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If $A\equiv B\pmod m$ then m|(A-B). What if you used a 2 coordinate congruence where $C\equiv D\pmod {m,n}$ then there exists R,S such that (C-D) = m R+n S, where R and S are coprime and |mR+nS| > gcd(m,n). So if $A\equiv B\pmod {m,n}$ then $A^2\equiv B^2\pmod {m,n}$. Is this a useful extension of congruences? (If $A\equiv B\pmod {m,n}$ then $Ar\equiv Br\pmod {m,n}$)

201044
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user128932
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1 Answers1

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The set of integers of the form $\{ Rm + Sn : R,S \in \mathbb{Z} \}$ consists of all integer multiples of $g=\mathrm{gcd}(m,n)$. Indeed, since $g$ divides both $m$ and $n$, it divides any integer of the form $Rm+Sn$. The converse follows from the well-known fact that $g = Rm+Sn$ for some integers $R,S$. You can see proofs on this Wikipedia page.

As a conclusion, we obtain that $a \equiv b \pmod{m,n}$ iff $a \equiv b \pmod{\mathrm{gcd}(m,n)}$.

Yuval Filmus
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  • If $a\equiv b\pmod {m,n}$ where (a-b) is a linear combination of m and n having integer coefficients that are not necessarily coprime ; gcd(m,n)= g , g divides (a-b)., this may seem trivial yet if g is coprime to w there exists v such that $w^v\equiv 1\pmod {m,n}$. – user128932 Sep 28 '14 at 04:07
  • JimmyK4542's point is that your definition reduces to the usual one. I don't see how what you mention changes that. – Yuval Filmus Sep 28 '14 at 04:09
  • It may reduce to the usual one but maybe in this form interesting congruence equations may result. The idea of 'preserving' the linear combination concept in representing congruences may be useful. For example $A^{(p-1)(q-1)}\equiv 1\pmod {p,q}$ for any odd primes p, q. – user128932 Sep 28 '14 at 04:17
  • I edited the question in Sept. 2014 I think ; and I received no response since then. If a question is initially thought of as having no meaning is it ignored from that moment on. If so why is the question still here? – 201044 Nov 25 '15 at 03:21