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Hello,
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?
Thanks!
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?
Thanks!