Modulo Arithmetic

If N≡a(modx) N\equiv a \pmod x and N≡b(mody) N\equiv b \pmod y such that x and y are co-prime . Consider N≡z(modxy) N\equiv z \pmod {xy} ,how can we relate a , b ,z.

Note by Deep Chanda
8 years, 4 months ago

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Comments

Solving problems like these hinge on a very important property of the gcd⁡\gcd of two numbers. If d=gcd⁡(x,y)d = \gcd(x,y), then there exists a,b∈Za,b \in \mathbb{Z} such that ax+by=dax+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+k1x=b+k2yN = a+k_1 x = b + k_2 y. Hence, we get that k1x−k2y=b−ak_1 x - k_2 y = b-a From Bezout's lemma, we have that there exists k1,k2∈Zk_1,k_2 \in \mathbb{Z} for k1x−k2y=b−ak_1x - k_2 y = b-a since gcd⁡(x,y)=1\gcd(x,y) = 1. Pick one such k1∗,k2∗∈Zk_1^{*},k_2^{*} \in \mathbb{Z} i.e. we have k1∗x−k2∗y=b−ak_1^* x - k_2^* y = b-a All the other solutions are given by k1=k1∗+ny\color{#D61F06}{k_1 = k_1^{*} + ny} and k2=k2∗+nx\color{#D61F06}{k_2 = k_2^{*} + nx} (since gcd⁡(x,y)=1 \gcd(x,y) = 1 ), where n∈Zn \in \mathbb{Z}. Hence, N=a+k1∗x+nxy=b+k2∗y+nxyN = a + k_1^{*} x + nxy = b + k_2^{*}y + nxy Note that the above is true since a+k1∗x=b+k2∗ya + k_1^{*} x = b + k_2^{*}y. Hence, N≡(a+k1∗x)(modxy)≡b+k2∗y(modxy)N \equiv (a+k_1^{*}x) \pmod{xy} \equiv b+k_2^{*}y \pmod{xy}

Marvis Narasakibma - 8 years, 4 months ago

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Much better.

Yong See Foo - 8 years, 4 months ago

It looks like the Chinese Remainder Theorem for 2 equations. I guess it is that there exists N≡z(modxy)N \equiv z \pmod{xy} which is a unique solution under Z/xyZ\mathbb{Z}/xy\mathbb{Z}, for N≡a(modx)N \equiv a \pmod {x} and N≡b(mody)N \equiv b \pmod {y} simultaneously if not mistaken.

Yong See Foo - 8 years, 4 months ago
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