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Legendre symbol proof for (-5/p)

  1. Mar 10, 2005 #1
    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 4|p-1 and congruent to 1 mod 3 for 3|p-1, so 12|p-1, 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. jcsd
  3. Mar 10, 2005 #2


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    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.
  4. Mar 10, 2005 #3
    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*
    Last edited: Mar 10, 2005
  5. Nov 7, 2009 #4
    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?
    Last edited: Nov 7, 2009
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