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Counting the distinct values of a modular mapping

  1. Jun 22, 2014 #1


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    first of all, sorry if my question is either trivial or imprecise, I'm from the engineering domain :)

    I need to know how many different values the following pair can take:

    [itex]\left(a\cdot i + b\cdot j\right) \bmod n_1[/itex]
    [itex]\left(c\cdot i + d\cdot j\right) \bmod n_2[/itex]

    as [itex](i,j)[/itex] spans [itex]\mathbb{Z}^2[/itex], with given integers [itex]a[/itex], [itex]b[/itex], [itex]c[/itex], [itex]d[/itex], [itex]n_1[/itex], [itex]n_2[/itex].

    I know that, in case I had a single expression, i.e.

    [itex]\left(a\cdot i + b\cdot j\right) \bmod n[/itex]

    the answer would be [itex]\frac{n}{\gcd(\gcd(a,b),n)}[/itex].

    I suspect that the answer to my question looks similar.
    In particular, I was trying to establish an isomorphism between [itex]\left(\mathbb{Z}_{n_1},\mathbb{Z}_{n_2}\right)[/itex] and [itex]\mathbb{Z}_{n_1\cdot n_2}[/itex], obtaining something looking like

    [itex]\left(u\cdot i + v\cdot j\right) \bmod \left( n_1\cdot n_2\right)[/itex]

    so as to exploit the same result, but so far I didn't come out with anything useful.

    Any clues?
  2. jcsd
  3. Jun 23, 2014 #2


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    But that's not true in general. For example [itex]\left(\mathbb{Z}_2,\mathbb{Z}_2\right)[/itex] is not isomorphic to [itex]\mathbb{Z}_4[/itex]. In the former, all nonzero elements have order 2, while the latter has a generator of order 2.

    (Unless I misunderstand your notation.)
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