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# Modulo Arithmetic

If $$N\equiv a \pmod x$$ and $$N\equiv b \pmod y$$ such that x and y are co-prime . Consider $$N\equiv z \pmod {xy}$$ ,how can we relate a , b ,z.

Note by Deep Chanda
4 years, 1 month ago

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Solving problems like these hinge on a very important property of the $$\gcd$$ of two numbers. If $$d = \gcd(x,y)$$, then there exists $$a,b \in \mathbb{Z}$$ such that $ax+by = d$ This is termed as Bezout's identity/lemma. The wiki link provides more details on this identity. In our case, we have $$N = a+k_1 x = b + k_2 y$$. Hence, we get that $k_1 x - k_2 y = b-a$ From Bezout's lemma, we have that there exists $$k_1,k_2 \in \mathbb{Z}$$ for $k_1x - k_2 y = b-a$ since $$\gcd(x,y) = 1$$. Pick one such $$k_1^{*},k_2^{*} \in \mathbb{Z}$$ i.e. we have $k_1^* x - k_2^* y = b-a$ All the other solutions are given by $$\color{red}{k_1 = k_1^{*} + ny}$$ and $$\color{red}{k_2 = k_2^{*} + nx}$$ (since $$\gcd(x,y) = 1$$ ), where $$n \in \mathbb{Z}$$. Hence, $N = a + k_1^{*} x + nxy = b + k_2^{*}y + nxy$ Note that the above is true since $$a + k_1^{*} x = b + k_2^{*}y$$. Hence, $N \equiv (a+k_1^{*}x) \pmod{xy} \equiv b+k_2^{*}y \pmod{xy}$ · 4 years, 1 month ago

Much better. · 4 years, 1 month ago

It looks like the Chinese Remainder Theorem for 2 equations. I guess it is that there exists $$N \equiv z \pmod{xy}$$ which is a unique solution under $$\mathbb{Z}/xy\mathbb{Z}$$, for $$N \equiv a \pmod {x}$$ and $$N \equiv b \pmod {y}$$ simultaneously if not mistaken. · 4 years, 1 month ago