- #1
T.Rex
- 62
- 0
Hi,
I'd like to know which property proves the following simple result.
Let p be a prime greatest than 3.
r is a quadratic residue of p if there exists a such that: [tex]a^2 \equiv r \pmod{p}[/tex].
Since p is prime, there are [tex]\frac{p-1}{2}[/tex] different residues (not counting 0).
Now, if you sum them all, you find: [tex]S_p=\sum_{i=1}^{\frac{p-1}{2}} r_i \equiv 0 \pmod{p}[/tex].
I cannot find any explanation in my Maths books (and remind I'm just an amateur).
If p is not prime, this property is false in general. Often, [tex]d \mid p \rightarrow d \mid S_p[/tex]. And sometimes the property is true.
Examples.
p=7 S7=7
p=10 S10=2*10
p=22 S22=9*11
p=33 S33=22*11
p=35 S35=7*35
p=43 S43=10*43
p=47 S47=9*47
p=48 S48=340
The PARI/gp program I use is:
for(p=5,50,S=0;for(i=1,(p-1)/2,S=S+(i^2%p));print(p," ",S," ",S/p)
wich uses the property that [tex]a^2 \equiv (p-a)^2 \pmod{p}[/tex] .
Thanks,
Tony
I'd like to know which property proves the following simple result.
Let p be a prime greatest than 3.
r is a quadratic residue of p if there exists a such that: [tex]a^2 \equiv r \pmod{p}[/tex].
Since p is prime, there are [tex]\frac{p-1}{2}[/tex] different residues (not counting 0).
Now, if you sum them all, you find: [tex]S_p=\sum_{i=1}^{\frac{p-1}{2}} r_i \equiv 0 \pmod{p}[/tex].
I cannot find any explanation in my Maths books (and remind I'm just an amateur).
If p is not prime, this property is false in general. Often, [tex]d \mid p \rightarrow d \mid S_p[/tex]. And sometimes the property is true.
Examples.
p=7 S7=7
p=10 S10=2*10
p=22 S22=9*11
p=33 S33=22*11
p=35 S35=7*35
p=43 S43=10*43
p=47 S47=9*47
p=48 S48=340
The PARI/gp program I use is:
for(p=5,50,S=0;for(i=1,(p-1)/2,S=S+(i^2%p));print(p," ",S," ",S/p)
wich uses the property that [tex]a^2 \equiv (p-a)^2 \pmod{p}[/tex] .
Thanks,
Tony