Legendre symbol identity: $sum_{a=1}^{p-1}a cdot (frac{a}{p} )$ and $sum_{a=1}^{p-1}2^a cdot (frac{a}{p} )$
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I am trying to solve the following problems ($p$ is an odd prime).
- Find the sum $$sum_{a=1}^{p-1}a cdot left (frac{a}{p} right),$$
- Find the sum $$sum_{a=1}^{p-1} 2^a cdot left (frac{a}{p} right).$$
Some thoughts :
- I reduced the sum to $2S_P-frac{p(p-1)}{2}$ where $S_p$ is the sum of the quadratic residues modulo $p$ but I don't know how to evaluate it .
- Nothing so far but more generally what can we say about the polynomial :
$$f(x)=sum_{a=1}^{p-1} x^a cdot left (frac{a}{p} right)$$
Is this polynomial interesting in any way ?
Thanks for all the help .
number-theory elementary-number-theory polynomials prime-numbers legendre-symbol
edited Nov 21 at 11:59
amWhy
191k27223438
add a comment |
up vote
3
down vote
favorite
I am trying to solve the following problems ($p$ is an odd prime).
- Find the sum $$sum_{a=1}^{p-1}a cdot left (frac{a}{p} right),$$
- Find the sum $$sum_{a=1}^{p-1} 2^a cdot left (frac{a}{p} right).$$
Some thoughts :
- I reduced the sum to $2S_P-frac{p(p-1)}{2}$ where $S_p$ is the sum of the quadratic residues modulo $p$ but I don't know how to evaluate it .
- Nothing so far but more generally what can we say about the polynomial :
$$f(x)=sum_{a=1}^{p-1} x^a cdot left (frac{a}{p} right)$$
Is this polynomial interesting in any way ?
Thanks for all the help .
number-theory elementary-number-theory polynomials prime-numbers legendre-symbol
edited Nov 21 at 11:59
amWhy
191k27223438
How did you reduce such sum?
– Paolo Leonetti
Aug 24 '15 at 21:47
1
Are you interested in computing the sums $pmod{p}$ or in finding their exact values? In the first case, notice that $$sum aleft(frac{a}{p}right) = sum a^{frac{p+1}{2}}.$$
– Jack D'Aurizio
Aug 24 '15 at 21:56
For the case when $p equiv 1 pmod{4}$ it's easy to get (using a little symmetry) $S_p=frac{p(p-1)}{4}$ so the sum is $0$ but I can't find a nice answer for $p equiv 3 pmod{4} $ .
– user252450
Aug 24 '15 at 21:57
@ Jack D'Aurizio I am asking for the exact values . As for number $2$ I think I am too optimistic to think there is a nice closed form . The polynomial looks interesting and maybe has some general nice properties . What do you think ?
– user252450
Aug 24 '15 at 22:01
and you should look at en.wikipedia.org/wiki/Proofs_of_quadratic_reciprocity
– reuns
Mar 31 '16 at 16:57
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I am trying to solve the following problems ($p$ is an odd prime).
- Find the sum $$sum_{a=1}^{p-1}a cdot left (frac{a}{p} right),$$
- Find the sum $$sum_{a=1}^{p-1} 2^a cdot left (frac{a}{p} right).$$
Some thoughts :
- I reduced the sum to $2S_P-frac{p(p-1)}{2}$ where $S_p$ is the sum of the quadratic residues modulo $p$ but I don't know how to evaluate it .
- Nothing so far but more generally what can we say about the polynomial :
$$f(x)=sum_{a=1}^{p-1} x^a cdot left (frac{a}{p} right)$$
Is this polynomial interesting in any way ?
Thanks for all the help .
number-theory elementary-number-theory polynomials prime-numbers legendre-symbol
edited Nov 21 at 11:59
amWhy
191k27223438
I am trying to solve the following problems ($p$ is an odd prime).
- Find the sum $$sum_{a=1}^{p-1}a cdot left (frac{a}{p} right),$$
- Find the sum $$sum_{a=1}^{p-1} 2^a cdot left (frac{a}{p} right).$$
Some thoughts :
- I reduced the sum to $2S_P-frac{p(p-1)}{2}$ where $S_p$ is the sum of the quadratic residues modulo $p$ but I don't know how to evaluate it .
- Nothing so far but more generally what can we say about the polynomial :
$$f(x)=sum_{a=1}^{p-1} x^a cdot left (frac{a}{p} right)$$
Is this polynomial interesting in any way ?
Thanks for all the help .
number-theory elementary-number-theory polynomials prime-numbers legendre-symbol
number-theory elementary-number-theory polynomials prime-numbers legendre-symbol
edited Nov 21 at 11:59
amWhy
191k27223438
edited Nov 21 at 11:59
amWhy
191k27223438
edited Nov 21 at 11:59
amWhy
191k27223438
edited Nov 21 at 11:59
amWhy
191k27223438
edited Nov 21 at 11:59
amWhy
191k27223438
191k27223438
asked Aug 24 '15 at 21:41
user252450
How did you reduce such sum?
