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Chinese Remainder Theorem (I think?)

  • Thread starter PhDorBust
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Given system of congruences,

[tex]x^3 \equiv y_1 \mod n_1[/tex]
[tex]x^3 \equiv y_2 \mod n_2[/tex]
[tex]x^3 \equiv y_3 \mod n_3[/tex]

You are given [tex]y_1, y_2, y_3, n_1, n_2, n_3[/tex]. The [tex]n_i[/tex]'s are pairwise relatively prime. Solve for x.

I think there might be a connection between the fact that the exponent of x is 3 and there are 3 congruences given. But I haven't the slightest idea how to approach.

For each of them you also know there exists a d such that [tex]y_i^d \equiv x \mod n_i[/tex] but I don't think that is so useful.
 

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