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Index arithmetic

  1. Mar 28, 2006 #1
    How would i go about solving the problem of for which values of a is the congruence ax^4≡2(mod 13) solvable? I think it might have something to do with power residues but i'm not sure.
  2. jcsd
  3. Mar 30, 2006 #2
    I would first find all values of x^2 mod 13 for x = 1 to 6 since the possible values just repeat for x > 6. For x = 1,2 and 4 they are 1,4 and 3, respectively. For x = 3,5 and 6 they are -4, -1 and -3. Squaring those values give just three possible values of x^4, i.e., 1,9,3 having respective [tex]a[/tex] values of 2, 6 and 5.
    If there is an easilier way, let someone else post it. Note that 5*3 = 2 mod 13, 2*3 = 6 mod 13 and 6*3 = 5 mod 13 so I guess that powers of 3 are significant here.
  4. May 23, 2006 #3
    oops,, what is that "ax^4≡2(mod 13)" means ??
    ax^4=13n+2 !?
    how do you use this "mod" stuff,, I don't use it in this format,, because excel and VBA not in this format... ...
  5. May 24, 2006 #4


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    [tex]a \equiv b ~(mod~n) [/tex] means n divides a-b.

    Congruences modulo the same number (n above) can be added, subtracted or multiplied together, just like regular equations.

    Can't say I understand your "format" question.

    Many programming languages (and possibly Excel) include a function along the lines of "mod(a,n)" which usually returns the smallest positive b, such that b == a (mod n).
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