– Paolo Leonetti
Aug 24 '15 at 21:47
1
Are you interested in computing the sums $pmod{p}$ or in finding their exact values? In the first case, notice that $$sum aleft(frac{a}{p}right) = sum a^{frac{p+1}{2}}.$$
– Jack D'Aurizio
Aug 24 '15 at 21:56
For the case when $p equiv 1 pmod{4}$ it's easy to get (using a little symmetry) $S_p=frac{p(p-1)}{4}$ so the sum is $0$ but I can't find a nice answer for $p equiv 3 pmod{4} $ .
– user252450
Aug 24 '15 at 21:57
@ Jack D'Aurizio I am asking for the exact values . As for number $2$ I think I am too optimistic to think there is a nice closed form . The polynomial looks interesting and maybe has some general nice properties . What do you think ?
– user252450
Aug 24 '15 at 22:01
and you should look at en.wikipedia.org/wiki/Proofs_of_quadratic_reciprocity
– reuns
Mar 31 '16 at 16:57
add a comment |
How did you reduce such sum?
– Paolo Leonetti
Aug 24 '15 at 21:47
1
Are you interested in computing the sums $pmod{p}$ or in finding their exact values? In the first case, notice that $$sum aleft(frac{a}{p}right) = sum a^{frac{p+1}{2}}.$$
– Jack D'Aurizio
Aug 24 '15 at 21:56
For the case when $p equiv 1 pmod{4}$ it's easy to get (using a little symmetry) $S_p=frac{p(p-1)}{4}$ so the sum is $0$ but I can't find a nice answer for $p equiv 3 pmod{4} $ .
– user252450
Aug 24 '15 at 21:57
@ Jack D'Aurizio I am asking for the exact values . As for number $2$ I think I am too optimistic to think there is a nice closed form . The polynomial looks interesting and maybe has some general nice properties . What do you think ?
– user252450
Aug 24 '15 at 22:01
and you should look at en.wikipedia.org/wiki/Proofs_of_quadratic_reciprocity
– reuns
Mar 31 '16 at 16:57
How did you reduce such sum?
– Paolo Leonetti
Aug 24 '15 at 21:47
How did you reduce such sum?
– Paolo Leonetti
Aug 24 '15 at 21:47
1
1
Are you interested in computing the sums $pmod{p}$ or in finding their exact values? In the first case, notice that $$sum aleft(frac{a}{p}right) = sum a^{frac{p+1}{2}}.$$
– Jack D'Aurizio
Aug 24 '15 at 21:56
Are you interested in computing the sums $pmod{p}$ or in finding their exact values? In the first case, notice that $$sum aleft(frac{a}{p}right) = sum a^{frac{p+1}{2}}.$$
– Jack D'Aurizio
Aug 24 '15 at 21:56
For the case when $p equiv 1 pmod{4}$ it's easy to get (using a little symmetry) $S_p=frac{p(p-1)}{4}$ so the sum is $0$ but I can't find a nice answer for $p equiv 3 pmod{4} $ .
– user252450
Aug 24 '15 at 21:57
For the case when $p equiv 1 pmod{4}$ it's easy to get (using a little symmetry) $S_p=frac{p(p-1)}{4}$ so the sum is $0$ but I can't find a nice answer for $p equiv 3 pmod{4} $ .
– user252450
Aug 24 '15 at 21:57
@ Jack D'Aurizio I am asking for the exact values . As for number $2$ I think I am too optimistic to think there is a nice closed form . The polynomial looks interesting and maybe has some general nice properties . What do you think ?
– user252450
Aug 24 '15 at 22:01
@ Jack D'Aurizio I am asking for the exact values . As for number $2$ I think I am too optimistic to think there is a nice closed form . The polynomial looks interesting and maybe has some general nice properties . What do you think ?
– user252450
Aug 24 '15 at 22:01
and you should look at en.wikipedia.org/wiki/Proofs_of_quadratic_reciprocity
– reuns
Mar 31 '16 at 16:57
and you should look at en.wikipedia.org/wiki/Proofs_of_quadratic_reciprocity
– reuns
Mar 31 '16 at 16:57
add a comment |
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How did you reduce such sum?
– Paolo Leonetti
Aug 24 '15 at 21:47
1
Are you interested in computing the sums $pmod{p}$ or in finding their exact values? In the first case, notice that $$sum aleft(frac{a}{p}right) = sum a^{frac{p+1}{2}}.$$
– Jack D'Aurizio
Aug 24 '15 at 21:56
For the case when $p equiv 1 pmod{4}$ it's easy to get (using a little symmetry) $S_p=frac{p(p-1)}{4}$ so the sum is $0$ but I can't find a nice answer for $p equiv 3 pmod{4} $ .
– user252450
Aug 24 '15 at 21:57
@ Jack D'Aurizio I am asking for the exact values . As for number $2$ I think I am too optimistic to think there is a nice closed form . The polynomial looks interesting and maybe has some general nice properties . What do you think ?
– user252450
Aug 24 '15 at 22:01
and you should look at en.wikipedia.org/wiki/Proofs_of_quadratic_reciprocity
– reuns
Mar 31 '16 at 16:57