
#1
Mar1005, 05:40 PM

P: 54

I am working on some homework that I already handed in, but I cant get one of the problems. The fourth problem on the HW was to prove the forms of (1/p), (2/p), (3/p), (5/p), and (7/p).
I did this for 1 and 2 using the quadratic residues and generalizing a form for them. for 3 and 7 i used QRL, since they are both 1 mod 4, can i use QRL for the proof of 5 too? i know i got at least 80% on this problem, and thats a B+, so i should be fine on this problem. could someone please guide me on the first steps of this proof so that i can understand it? 3 and 7 were pretty easy, but im not sure i got 7 right. most of it was in the book by David Burton that we use. BTW, Im a sophomore in math, so this class is really hard for me. thats why im coming here for more understanding, that and my profs office hours are short and i use them for linear algebra. for 3, i showed p congruent to 1 mod 4 for 4p1 and congruent to 1 mod 3 for 3p1, so 12p1, the forms of this p congruent to 3 mod 4 are 3 mod 12, 7 mod 12, 11 mod 12, and p congruent to 2 mod 3, if p congruent to 2 mod 12, 5 mod 12, 8 mod 12, 11 mod 12. the common solutions are p congruent to 1 and 11 mod 12, so its + 1 mod 12, (3/p)=1, and since 8 is 0 mod 4, toss it, 3 and 9 are 0 mod 3, toss em, so 5,7 yield + 5 mod 12, (3/p)=1 can someone lead me through this for 5 now? sorry for type settting, it wasnt really that necessary for this problem, and im in a lab where i dont have much time left. sorry for long paragraphs too! 



#2
Mar1005, 06:16 PM

Sci Advisor
HW Helper
P: 1,996

First break it into cases depending on what (1/p) is, which you should know everything about. Then use quadratic reciprocity to determine (5/p), according to what p is mod 5.




#3
Mar1005, 06:25 PM

P: 54

ok, i see. im needing to break (5/p) into (5/p)(p/5)(1/p) and solve for all the common congruences?
i get (5/p) = {1 for p congruent to 1,9 (mod 20) and 1 for p congruent to 1,9 (mod 20)} is that right? *EDIT* oops, dont i need to hit 3,7,13,17? *works on second half* *EDIT* *2nd EDIT* so, for p congruent to 3 (mod 20), both (1/p) and (5/p) are 1, so 3 goes in the 1s, p congruent to 7 (mod 20), both (1/p) and (5/p) are 1, so 7 goes in the 1s too, p congruent to 13, (1/p) is 1, so its a 1s, same with 17... so its (5/p)={1 if p congruent to 1,3,7,9 (mod 20), 1 if p congruent to 1,3,7,9 (mod 20)} *2nd EDIT* 



#4
Nov709, 11:46 AM

P: 1

legendre symbol proof for (5/p)
what would the proof be if it was to be (3/p)? i know it is suppose to end up as =1 if p == 1 mod 6, and 1 if p == 1 mod 6, but why?



Register to reply 
Related Discussions  
Proof of orthogonality of associated Legendre polynomial  Calculus  10  
Proof using LeviCivita symbol  Introductory Physics Homework  5  
Legendre polynomials proof question.Help  Advanced Physics Homework  31  
Legendre poynomials proof question.Help  Calculus & Beyond Homework  2 