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I am working on my own version of a proof of RSA and have come do a conclusion based on these simplified statements.
Given: N = pq
X ≡ 1 (mod p)
X ≡ 1 (mod q)
X ≡ 1 (mod N) by chinese remainder theorem

I have the original expression: MX ≡ M' (mod N)

When substituting X ≡ 1 (mod N) into this equation, is it correct to say that M = M' (removing the modulus) or must I say M ≡ M' (mod N)?

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  • $\begingroup$ Have you created a few examples and test cases to see? $\endgroup$ – Matthew Conroy Jan 2 '11 at 21:40
  • $\begingroup$ i guess that would be a good thing to start with $\endgroup$ – dhatch387 Jan 2 '11 at 21:42
  • $\begingroup$ hm. it seems like my train of thought here is totally inaccurate... $\endgroup$ – dhatch387 Jan 2 '11 at 21:47
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HINT $\ $ No,$\ $ e.g.$\ $ let $\rm\ M'\: =\: M + N\:.\ $ But the inference that $\rm\ M'\equiv M\ (mod\ N)\ $ is correct since it's a special case of the congruence product rule, viz. $\rm\ X\equiv 1\ \Rightarrow\ M\:X \equiv M\cdot 1 \equiv M\:,\:$ i.e. $\rm\ M'\equiv M\:$.

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  • $\begingroup$ but I could say that M ≡ M' (mod N)? $\endgroup$ – dhatch387 Jan 2 '11 at 22:19
  • $\begingroup$ Yes, see my link to the congruence product rule. $\endgroup$ – Bill Dubuque Jan 2 '11 at 22:32

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