-2}}right) $$
but this does not bring me further .
Thanks for helping .
number-theory quadratic-forms
edited Dec 13 '18 at 17:23
Larry
2,41331129
asked Dec 13 '18 at 16:52
MatilloMatillo
107
$endgroup$
Proof that $$ G(n,p^k)=pG(n,p^{k-2}) if k ≥ 2 and p is an odd prime number or if k ≥ 4 and p = 2. $$
Where $$ G(n,m)=sum_{x=0}^{m-1}eleft(frac{nx^2}{m}right) , for gcd(n,m)=1$$
My idea is to split up the sum and consider , say $$
G(n,2^k)=sum_{x=0}^{2^k-1}eleft(frac{nx^2}{2^k}right)=
sum_{x=0}^{2^{k-2}-1}eleft(frac{frac{1}{4}nx^2}{2^{k-2}}right)+sum_{x=2^{k-2}}^{2^{k-1}-1}eleft(frac{frac{1}{4}nx^2}{2^{k-2}}right)+sum_{x=2^{k-1}}^{3*2^{k-2}-1}eleft(frac{frac{1}{4}nx^2}{2^{k-2}}right)+sum_{x=3*2^{k-2}}^{2^{k}-1}eleft(frac{frac{1}{4}nx^2}{2^{k-2}}right)$$
Due to the periodicity this is equal to $$ 4sum_{x=0}^{2^{k-2}-1}eleft(frac{frac{1}{4}nx^2}{2^{k-2}}right) $$
but this does not bring me further .
Thanks for helping .
number-theory quadratic-forms
number-theory quadratic-forms
edited Dec 13 '18 at 17:23
Larry
2,41331129
asked Dec 13 '18 at 16:52
MatilloMatillo
107
edited Dec 13 '18 at 17:23
Larry
2,41331129
asked Dec 13 '18 at 16:52
MatilloMatillo
107
edited Dec 13 '18 at 17:23
Larry
2,41331129
edited Dec 13 '18 at 17:23
Larry
2,41331129
edited Dec 13 '18 at 17:23
Larry
2,41331129
2,41331129
asked Dec 13 '18 at 16:52
MatilloMatillo
107
asked Dec 13 '18 at 16:52
MatilloMatillo
107
asked Dec 13 '18 at 16:52
MatilloMatillo
107
107
$begingroup$
For $p$ odd, $y$ is an inversible square $bmod p^k$ iff it is square non-zero $bmod p$. So $sum_{x=1}^{p^k} e(frac{nx^2}{p^k}) =2 sum_{l=1}^{k/2} sum_{a=1}^{p-1}sum_{r=1}^{p^{k-2l-1}} e(frac{n p^{2l}(a^2+r p)}{p^{k}})$
$endgroup$
– reuns
Dec 13 '18 at 21:03
add a comment |
$begingroup$
For $p$ odd, $y$ is an inversible square $bmod p^k$ iff it is square non-zero $bmod p$. So $sum_{x=1}^{p^k} e(frac{nx^2}{p^k}) =2 sum_{l=1}^{k/2} sum_{a=1}^{p-1}sum_{r=1}^{p^{k-2l-1}} e(frac{n p^{2l}(a^2+r p)}{p^{k}})$
$endgroup$
– reuns
Dec 13 '18 at 21:03
$begingroup$
For $p$ odd, $y$ is an inversible square $bmod p^k$ iff it is square non-zero $bmod p$. So $sum_{x=1}^{p^k} e(frac{nx^2}{p^k}) =2 sum_{l=1}^{k/2} sum_{a=1}^{p-1}sum_{r=1}^{p^{k-2l-1}} e(frac{n p^{2l}(a^2+r p)}{p^{k}})$
$endgroup$
– reuns
Dec 13 '18 at 21:03
$begingroup$
For $p$ odd, $y$ is an inversible square $bmod p^k$ iff it is square non-zero $bmod p$. So $sum_{x=1}^{p^k} e(frac{nx^2}{p^k}) =2 sum_{l=1}^{k/2} sum_{a=1}^{p-1}sum_{r=1}^{p^{k-2l-1}} e(frac{n p^{2l}(a^2+r p)}{p^{k}})$
$endgroup$
– reuns
Dec 13 '18 at 21:03
add a comment |
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$begingroup$
For $p$ odd, $y$ is an inversible square $bmod p^k$ iff it is square non-zero $bmod p$. So $sum_{x=1}^{p^k} e(frac{nx^2}{p^k}) =2 sum_{l=1}^{k/2} sum_{a=1}^{p-1}sum_{r=1}^{p^{k-2l-1}} e(frac{n p^{2l}(a^2+r p)}{p^{k}})$
$endgroup$
– reuns
Dec 13 '18 at 21:03